{"id":120407,"date":"2022-05-10T12:24:09","date_gmt":"2022-05-10T06:54:09","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=120407"},"modified":"2022-05-10T12:24:09","modified_gmt":"2022-05-10T06:54:09","slug":"chapter-7-alternating-current-questions-and-answers-ncert-solutions-for-class-12-physics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-7-alternating-current-questions-and-answers-ncert-solutions-for-class-12-physics","title":{"rendered":"Chapter 7 \u2013 Alternating Current Questions and Answers: NCERT Solutions for Class 12 Physics"},"content":{"rendered":"<h2>1. A 100\u03a9100\u03a9 resistor is connected to a 220V220V, 50Hz50Hz ac supply.<br \/>\na) What is the rms value of current in the circuit?<\/h2>\n<p>Ans: It is given that,<br \/>\nResistance, R=100\u03a9R=100\u03a9<br \/>\nVoltage, V=220VV=220V<br \/>\nFrequency, f=50Hzf=50Hz<br \/>\nIt is known that,<br \/>\nIrms=VrmsRIrms=VrmsR<br \/>\n\u21d2Irms=220100=2.2A\u21d2Irms=220100=2.2A<br \/>\nTherefore, the rms value of current in the circuit isIrms=2.2AIrms=2.2A.<\/p>\n<h2>b) What is the net power consumed over a full cycle?<\/h2>\n<h3>Ans: It is known that,<br \/>\nPower=V\u00d7IPower=V\u00d7I<br \/>\n\u21d2Power=220\u00d72.2\u21d2Power=220\u00d72.2<br \/>\n\u21d2Power=484W\u21d2Power=484W<br \/>\nTherefore, the net power consumed over a full cycle is484W484W.<\/h3>\n<h2>2.<br \/>\na) The peak voltage of an ac supply is 300V300V. What is the rms voltage?<\/h2>\n<h3>Ans: It is given that,<br \/>\nPeak voltage of the ac supply, V0=300VV0=300V<br \/>\nIt is known that,<br \/>\nVrms=V02\u221aVrms=V02<br \/>\n\u21d2Vrms=3002\u221a\u21d2Vrms=3002<br \/>\n\u21d2Vrms=212.1V\u21d2Vrms=212.1V<br \/>\nTherefore, the rms voltage is 212.1V212.1V.<\/h3>\n<h2>b) The rms value of current in an ac circuit is 10A10A. What is the peak current?<\/h2>\n<h3>Ans: It is given that,<br \/>\nRms value of current in an ac circuit, Irms=10AIrms=10A<br \/>\nIt is known that,<br \/>\nI0=2\u2013\u221a\u00d7IrmsI0=2\u00d7Irms<br \/>\n\u21d2I0=1.414\u00d710\u21d2I0=1.414\u00d710<br \/>\n\u21d2I0=14.14A\u21d2I0=14.14A<br \/>\nTherefore, the peak current is 14.14A14.14A.<\/h3>\n<h2>3. A 44mH44mH inductor is connected to 220V220V,50Hz50Hz ac supply. Determine the rms value of the current in the circuit.<\/h2>\n<h3>Ans: It is known that,<br \/>\nInductance, L=44mH=44\u00d710\u22123HL=44mH=44\u00d710\u22123H<br \/>\nVoltage,V=220VV=220V<br \/>\nFrequency, fL=50HzfL=50Hz<br \/>\nAngular frequency, \u03c9L=2\u03c0fL\u03c9L=2\u03c0fL<br \/>\nIt is known that,<br \/>\nInductive reactance, XL=\u03c9LL=2\u03c0fLLXL=\u03c9LL=2\u03c0fLL<br \/>\n\u21d2XL=2\u00d73.14\u00d750\u00d744\u00d710\u22123\u03a9\u21d2XL=2\u00d73.14\u00d750\u00d744\u00d710\u22123\u03a9<br \/>\n\u21d2XL=13.8\u03a9\u21d2XL=13.8\u03a9<br \/>\nIrms=VXLIrms=VXL<br \/>\n\u21d2Irms=22013.82\u21d2Irms=22013.82<br \/>\n\u21d2Irms=15.92A\u21d2Irms=15.92A<br \/>\nTherefore, the rms value of the current in the circuit is 15.92A15.92A.<\/h3>\n<h2>4. A 60\u03bcF60\u03bcF capacitor is connected to a 110V110V,60Hz60Hz ac supply. Determine the rms value of the current in the circuit.<\/h2>\n<h3>Ans: It is given that,<br \/>\nCapacitance, C=60\u03bcF=60\u00d710\u22126FC=60\u03bcF=60\u00d710\u22126F<br \/>\nVoltage, V=110VV=110V<br \/>\nFrequency, fC=60HzfC=60Hz<br \/>\nIt is known that,<br \/>\nIrms=VXCIrms=VXC<br \/>\nXC=1\u03c9CC=12\u03c0fCCXC=1\u03c9CC=12\u03c0fCC<br \/>\n\u21d2XC=12\u00d73.14\u00d760\u00d760\u00d710\u22126\u21d2XC=12\u00d73.14\u00d760\u00d760\u00d710\u22126<br \/>\n\u21d2XC=44.248\u03a9\u21d2XC=44.248\u03a9<br \/>\n\u21d2Irms=11044.28\u21d2Irms=11044.28<br \/>\n\u21d2Irms=2.488A\u21d2Irms=2.488A<br \/>\nTherefore, the rms value of the current in the circuit is 2.488A2.488A.<\/h3>\n<h2>5. In exercises 44 and 55 What is the net power absorbed by each circuit over a complete cycle? Explain your answer.<\/h2>\n<h3>Ans: From the inductive circuit,<br \/>\nRms value of current, Irms=15.92AIrms=15.92A<br \/>\nRms value of voltage, Vrms=220VVrms=220V<br \/>\nIt is known that,<br \/>\nNet power absorbed, P=Vrms\u00d7Irmscos\u03d5P=Vrms\u00d7Irmscos\u2061\u03d5<br \/>\nWhere,<br \/>\n\u03d5\u03d5 is the phase difference between voltage and current<br \/>\nFor a pure inductive circuit, the phase difference between alternating voltage and current is 900900i.e., \u03d5=900\u03d5=900<br \/>\n\u21d2P=220\u00d715.92cos900=0\u21d2P=220\u00d715.92cos\u2061900=0<br \/>\nTherefore, net power absorbed is zero in a pure inductive circuit.<br \/>\nIn a capacitive circuit,<br \/>\nRms value of current, Irms=2.49AIrms=2.49A<br \/>\nRms value of voltage, Vrms=110VVrms=110V<br \/>\nIt is known that,<br \/>\nNet power absorbed, P=Vrms\u00d7Irmscos\u03d5P=Vrms\u00d7Irmscos\u2061\u03d5<br \/>\nWhere,<br \/>\n\u03d5\u03d5 is the phase difference between voltage and current<br \/>\nFor a pure capacitive circuit, the phase difference between alternating voltage and current is 900900i.e., \u03d5=900\u03d5=900<br \/>\n\u21d2P=110\u00d72.49cos900=0\u21d2P=110\u00d72.49cos\u2061900=0<br \/>\nTherefore, net power absorbed is zero in a pure capacitive circuit.<\/h3>\n<h2>6. Obtain the resonant frequency \u03c9r\u03c9r of a series LCR circuit with L=2.0HL=2.0H, C=32\u03bcFC=32\u03bcF and R=10\u03a9R=10\u03a9 . What is the Q-value of this current?<\/h2>\n<h3>Ans: It is given that,<br \/>\nInductance, L=2HL=2H<br \/>\nCapacitance,<br \/>\nC=32\u03bcF=32\u00d710\u22126FC=32\u03bcF=32\u00d710\u22126F<br \/>\nR=10\u03a9R=10\u03a9<br \/>\nIt is known that,<br \/>\nResonant frequency, \u03c9r=1LC\u221a\u03c9r=1LC<br \/>\n\u21d2\u03c9r=12\u00d732\u00d710\u22126\u221a\u21d2\u03c9r=12\u00d732\u00d710\u22126<br \/>\n\u21d2\u03c9r=18\u00d710\u22123\u21d2\u03c9r=18\u00d710\u22123<br \/>\n\u21d2\u03c9r=125rad\/s\u21d2\u03c9r=125rad\/s<br \/>\nQ\u2212value=\u03c9rLRQ\u2212value=\u03c9rLR<br \/>\n\u21d2Q=1RLC\u2212\u2212\u221a\u21d2Q=1RLC<br \/>\n\u21d2Q=110232\u00d710\u22126\u2212\u2212\u2212\u2212\u2212\u2212\u221a\u21d2Q=110232\u00d710\u22126<br \/>\n\u21d2Q=110\u00d74\u00d710\u22123\u21d2Q=110\u00d74\u00d710\u22123<br \/>\n\u21d2Q=25\u21d2Q=25<br \/>\nTherefore, the resonant frequency is 125rad\/s125rad\/s and Q-value is 2525.<\/h3>\n<h2>7. A charged 30\u03bcF30\u03bcF capacitor is connected to a 27mH27mH inductor. What is the angular frequency of free oscillations of the circuit?<\/h2>\n<h3>Ans: It is given that,<br \/>\nCapacitance, C=30\u03bcF=30\u00d710\u22126FC=30\u03bcF=30\u00d710\u22126F<br \/>\nInductance, L=27mH=27\u00d710\u22123HL=27mH=27\u00d710\u22123H<br \/>\nIt is known that,<br \/>\nAngular frequency of free oscillations, \u03c9r=1LC\u221a\u03c9r=1LC<br \/>\n\u21d2\u03c9r=127\u00d710\u22123\u00d730\u00d710\u22126\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a\u21d2\u03c9r=127\u00d710\u22123\u00d730\u00d710\u22126<br \/>\n\u21d2\u03c9r=19\u00d710\u22124\u21d2\u03c9r=19\u00d710\u22124<br \/>\n\u21d2\u03c9r=1.11\u00d7103rad\/s\u21d2\u03c9r=1.11\u00d7103rad\/s<br \/>\nTherefore, the angular frequency of free oscillations of the circuit is<br \/>\n1.11\u00d7103rad\/s1.11\u00d7103rad\/s<\/h3>\n<h2>8. Suppose the initial charge on the capacitor in exercise 77 is 6mC6mC . What is the total energy stored in the circuit initially? What is the total energy at a later time?<\/h2>\n<h3>Ans: It is known that,<br \/>\nCapacitance of the capacitor, C=30\u03bcF=30\u00d710\u22126FC=30\u03bcF=30\u00d710\u22126F<br \/>\nInductance of the capacitor, L=27mH=27\u00d710\u22123HL=27mH=27\u00d710\u22123H<br \/>\nCharge on the capacitor, Q=6mC=6\u00d710\u22123CQ=6mC=6\u00d710\u22123C<br \/>\nIt is known that,<br \/>\nEnergy,E=12Q2CE=12Q2C<br \/>\n\u21d2E=12(6\u00d710\u22123)230\u00d710\u22126\u21d2E=12(6\u00d710\u22123)230\u00d710\u22126<br \/>\n\u21d2E=610=0.6J\u21d2E=610=0.6J<br \/>\nTherefore, the energy stored in the circuit initially is E=0.6JE=0.6J.<br \/>\nTotal energy at later time will remain same as the initially stored i.e., 0.6J0.6J because energy is shared between the capacitor and the inductor.<\/h3>\n<h2>9. A series LCR circuit with R=20\u03a9R=20\u03a9, L=1.5HL=1.5H and C=35\u03bcFC=35\u03bcF is connected to a variable frequency 200V200V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?<\/h2>\n<h3>Ans: It is known that,<br \/>\nResistance, R=20\u03a9R=20\u03a9<br \/>\nInductance, L=1.5HL=1.5H<br \/>\nCapacitance, C=35\u03bcF=35\u00d710\u22126FC=35\u03bcF=35\u00d710\u22126F<br \/>\nVoltage, V=200VV=200V<br \/>\nIt is known that,<br \/>\nImpedance,Z=R2+(XL\u2212XC)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221aZ=R2+(XL\u2212XC)2<br \/>\nAt resonance, XL=XCXL=XC<br \/>\n\u21d2Z=R=20\u03a9\u21d2Z=R=20\u03a9<br \/>\nI=VZ=20020I=VZ=20020<br \/>\n\u21d2I=10A\u21d2I=10A<br \/>\nAverage power, P=I2RP=I2R<br \/>\n\u21d2P=102\u00d720\u21d2P=102\u00d720<br \/>\n\u21d2P=2000W\u21d2P=2000W<br \/>\nTherefore, the average power transferred is 2000W2000W.<\/h3>\n<h2>10. A radio can tune over the frequency range of a portion of MWMW broadcast band: (800kHz800kHz to 1200kHz1200kHz). If its LC circuit has an effective inductance of 200\u03bcH200\u03bcH, what must be the range of its variable capacitor?<br \/>\n(Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave.)<\/h2>\n<h3>Ans: It is given that,<br \/>\nThe range of frequency(f)(f) of a radio is 800kHz800kHz to 1200kHz1200kHz.<br \/>\nEffective inductance of the circuit, L=200\u03bcH=200\u00d710\u22126HL=200\u03bcH=200\u00d710\u22126H<br \/>\nIt is known that,<br \/>\nCapacitance of variable capacitor for f1f1 is C1=1\u03c912LC1=1\u03c912L<br \/>\nWhere,<br \/>\n\u03c91\u03c91 is the angular frequency for capacitor for f1=2\u03c0f1f1=2\u03c0f1<br \/>\n\u21d2\u03c91=2\u00d73.14\u00d7800\u00d7103rad\/s\u21d2\u03c91=2\u00d73.14\u00d7800\u00d7103rad\/s<br \/>\n\u21d2C1=1(2\u00d73.14\u00d7800\u00d7103)2\u00d7200\u00d710\u22126\u21d2C1=1(2\u00d73.14\u00d7800\u00d7103)2\u00d7200\u00d710\u22126<br \/>\n\u21d2C1=1.9809\u00d710\u221210F\u21d2C1=1.9809\u00d710\u221210F<br \/>\n\u21d2C1=198.1pF\u21d2C1=198.1pF<br \/>\nC2=1\u03c922LC2=1\u03c922L<br \/>\n\u21d2C2=1(2\u00d73.14\u00d71200\u00d7103)2\u00d7200\u00d710\u22126\u21d2C2=1(2\u00d73.14\u00d71200\u00d7103)2\u00d7200\u00d710\u22126<br \/>\n\u21d2C2=0.8804\u00d710\u221210F\u21d2C2=0.8804\u00d710\u221210F<br \/>\n\u21d2C2=88.04pF\u21d2C2=88.04pF<br \/>\nTherefore, the range of the variable capacitor is from 88.04pF88.04pF to 198.1pF198.1pF.<\/h3>\n<h2>11. Figure shows a series LCR circuit connected to a variable frequency 230V230V source. L=5.0HL=5.0H, C=80\u03bcFC=80\u03bcF, R=40\u03a9R=40\u03a9 .<br \/>\n(Image will be Uploaded Soon)<br \/>\na) Determine the source frequency which drives the circuit in resonance.<\/h2>\n<h3>Ans: It is given that,<br \/>\nVoltage, V=230VV=230V<br \/>\nInductance, L=5.0HL=5.0H<br \/>\nCapacitance, C=80\u03bcF=80\u00d710\u22126FC=80\u03bcF=80\u00d710\u22126F<br \/>\nResistance, R=40\u03a9R=40\u03a9<br \/>\nIt is known that,<br \/>\nSource frequency at resonance=1LC\u221a=1LC<br \/>\n\u21d215\u00d780\u00d710\u22126\u221a=50rad\/s\u21d215\u00d780\u00d710\u22126=50rad\/s<br \/>\nTherefore, the source frequency of the circuit in resonance is 50rad\/s50rad\/s.<\/h3>\n<h2>b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.<\/h2>\n<h3>Ans: It is known that,<br \/>\nAt resonance, Impedance,Z=Z= Resistance,RR<br \/>\n\u21d2Z=R=40\u03a9\u21d2Z=R=40\u03a9<br \/>\nI=VZI=VZ<br \/>\n\u21d2I=23040=5.75A\u21d2I=23040=5.75A<br \/>\nAmplitude, I0=1.414\u00d7II0=1.414\u00d7I<br \/>\n\u21d2I0=1.414\u00d75.75\u21d2I0=1.414\u00d75.75<br \/>\n\u21d2I0=8.13A\u21d2I0=8.13A<br \/>\nTherefore, the impedance of the circuit is 40\u03a940\u03a9 and the amplitude of current at resonating frequency is 8.13A8.13A.<\/h3>\n<h2>c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.<\/h2>\n<h3>Ans: It is known that,<br \/>\nPotential drop, V=IRV=IR<br \/>\nAcross resistor, VR=IRVR=IR<br \/>\n\u21d2VR=5.75\u00d740=230V\u21d2VR=5.75\u00d740=230V<br \/>\nAcross capacitor, VC=IXC=I\u03c9CVC=IXC=I\u03c9C<br \/>\n\u21d2VC=5.75\u00d7150\u00d780\u00d710\u22126\u21d2VC=5.75\u00d7150\u00d780\u00d710\u22126<br \/>\n\u21d2VC=1437.5V\u21d2VC=1437.5V<br \/>\nAcross Inductor, VL=IXL=I\u03c9LVL=IXL=I\u03c9L<br \/>\n\u21d2VL=5.75\u00d750\u00d75\u21d2VL=5.75\u00d750\u00d75<br \/>\n\u21d2VL=1437.5V\u21d2VL=1437.5V<br \/>\nAcross LC combination, VLC=I(XL\u2212XC)VLC=I(XL\u2212XC)<br \/>\nAt resonance, XL=XCXL=XC<br \/>\n\u21d2VLC=0\u21d2VLC=0<br \/>\nTherefore, the rms potential drop across Resistor is 230V230V, Capacitor is 1437.5V1437.5V, Inductor is 1437.5V1437.5V and the potential drop across LC combination is zero at resonating frequency.<\/h3>\n<h2>12. An LC circuit contains a 20mH20mH inductor and a 50\u03bcF50\u03bcF capacitor with an initial charge of 10mC10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0t=0.<br \/>\na) What is the total energy stored initially? Is it conserved during LC oscillations?<\/h2>\n<h3>Ans: It is given that,<br \/>\nInductance of the inductor, L=20mH=20\u00d710\u22123HL=20mH=20\u00d710\u22123H<br \/>\nCapacitance of the capacitor, C=50\u03bcF=50\u00d710\u22126FC=50\u03bcF=50\u00d710\u22126F<br \/>\nInitial charge on the capacitor, Q=10mC=10\u00d710\u22123CQ=10mC=10\u00d710\u22123C<br \/>\nIt is known that,<br \/>\nTotal energy stored initially in the circuit, E=12Q2CE=12Q2C<br \/>\n\u21d2E=(10\u00d710\u22123)22\u00d750\u00d710\u22126=1J\u21d2E=(10\u00d710\u22123)22\u00d750\u00d710\u22126=1J<br \/>\nTherefore, the total energy stored in the LC circuit will be conserved because there is no resistor (R=0)(R=0) connected in the circuit.<\/h3>\n<h2>b) What is the natural frequency of the circuit?<\/h2>\n<h3>Ans: It is known that,<br \/>\nNatural frequency of the circuit, \u03bd=12\u03c0LC\u221a\u03bd=12\u03c0LC<br \/>\n\u21d2\u03bd=12\u03c020\u00d710\u22123\u00d750\u00d710\u22126\u221a\u21d2\u03bd=12\u03c020\u00d710\u22123\u00d750\u00d710\u22126<br \/>\n\u21d2\u03bd=1032\u03c0=159.24Hz\u21d2\u03bd=1032\u03c0=159.24Hz<br \/>\nNatural angular frequency, \u03c9r=1LC\u221a\u03c9r=1LC<br \/>\n\u21d2\u03c9r=120\u00d710\u22123\u00d750\u00d710\u22126\u221a\u21d2\u03c9r=120\u00d710\u22123\u00d750\u00d710\u22126<br \/>\n\u21d2\u03c9r=110\u22126\u221a=103rad\/s\u21d2\u03c9r=110\u22126=103rad\/s<br \/>\nTherefore, the natural frequency is 159.24Hz159.24Hz and the natural angular frequency is 103rad\/s103rad\/s.<\/h3>\n<h2>c) At what time is the energy stored (i)(i) completely electrical (i.e., stored in the capacitor)? (ii)(ii) completely magnetic (i.e., stored in the inductor)?<\/h2>\n<h3>Ans:<br \/>\n(i) Completely electrical:<br \/>\nIt is known that,<br \/>\nTime period for LC oscillations, T=1\u03bdT=1\u03bd<br \/>\n\u21d2T=1159.24=6.28ms\u21d2T=1159.24=6.28ms<br \/>\nTotal charge on the capacitor at time tt, Q\u2032=Qcos(2\u03c0Tt)Q\u2032=Qcos\u2061(2\u03c0Tt)<br \/>\nIf energy stored is electrical, Q\u2032=\u00b1QQ\u2032=\u00b1Q<br \/>\nTherefore, it can be inferred that the energy stored in the capacitor is completely electrical at time, t=0,T2,T,3T2,&#8230;&#8230;&#8230;&#8230;..t=0,T2,T,3T2,&#8230;&#8230;&#8230;&#8230;.. where, T=6.3msT=6.3ms.<br \/>\n(ii) Completely magnetic:<br \/>\nMagnetic energy is maximum, when electrical energy Q\u2032Q\u2032 is equal to 00.<br \/>\nTherefore, it can be inferred that the energy stored is completely magnetic at time, t=T4,3T4,5T4,&#8230;&#8230;&#8230;..t=T4,3T4,5T4,&#8230;&#8230;&#8230;.. where, T=6.3msT=6.3ms.<\/h3>\n<h2>d) At what times is the total energy shared equally between the inductor and the capacitor?<\/h2>\n<h3>Ans: Consider, Q\u2032Q\u2032 be the charge on capacitor when total energy is equally shared between the capacitor and the inductor at time tt.<br \/>\nWhen total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor=1\/2(=1\/2(maximum energy)).<br \/>\n\u21d212(Q\u2032)2C=12(12Q2C)\u21d212(Q\u2032)2C=12(12Q2C)<br \/>\n\u21d212(Q\u2032)2C=14Q2C\u21d212(Q\u2032)2C=14Q2C<br \/>\n\u21d2Q\u2032=Q2\u221a\u21d2Q\u2032=Q2<br \/>\nIt is known that, Q\u2032=Qcos2\u03c0TtQ\u2032=Qcos\u20612\u03c0Tt<br \/>\n\u21d2Q2\u221a=Qcos2\u03c0Tt\u21d2Q2=Qcos\u20612\u03c0Tt<br \/>\n\u21d2cos2\u03c0Tt=12\u221a=cos(2n+1)\u03c04;\u21d2cos\u20612\u03c0Tt=12=cos\u2061(2n+1)\u03c04; n=0,1,2,3&#8230;..n=0,1,2,3&#8230;..<br \/>\n\u21d2t=(2n+1)T8\u21d2t=(2n+1)T8<br \/>\nTherefore, the total energy is equally shared between the inductor and the capacitor at the time,t=T8,3T8,5T8&#8230;&#8230;&#8230;..t=T8,3T8,5T8&#8230;&#8230;&#8230;&#8230;<\/h3>\n<h2>e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?<\/h2>\n<h3>Ans: If a resistor is included in the circuit, then the total initial energy gets dissipated as heat energy in the circuit. The LC oscillation gets damped due to the resistance.<\/h3>\n<h2>13. A coil of inductance 0.5H0.5H and resistance 100\u03a9100\u03a9 is connected to a 240V,50Hz240V,50Hz ac supply.<br \/>\na) What is the maximum current in the coil?<\/h2>\n<h3>Ans: It is given that,<br \/>\nInductance of the inductor, L=0.5HL=0.5H<br \/>\nResistance of the resistor, R=100\u03a9R=100\u03a9<br \/>\nPotential of the supply voltage, V=240VV=240V<br \/>\nFrequency of the supply, \u03bd=50Hz\u03bd=50Hz<br \/>\nIt is known that,<br \/>\nPeak voltage, V0=2\u2013\u221aVV0=2V<br \/>\n\u21d2V0=2\u2013\u221a\u00d7240\u21d2V0=2\u00d7240<br \/>\n\u21d2V0=339.41V\u21d2V0=339.41V<br \/>\nAngular frequency of the supply, \u03c9=2\u03c0\u03bd\u03c9=2\u03c0\u03bd<br \/>\n\u21d2\u03c9=2\u03c0\u00d750=100\u03c0rad\/s\u21d2\u03c9=2\u03c0\u00d750=100\u03c0rad\/s<br \/>\nMaximum current in the circuit, I0=V0R2+\u03c92L2\u221aI0=V0R2+\u03c92L2<br \/>\n\u21d2I0=339.41(100)2+(100\u03c0)2(0.50)2\u221a=1.82A\u21d2I0=339.41(100)2+(100\u03c0)2(0.50)2=1.82A<br \/>\nTherefore, the maximum current in the coil is 1.82A1.82A.<\/h3>\n<h2>b) What is the time lag between the voltage maximum and the current maximum?<\/h2>\n<h3>Ans: It is known that,<br \/>\nEquation for voltage, V=V0cos\u03c9tV=V0cos\u2061\u03c9t<br \/>\nEquation for current, I=I0cos(\u03c9t\u2212\u03d5)I=I0cos\u2061(\u03c9t\u2212\u03d5)<br \/>\nWhere,<br \/>\n\u03d5\u03d5 is the phase difference between voltage and current.<br \/>\nAt time t=0t=0, V=V0V=V0 (voltage is maximum)<br \/>\nIf \u03c9t\u2212\u03d5=0\u03c9t\u2212\u03d5=0 i.e., at t=\u03d5\u03c9t=\u03d5\u03c9 , I=I0I=I0 (current is maximum)<br \/>\nTherefore, the time lag between maximum voltage and maximum current is \u03d5\u03c9\u03d5\u03c9 .<br \/>\n\u21d2tan\u03d5=\u03c9LR\u21d2tan\u2061\u03d5=\u03c9LR<br \/>\n\u21d2tan\u03d5=2\u03c0\u00d750\u00d70.5100=1.57\u21d2tan\u2061\u03d5=2\u03c0\u00d750\u00d70.5100=1.57<br \/>\n\u21d2\u03d5=tan\u22121(1.57)\u21d2\u03d5=tan\u22121(1.57)<br \/>\n\u21d2\u03d5=57.5\u2218=57.5\u03c0180rad\u21d2\u03d5=57.5\u2218=57.5\u03c0180rad<br \/>\nTime lag, t=\u03d5\u03c9t=\u03d5\u03c9<br \/>\n\u21d2t=57.5\u03c0180\u00d72\u03c0\u00d750\u21d2t=57.5\u03c0180\u00d72\u03c0\u00d750<br \/>\n\u21d2t=3.19\u00d710\u22123s\u21d2t=3.19\u00d710\u22123s<br \/>\n\u21d2t=3.2ms\u21d2t=3.2ms<br \/>\nTherefore, the time lag between the maximum voltage and maximum current is 3.2ms3.2ms.<\/h3>\n<h2>14. Obtain the answers (a)(a) to (b)(b) in Exercise 1313 if the circuit is connected to a high frequency supply (240V,10kHz)(240V,10kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?<\/h2>\n<h3>Ans: It is given that,<br \/>\nInductance of the inductor, L=0.5HL=0.5H<br \/>\nResistance of the resistor, R=100\u03a9R=100\u03a9<br \/>\nPotential of the supply voltage, V=240VV=240V<br \/>\nFrequency of the supply, \u03bd=10kHz=104Hz\u03bd=10kHz=104Hz<br \/>\nAngular frequency, \u03c9=2\u03c0\u03bd=2\u03c0\u00d7104rad\/s\u03c9=2\u03c0\u03bd=2\u03c0\u00d7104rad\/s<br \/>\nPeak Voltage, V0=V2\u2013\u221a=1102\u2013\u221aVV0=V2=1102V<br \/>\n1. Maximum current, I0=V0R2+\u03c92C2\u221aI0=V0R2+\u03c92C2<br \/>\n\u21d2I0=2402\u221a(100)2+(2\u03c0\u00d7104)2\u00d7(0.5)2\u221a=1.1\u00d710\u22122A\u21d2I0=2402(100)2+(2\u03c0\u00d7104)2\u00d7(0.5)2=1.1\u00d710\u22122A<br \/>\nTherefore, the maximum current in the coil is 1.1\u00d710\u22122A1.1\u00d710\u22122A.<br \/>\n2. The time lag between maximum voltage and maximum current is \u03d5\u03c9\u03d5\u03c9 .<br \/>\nFor phase difference \u03d5\u03d5:tan\u03d5=\u03c9LRtan\u2061\u03d5=\u03c9LR<br \/>\n\u21d2tan\u03d5=2\u03c0\u00d7104\u00d70.5100=100\u03c0\u21d2tan\u2061\u03d5=2\u03c0\u00d7104\u00d70.5100=100\u03c0<br \/>\n\u21d2\u03d5=tan\u22121(100\u03c0)\u21d2\u03d5=tan\u22121(100\u03c0)<br \/>\n\u21d2\u03d5=89.82\u2218=89.82\u03c0180rad\u21d2\u03d5=89.82\u2218=89.82\u03c0180rad<br \/>\nTime lag, t=\u03d5\u03c9t=\u03d5\u03c9<br \/>\n\u21d2t=89.82\u03c0180\u00d72\u03c0\u00d7104\u21d2t=89.82\u03c0180\u00d72\u03c0\u00d7104<br \/>\n\u21d2t=25\u00d710\u22126s\u21d2t=25\u00d710\u22126s<br \/>\n\u21d2t=25\u03bcs\u21d2t=25\u03bcs<br \/>\nTherefore, the time lag between the maximum voltage and maximum current is 25\u03bcs25\u03bcs.<br \/>\nIt can be observed that I0I0 is very small in this case.<br \/>\nThus, at high frequencies, the inductor amounts to an open circuit.<br \/>\nIn a dc circuit, after a steady state is achieved, \u03c9=0\u03c9=0. Thus, inductor LL behaves like a pure conducting object.<\/h3>\n<h2>15. A 100\u03bcF100\u03bcF capacitor in series with a 40\u03a940\u03a9 resistance is connected to a 110V,60Hz110V,60Hz supply.<br \/>\na) What is the maximum current in the circuit?<\/h2>\n<h3>Ans: It is given that,<br \/>\nCapacitance of the capacitor, C=100\u03bcF=100\u00d710\u22126FC=100\u03bcF=100\u00d710\u22126F<br \/>\nResistance of the resistor, R=40\u03a9R=40\u03a9<br \/>\nSupply voltage, V=110VV=110V<br \/>\nFrequency oscillations, \u03bd=60Hz\u03bd=60Hz<br \/>\nAngular frequency, \u03c9=2\u03c0\u03bd=2\u03c0\u00d760rad\/s\u03c9=2\u03c0\u03bd=2\u03c0\u00d760rad\/s<br \/>\nIt is known that,<br \/>\nFor a RC circuit, Impedance: Z=R2+1\u03c92C2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221aZ=R2+1\u03c92C2<br \/>\nPeak Voltage, V0=V2\u2013\u221a=1102\u2013\u221aVV0=V2=1102V<br \/>\nMaximum current; I0=V0ZI0=V0Z<br \/>\n\u21d2I0=V0R2+1\u03c92C2\u221a\u21d2I0=V0R2+1\u03c92C2<br \/>\n\u21d2I0=1102\u221a(40)2+1(120\u03c0)2(10\u22124)2\u221a\u21d2I0=1102(40)2+1(120\u03c0)2(10\u22124)2<br \/>\n\u21d2I0=1102\u2013\u221a1600+1(120\u03c0)2(10\u22124)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a=3.24A\u21d2I0=11021600+1(120\u03c0)2(10\u22124)2=3.24A<br \/>\nTherefore, the maximum current in the circuit is 3.24A3.24A.<\/h3>\n<h2>b) What is the time lag between the current maximum and the voltage maximum?<\/h2>\n<h3>Ans: It is known that,<br \/>\nIn a capacitor circuit, the voltage lags behind the current by a phase angle of \u03d5\u03d5.<br \/>\ntan\u03d5=1\u03c9CR=1\u03c9CRtan\u2061\u03d5=1\u03c9CR=1\u03c9CR<br \/>\n\u21d2tan\u03d5=1120\u03c0\u00d710\u22124\u00d740=0.6635\u21d2tan\u2061\u03d5=1120\u03c0\u00d710\u22124\u00d740=0.6635<br \/>\n\u21d2\u03d5=tan\u22121(0.6635)\u21d2\u03d5=tan\u22121(0.6635)<br \/>\n\u21d2\u03d5=33.56\u2218=33.56\u03c0180rad\u21d2\u03d5=33.56\u2218=33.56\u03c0180rad<br \/>\nIt is known that,<br \/>\nTime lag, t=\u03d5\u03c9t=\u03d5\u03c9<br \/>\n\u21d2t=33.56\u03c0180\u00d7120\u03c0\u21d2t=33.56\u03c0180\u00d7120\u03c0<br \/>\n\u21d2t=1.55\u00d710\u22123s\u21d2t=1.55\u00d710\u22123s<br \/>\n\u21d2t=1.55ms\u21d2t=1.55ms<br \/>\nTherefore, the time lag between maximum current and maximum voltage is 1.55ms1.55ms.<\/h3>\n<h2>16. Obtain the answers to (a)(a) and (b)(b) in Exercise 1515 if the circuit is connected to a<br \/>\n110V,12kHz110V,12kHz<br \/>\nsupply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.<\/h2>\n<h3>Ans: It is given that,<br \/>\nCapacitance of the capacitor, C=100\u03bcF=100\u00d710\u22126FC=100\u03bcF=100\u00d710\u22126F<br \/>\nResistance of the resistor, R=40\u03a9R=40\u03a9<br \/>\nSupply voltage, V=110VV=110V<br \/>\nFrequency oscillations, \u03bd=12kHz=12\u00d7103Hz\u03bd=12kHz=12\u00d7103Hz<br \/>\nAngular frequency, \u03c9=2\u03c0\u03bd=2\u03c0\u00d712\u00d7103rad\/s=24\u03c0\u00d7103rad\/s\u03c9=2\u03c0\u03bd=2\u03c0\u00d712\u00d7103rad\/s=24\u03c0\u00d7103rad\/s<br \/>\nPeak Voltage, V0=V2\u2013\u221a=1102\u2013\u221aVV0=V2=1102V<br \/>\n1. It is known that,<br \/>\nFor a RC circuit, Impedance: Z=R2+1\u03c92C2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221aZ=R2+1\u03c92C2<br \/>\nMaximum current; I0=V0ZI0=V0Z<br \/>\n\u21d2I0=V0R2+1\u03c92C2\u221a\u21d2I0=V0R2+1\u03c92C2<br \/>\n\u21d2I0=1102\u221a(40)2+1(24\u03c0\u00d7103)2(10\u22124)2\u221a\u21d2I0=1102(40)2+1(24\u03c0\u00d7103)2(10\u22124)2<br \/>\n\u21d2I0=1102\u2013\u221a1600+(1024\u03c0)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a=3.9A\u21d2I0=11021600+(1024\u03c0)2=3.9A<br \/>\nTherefore, the maximum current in the circuit is 3.9A3.9A.<br \/>\n2. It is known that,<br \/>\nIn a capacitor circuit, the voltage lags behind the current by a phase angle of \u03d5\u03d5.<br \/>\ntan\u03d5=1\u03c9CR=1\u03c9CRtan\u2061\u03d5=1\u03c9CR=1\u03c9CR<br \/>\n\u21d2tan\u03d5=124\u03c0\u00d7103\u00d710\u22124\u00d740=196\u03c0\u21d2tan\u2061\u03d5=124\u03c0\u00d7103\u00d710\u22124\u00d740=196\u03c0<br \/>\n\u21d2\u03d5=tan\u22121(196\u03c0)\u21d2\u03d5=tan\u22121(196\u03c0)<br \/>\n\u21d2\u03d5=0.2\u2218=0.2\u03c0180rad\u21d2\u03d5=0.2\u2218=0.2\u03c0180rad<br \/>\nIt is known that,<br \/>\nTime lag, t=\u03d5\u03c9t=\u03d5\u03c9<br \/>\n\u21d2t=0.2\u03c0180\u00d724\u03c0\u00d7103\u21d2t=0.2\u03c0180\u00d724\u03c0\u00d7103<br \/>\n\u21d2t=0.04\u00d710\u22126s\u21d2t=0.04\u00d710\u22126s<br \/>\n\u21d2t=0.04\u03bcs\u21d2t=0.04\u03bcs<br \/>\nTherefore, the time lag between maximum current and maximum voltage is 0.04\u03bcs0.04\u03bcs.<br \/>\nIt can be concluded that \u03d5\u03d5 tends to become zero at high frequencies. At a high frequency, capacitor CC acts as a conductor.<br \/>\nIn a dc circuit, after the steady state is achieved, \u03c9=0\u03c9=0. Therefore, capacitor CC amounts to an open circuit.<\/h3>\n<h2>17. Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if three elements, L,C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 11 for this frequency.<\/h2>\n<h3>Ans: It is given that,<br \/>\nAn inductor (L)(L), a capacitor (C)(C) and a resistor (R)(R) is connected in parallel with each other in a circuit where,<br \/>\nInductance, L=5.0HL=5.0H<br \/>\nCapacitance, C=80\u03bcF=80\u00d710\u22126FC=80\u03bcF=80\u00d710\u22126F<br \/>\nResistance, R=40\u03a9R=40\u03a9<br \/>\nPotential of the voltage source, V=230VV=230V<br \/>\nIt is known that,<br \/>\nImpedance (Z)(Z) of the given LCR circuit is given as:<br \/>\n1Z=1R2+(1\u03c9L\u2212\u03c9C)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a1Z=1R2+(1\u03c9L\u2212\u03c9C)2<br \/>\nWhere,<br \/>\n\u03c9\u03c9 is the angular frequency<br \/>\nAt resonance: 1\u03c9L\u2212\u03c9C=01\u03c9L\u2212\u03c9C=0<br \/>\n\u21d2\u03c9=1LC\u221a\u21d2\u03c9=1LC<br \/>\n\u21d2\u03c9=15\u00d780\u00d710\u22126\u221a=50rad\/s\u21d2\u03c9=15\u00d780\u00d710\u22126=50rad\/s<br \/>\nTherefore, the magnitude of ZZ is maximum at 50rad\/s50rad\/s and the total current is minimum.<br \/>\nRms current flowing through inductor LL: IL=V\u03c9LIL=V\u03c9L<br \/>\n\u21d2IL=23050\u00d75=0.92A\u21d2IL=23050\u00d75=0.92A<br \/>\nRms current flowing through capacitor CC: IC=V1\u03c9C=\u03c9CVIC=V1\u03c9C=\u03c9CV<br \/>\n\u21d2IC=50\u00d780\u00d710\u22126\u00d7230=0.92A\u21d2IC=50\u00d780\u00d710\u22126\u00d7230=0.92A<br \/>\nRms current flowing through resistor RR: IR=VRIR=VR<br \/>\n\u21d2IR=23040=5.75A\u21d2IR=23040=5.75A<br \/>\nThe current RMS value in the inductor is 0.92A0.92A, in the capacitor is 0.92A0.92A and in the resistor is 5.75A5.75A.<\/h3>\n<h2>18. A circuit containing an 80mH80mH inductor and a 60\u03bcF60\u03bcF capacitor in series is connected to a<br \/>\n230V,50Hz230V,50Hz<br \/>\nsupply. The resistance of the circuit is negligible.<br \/>\na) Obtain the current amplitude and rms values.<\/h2>\n<h3>Ans: It is given that,<br \/>\nInductance, L=80mH=80\u00d710\u22123HL=80mH=80\u00d710\u22123H<br \/>\nCapacitance, C=60\u03bcF=60\u00d710\u22126FC=60\u03bcF=60\u00d710\u22126F<br \/>\nSupply voltage, V=230VV=230V<br \/>\nFrequency, \u03bd=50Hz\u03bd=50Hz<br \/>\nAngular frequency, \u03c9=2\u03c0\u03bd=100\u03c0rad\/s\u03c9=2\u03c0\u03bd=100\u03c0rad\/s<br \/>\nPeak voltage, V0=V2\u2013\u221a=2302\u2013\u221aVV0=V2=2302V<br \/>\nIt is known that,<br \/>\nMaximum current: I0=V0(\u03c9L\u22121\u03c9C)I0=V0(\u03c9L\u22121\u03c9C)<br \/>\n\u21d2I0=2303\u2013\u221a(100\u03c0\u00d780\u00d710\u22123\u22121100\u03c0\u00d760\u00d710\u22126)\u21d2I0=2303(100\u03c0\u00d780\u00d710\u22123\u22121100\u03c0\u00d760\u00d710\u22126)<br \/>\n\u21d2I0=2303\u2013\u221a(8\u03c0\u221210006\u03c0)=\u221211.63A\u21d2I0=2303(8\u03c0\u221210006\u03c0)=\u221211.63A<br \/>\nThe negative sign is because \u03c9L&lt;1\u03c9C\u03c9L&lt;1\u03c9C<br \/>\nAmplitude of maximum current, |I0|=11.63A|I0|=11.63A<br \/>\n\u21d2I=I02\u221a=\u221211.632\u221a\u21d2I=I02=\u221211.632<br \/>\n\u21d2I=\u22128.22A\u21d2I=\u22128.22A, which is the rms value of current.<\/h3>\n<h2>b) Obtain the rms values of potential drops across each element.<\/h2>\n<h3>Ans: It is known that,<br \/>\nPotential difference across the inductor, VL=I\u00d7\u03c9LVL=I\u00d7\u03c9L<br \/>\n\u21d2VL=8.22\u00d7100\u03c0\u00d780\u00d710\u22123\u21d2VL=8.22\u00d7100\u03c0\u00d780\u00d710\u22123<br \/>\n\u21d2VL=206.61V\u21d2VL=206.61V<br \/>\nPotential difference across the capacitor, VC=I\u00d71\u03c9CVC=I\u00d71\u03c9C<br \/>\n\u21d2VC=8.22\u00d71100\u03c0\u00d760\u00d710\u22126\u21d2VC=8.22\u00d71100\u03c0\u00d760\u00d710\u22126<br \/>\n\u21d2VC=436.3V\u21d2VC=436.3V, which is the rms value of potential drop.<\/h3>\n<h2>c) What is the average power transferred to the inductor?<\/h2>\n<h3>Ans: Average power transferred to the inductor is zero as actual voltage leads the current by \u03c02\u03c02.<\/h3>\n<h2>d) What is the average power transferred to the capacitor?<\/h2>\n<h3>Ans: Average power transferred to the capacitor is zero as actual voltage lags the current by \u03c02\u03c02.<\/h3>\n<h2>e) What is the total average power absorbed by the circuit? (\u2018Average\u2019 implies \u2018averaged over one cycle\u2019.)<\/h2>\n<h3>Ans: The total average power absorbed (averaged over one cycle) is zero.<\/h3>\n<h2>19. Suppose the circuit in Exercise 1818 has a resistance of 15\u03a915\u03a9 . Obtain the average power transferred to each element of the circuit, and the total power absorbed.<\/h2>\n<h3>Ans: It is given that,<br \/>\nAverage power transferred to the resistor<br \/>\n=788.44W=788.44W<br \/>\nAverage power transferred to the capacitor =0W=0W<br \/>\nTotal power absorbed by the circuit<br \/>\n=788.44W=788.44W<br \/>\nInductance of inductor,<br \/>\nL=80mH=80\u00d710\u22123HL=80mH=80\u00d710\u22123H<br \/>\nCapacitance of capacitor,<br \/>\nC=60\u03bcF=60\u00d710\u22126FC=60\u03bcF=60\u00d710\u22126F<br \/>\nResistance of resistor,<br \/>\nR=15\u03a9R=15\u03a9<br \/>\nPotential of voltage supply,<br \/>\nV=230VV=230V<br \/>\nFrequency of signal,<br \/>\n\u03bd=50Hz\u03bd=50Hz<br \/>\nAngular frequency of signal,<br \/>\n\u03c9=2\u03c0\u03bd=2\u03c0\u00d7(50)=100\u03c0rad\/s\u03c9=2\u03c0\u03bd=2\u03c0\u00d7(50)=100\u03c0rad\/s<br \/>\nIt is known that,<br \/>\nImpedance, Z=R2+(\u03c9L\u22121\u03c9C)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221aZ=R2+(\u03c9L\u22121\u03c9C)2<br \/>\n\u21d2Z=(15)2+(100\u03c0(80\u00d710\u22123)\u22121(100\u03c0\u00d760\u00d710\u22126))2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a\u21d2Z=(15)2+(100\u03c0(80\u00d710\u22123)\u22121(100\u03c0\u00d760\u00d710\u22126))2<br \/>\n\u21d2Z=(15)2+(25.12\u221253.08)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a=31.728\u03a9\u21d2Z=(15)2+(25.12\u221253.08)2=31.728\u03a9<br \/>\nNow,<br \/>\nI=VZI=VZ<br \/>\n\u21d2I=23031.728=7.25A\u21d2I=23031.728=7.25A<br \/>\nThe elements are connected in series to each other. Therefore, impedance of the circuit is given as current flowing in the circuit,<br \/>\nAverage power transferred to resistance is given as: PR=I2RPR=I2R<br \/>\n\u21d2PR=(7.25)2\u00d715=788.44W\u21d2PR=(7.25)2\u00d715=788.44W<br \/>\nAverage power transferred to capacitor,PC=PC= Average power transferred to inductor, PL=0PL=0<br \/>\nTotal power absorbed by the circuit: PT=PR+PC+PLPT=PR+PC+PL<br \/>\nPT=788.44+0+0=788.44WPT=788.44+0+0=788.44W<br \/>\nTherefore, the total power absorbed by the circuit is<br \/>\n788.44W788.44W<br \/>\n.<\/h3>\n<h2>20. A series LCR circuit with<br \/>\nL=0.12H,C=480nF,R=23 \u03a9L=0.12H,C=480nF,R=23 \u03a9<br \/>\nis connected to a 230V230V variable frequency supply.<br \/>\na) What is the source frequency for which current amplitude is maximum? Obtain this maximum value.<\/h2>\n<h3>Ans: It is given that,<br \/>\nInductance, L=0.12HL=0.12H<br \/>\nCapacitance, C=480nF=480\u00d710\u22129FC=480nF=480\u00d710\u22129F<br \/>\nResistance, R=23\u03a9R=23\u03a9<br \/>\nSupply voltage, V=230VV=230V<br \/>\nPeak voltage, V0=230\u00d72\u2013\u221a=325.22VV0=230\u00d72=325.22V<br \/>\nIt is known that,<br \/>\nCurrent flowing in the circuit, I0=V0R2+(\u03c9L\u22121\u03c9C)2\u221aI0=V0R2+(\u03c9L\u22121\u03c9C)2<br \/>\nWhere,<br \/>\nI0I0 is maximum at resonance.<br \/>\nAt resonance: \u03c9RL\u22121\u03c9RC=0\u03c9RL\u22121\u03c9RC=0<br \/>\nWhere,<br \/>\n\u03c9R\u03c9Ris the resonance angular frequency<br \/>\n\u03c9R=1LC\u221a\u03c9R=1LC<br \/>\n\u21d2\u03c9R=10.12\u00d7480\u00d710\u22129\u221a\u21d2\u03c9R=10.12\u00d7480\u00d710\u22129<br \/>\n\u21d2\u03c9R=4166.67rad\/s\u21d2\u03c9R=4166.67rad\/s<br \/>\nResonant frequency, \u03bdR=\u03c9R2\u03c0\u03bdR=\u03c9R2\u03c0<br \/>\n\u21d2\u03bdR=4166.672\u00d73.14=663.48Hz\u21d2\u03bdR=4166.672\u00d73.14=663.48Hz<br \/>\nMaximum current, (I0)Max=V0R(I0)Max=V0R<br \/>\n\u21d2(I0)Max=325.2223=14.14A\u21d2(I0)Max=325.2223=14.14A<\/h3>\n<h2>b) What is the source frequency for which average power absorbed by the circuit is maximum? Obtain the value of this maximum power.<\/h2>\n<h3>Ans: It is known that,<br \/>\nMaximum average power absorbed by the circuit; (PV)Max=12(I0)2MaxR(PV)Max=12(I0)Max2R<br \/>\n\u21d2(PV)Max=12\u00d7(14.14)2\u00d723\u21d2(PV)Max=12\u00d7(14.14)2\u00d723<br \/>\n\u21d2(PV)Max=2299.3W\u21d2(PV)Max=2299.3W<br \/>\nTherefore, the resonant frequency, \u03bdR=663.48Hz\u03bdR=663.48Hz<\/h3>\n<h2>c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?<\/h2>\n<h3>Ans: It is known that,<br \/>\nThe power transferred to the circuit is half the power at resonant frequency.<br \/>\nFrequencies at which power transferred is half, =\u03c9R\u00b1\u0394\u03c9=2\u03c0(\u03bdR\u00b1\u0394\u03bd)=\u03c9R\u00b1\u0394\u03c9=2\u03c0(\u03bdR\u00b1\u0394\u03bd)<br \/>\nWhere,<br \/>\n\u0394\u03c9=R2L\u0394\u03c9=R2L<br \/>\n\u21d2\u0394\u03c9=232\u00d70.12=95.83rad\/s\u21d2\u0394\u03c9=232\u00d70.12=95.83rad\/s<br \/>\nTherefore, the change in frequency, \u0394\u03bd=12\u03c0\u0394\u03c9\u0394\u03bd=12\u03c0\u0394\u03c9<br \/>\n\u0394\u03bd=95.832\u03c0=15.26Hz\u0394\u03bd=95.832\u03c0=15.26Hz<br \/>\n\u03bdR+\u0394\u03bd=663.48+15.26=678.74Hz\u03bdR+\u0394\u03bd=663.48+15.26=678.74Hz<br \/>\n\u03bdR\u2212\u0394\u03bd=663.48\u221215.26=648.22Hz\u03bdR\u2212\u0394\u03bd=663.48\u221215.26=648.22Hz<br \/>\nTherefore, at 648.22Hz648.22Hz and 678.74Hz678.74Hz frequencies, the power transferred is half.<br \/>\nAt these frequencies, current amplitude:I\u2032=12\u221a\u00d7(I0)MaxI\u2032=12\u00d7(I0)Max<br \/>\n\u21d2I\u2032=14.142\u221a=10A\u21d2I\u2032=14.142=10A<br \/>\nTherefore, the current amplitude is 10A10A.<\/h3>\n<h2>d) What is the Q-factor of the given circuit?<\/h2>\n<h3>Ans: It is known that,<br \/>\nQ-factor of the given circuit, Q=\u03c9rLRQ=\u03c9rLR<br \/>\n\u21d2Q=4166.67\u00d70.1223=21.74\u21d2Q=4166.67\u00d70.1223=21.74<br \/>\nTherefore, the Q-factor of the given circuit is 21.7421.74.<\/h3>\n<h2>21. Obtain the resonant frequency and Q-factor of a series LCR circuit with L=3.0HL=3.0H ,C=27\u03bcFC=27\u03bcF and R=7.4\u03a9R=7.4\u03a9. It is desired to improve the sharpness of the resonance of the circuit by reducing its \u2018full width at half maximum\u2019 by a factor of 22. Suggest a suitable way.<\/h2>\n<h3>Ans: It is given that,<br \/>\nInductance, L=3.0HL=3.0H<br \/>\nCapacitance, C=27\u03bcF=27\u00d710\u22126FC=27\u03bcF=27\u00d710\u22126F<br \/>\nResistance,R=7.4\u03a9R=7.4\u03a9<br \/>\nIt is known that,<br \/>\nAt resonance, angular frequency of the source for the given LCR series circuit is \u03c9r=1LC\u221a\u03c9r=1LC<br \/>\n\u21d2\u03c9r=13\u00d727\u00d710\u22126\u221a\u21d2\u03c9r=13\u00d727\u00d710\u22126<br \/>\n\u21d2\u03c9r=1039=111.11rad\/s\u21d2\u03c9r=1039=111.11rad\/s<br \/>\nTherefore, the resonant frequency is 111.11rad\/s111.11rad\/s.<br \/>\nQ-factor of the series, Q=\u03c9rLRQ=\u03c9rLR<br \/>\n\u21d2Q=111.11\u00d737.4=45.0446\u21d2Q=111.11\u00d737.4=45.0446<br \/>\nTherefore, the Q-factor is 45.044645.0446.<br \/>\nTo improve the sharpness of the resonance by reducing \u2018full width at half maximum\u2019 by a factor of 22without changing \u03c9r\u03c9r , reduce the resistance to half.<br \/>\n\u21d2R=7.42=3.7\u03a9\u21d2R=7.42=3.7\u03a9<br \/>\nTherefore, required resistance is 3.7\u03a93.7\u03a9.<\/h3>\n<h2>22. Answer the following questions:<br \/>\na) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?<\/h2>\n<h3>Ans: Yes, in any ac circuit, the applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit.<br \/>\nThe same is not true for rms voltage because voltages across different elements may not be in phase.<\/h3>\n<h2>b) A capacitor is used in the primary circuit of an induction coil.<\/h2>\n<h3>Ans: Yes, a capacitor is used in the primary circuit of an induction coil.<br \/>\nThis is because, when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.<\/h3>\n<h2>c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.<\/h2>\n<h3>Ans: The dc signal will appear across capacitor CC because for dc signals, the impedance of an inductor LL is negligible while the impedance of a capacitor CC is very high (almost infinite).<br \/>\nTherefore, a dc signal appears acrossCC.<br \/>\nFor an ac signal of high frequency, the impedance of LLis high and that of CC is very low.<br \/>\nThus, an ac signal of high frequency appears across LL.<\/h3>\n<h2>d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp\u2019s brightness. Predict the corresponding observations if the connection is to an ac line.<\/h2>\n<h3>Ans: When an iron core is inserted in the choke coil (which is in series with a lamp connected to an ac line), the lamp will glow dimly.<br \/>\nThis is because the choke coil and the iron core increase the impedance of the circuit.<\/h3>\n<h2>e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?<\/h2>\n<h3>Ans: As the choke coil reduces the voltage across the tube without wasting much power, it is used in the fluorescent tubes with ac mains. An ordinary resistor cannot be used instead of choke coil because it wastes power in the form of heat.<\/h3>\n<h2>23. A power transmission line feeds input power at 2300V2300V to a step-down transformer with its primary windings having 40004000 turns. What should be the number of turns in the secondary in order to get output power at 230V230V?<\/h2>\n<h3>Ans: It is given that,<br \/>\nInput voltage, V1=2300VV1=2300V<br \/>\nNumber of turns in primary coil, n1=4000n1=4000<br \/>\nOutput voltage, V2=230VV2=230V<br \/>\nNumber of turns in secondary coil ,n2=?n2=?<br \/>\nIt is known that,<br \/>\nVoltage is related to number of terms: V1V2=n1n2V1V2=n1n2<br \/>\n\u21d22300230=4000n2\u21d22300230=4000n2<br \/>\n\u21d2n2=4000\u00d72302300=400\u21d2n2=4000\u00d72302300=400<br \/>\nTherefore, the number of turns in the second winding is 400400.<\/h3>\n<h2>24. At a hydroelectric power plant, the water pressure head is at a height of 300m300m and the water flow available is 100m3\/s100m3\/s . If the turbine generator efficiency is 6060 , estimate the electric power available from the plant (g=9.8m\/s2g=9.8m\/s2 ).<\/h2>\n<h3>Ans: It is known that,<br \/>\nHeight of water pressure head,<br \/>\nh=300mh=300m<br \/>\nVolume of water flow per second, V=100m3\/sV=100m3\/s<br \/>\nEfficiency of turbine generator,<br \/>\nn=60n=60<br \/>\nAcceleration due to gravity, g=9.8m\/s2g=9.8m\/s2<br \/>\nDensity of water,<br \/>\n\u03c1=103kg\/m3\u03c1=103kg\/m3<br \/>\nIt is known that,<br \/>\nElectric power available from the plant=\u03b7\u00d7h\u03c1gV=\u03b7\u00d7h\u03c1gV<br \/>\n\u21d2P=0.6\u00d7300\u00d7103\u00d79.8\u00d7100\u21d2P=0.6\u00d7300\u00d7103\u00d79.8\u00d7100<br \/>\n\u21d2P=176.4\u00d7106W\u21d2P=176.4\u00d7106W<br \/>\n\u21d2P=176.4MW\u21d2P=176.4MW<br \/>\nTherefore, the estimated electric power available from the plant is 176.4MW176.4MW.<\/h3>\n<h2>25. A small town with a demand of 800kW800kW of electric power at 220V220V is situated 15km15km away from an electric plant generating power at 440V440V . The resistance of the two-wire line carrying power is 0.5\u03a9perkm0.5\u03a9perkm. The town gets power from the line through a 400\u2212220V400\u2212220V step-down transformer at a substation in the town.<br \/>\na) Estimate the line power loss in the form of heat.<\/h2>\n<h3>Ans: It is given that,<br \/>\nTotal electric power required,<br \/>\nP=800kW=800\u00d7103WP=800kW=800\u00d7103W<br \/>\nSupply voltage,<br \/>\nV=220VV=220V<br \/>\nVoltage at which electric plant is generating power, V\u2032=440VV\u2032=440V<br \/>\nDistance between the town and power generating station, d=15kmd=15km<br \/>\nResistance of the two wire lines carrying power=0.5\u03a9\/km=0.5\u03a9\/km<br \/>\nTotal resistance of the wires,<br \/>\nR=(15+15)0.5=15 \u03a9R=(15+15)0.5=15 \u03a9<br \/>\nA step-down transformer of rating<br \/>\n4000\u2212220V4000\u2212220V<br \/>\nis used in the sub-station.<br \/>\nInput voltage, V1=4000VV1=4000V<br \/>\nOutput voltage, V2=220VV2=220V<br \/>\nIt is known that,<br \/>\nRms current in the wire lines: I=PV1I=PV1<br \/>\n\u21d2I=800\u00d71034000=200A\u21d2I=800\u00d71034000=200A<br \/>\nLine power loss=I2R=I2R<br \/>\n\u21d2(200)2\u00d715\u21d2(200)2\u00d715<br \/>\n\u21d2600\u00d7103W=600kW\u21d2600\u00d7103W=600kW<br \/>\nTherefore, the line power loss is 600kW600kW.<\/h3>\n<h2>b) How much power must the plant supply, assuming there is negligible power loss due to leakage?<\/h2>\n<h3>Ans: Assuming that there is negligible power loss due to leakage of the current:<br \/>\nTotal power supplied by the plant=800kW+600kW=1400kW=800kW+600kW=1400kW<br \/>\nTherefore, the plant must supply 1400kW1400kW of power.<\/h3>\n<h2>c) Characterise the step up transformer at the plant.<\/h2>\n<h3>Ans: It is known that,<br \/>\nVoltage drop in the power line=IR=IR<br \/>\n\u21d2V=200\u00d715=3000V\u21d2V=200\u00d715=3000V<br \/>\nTotal voltage transmitted from the plant=3000+4000=7000V=3000+4000=7000V<br \/>\nThe power generated is 440V440V.<br \/>\nTherefore, the rating of the step-up transformer situated at the power plant is 440V\u22127000V440V\u22127000V.<\/h3>\n<h2>26.Do the same exercise as above with the replacement of the earlier transformer by a 40,000\u2212220V40,000\u2212220V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?<\/h2>\n<h3>Ans: It is given that,<br \/>\nTotal electric power required,<br \/>\nP=800kW=800\u00d7103WP=800kW=800\u00d7103W<br \/>\nSupply voltage,<br \/>\nV=220VV=220V<br \/>\nVoltage at which electric plant is generating power, V\u2032=440VV\u2032=440V<br \/>\nDistance between the town and power generating station, d=15kmd=15km<br \/>\nResistance of the two wire lines carrying power=0.5\u03a9\/km=0.5\u03a9\/km<br \/>\nTotal resistance of the wires,<br \/>\nR=(15+15)0.5=15 \u03a9R=(15+15)0.5=15 \u03a9<br \/>\nThe rating of a step-down transformer is<br \/>\n40000V\u2212220V40000V\u2212220V<br \/>\n.<br \/>\nInput voltage, V1=40000VV1=40000V<br \/>\nOutput voltage, V2=220VV2=220V<br \/>\n1. It is known that,<br \/>\nRms current in the wire lines: I=PV1I=PV1<br \/>\n\u21d2I=800\u00d710340000=20A\u21d2I=800\u00d710340000=20A<br \/>\nLine power loss=I2R=I2R<br \/>\n\u21d2(20)2\u00d715\u21d2(20)2\u00d715<br \/>\n\u21d26\u00d7103W=6kW\u21d26\u00d7103W=6kW<br \/>\nTherefore, the line power loss is 6kW6kW.<br \/>\n2. Assume that there is negligible power loss due to leakage of the current:<br \/>\nTotal power supplied by the plant=800kW+6kW=806kW=800kW+6kW=806kW<br \/>\nTherefore, the plant must supply 806kW806kW of power.<br \/>\n3. It is known that,<br \/>\nVoltage drop in the power line=IR=IR<br \/>\n\u21d2V=20\u00d715=300V\u21d2V=20\u00d715=300V<br \/>\nTotal voltage transmitted from the plant=300+40000=40300V=300+40000=40300V<br \/>\nThe power generated in the plant is generated at 440V440V.<br \/>\nTherefore, the rating of the step-up transformer situated at the power plant is 440V\u221240300V440V\u221240300V.<br \/>\nPower loss during transmission=6001400\u00d7100=42.8=6001400\u00d7100=42.8<br \/>\nIn previous exercise the power loss due to the same reason is =6806\u00d7100=0.744=6806\u00d7100=0.744<br \/>\nAs the power loss is less for a high voltage transmission, High voltage transmissions are preferred for this purpose.<\/h3>\n","protected":false},"excerpt":{"rendered":"<p>Class 12 Physics NCERT book solutions for Chapter 7 &#8211; Alternating Current Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":118347,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-120407","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/120407","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=120407"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/120407\/revisions"}],"predecessor-version":[{"id":120435,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/120407\/revisions\/120435"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/118347"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=120407"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=120407"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=120407"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}