1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is
0.4Ω0.4Ω
, what is the maximum current that can be drawn from the battery?
Ans: In the above question it is given that:
Emf of the battery, E=12VE=12V
Internal resistance of the battery, r=0.4Ωr=0.4Ω
Consider the maximum current drawn from the battery to be II.
Therefore, using Ohm’s law,
E=IrE=Ir
⇒I=Er⇒I=Er
⇒I=120.4⇒I=120.4
⇒I=30A⇒I=30A
Clearly, the maximum current drawn from the given battery is 30A30A .
2. A battery of emf 10 V and internal resistance 3Ω3Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Ans: In the above question it is given that:
Emf of the battery, E = 10 V
Internal resistance of the battery, r=3Ωr=3Ω
Current in the circuit, I=0.5AI=0.5A
Consider the resistance of the resistor to be RR.
Therefore, using Ohm’s law,
I=ER+rI=ER+r
R+r=EIR+r=EI
⇒R+r=100.5⇒R+r=100.5
⇒R+r=20⇒R+r=20
⇒R=20−3=17Ω⇒R=20−3=17Ω
Let the terminal voltage of the resistor be VV.
Using the Ohm’s law,
V=IRV=IR
⇒V=0.5×17=8.5V⇒V=0.5×17=8.5V
Thus, the resistance of the resistor is 17Ω17Ω and the terminal voltage is 8.5V8.5V .
3.
a) Three resistors 1Ω1Ω , 2Ω2Ω and 3Ω3Ω are combined in series. What is the total resistance of the combination?
Ans: In the above question it is given that three resistors of resistances 1Ω1Ω , 2Ω2Ω and 3Ω3Ω are combined in series.
The total resistance of a series combination of resistors is the algebraic sum of individual resistances.
Hence the total resistance is given by:
Total Resistance =1+2+3=6Ω=1+2+3=6Ω
b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Ans: Consider the current flowing through the circuit to be II.
Emf of the battery, E=12VE=12V
Total resistance of the circuit, R=6ΩR=6Ω
The relation for current using Ohm’s law is given by:
I=ERI=ER
⇒I=126=2A⇒I=126=2A
Consider potential drop across 1Ω1Ω resistor to be V1V1 .
Using Ohm’s law, the value of V1V1 can be obtained as:
V1=2×1=2VV1=2×1=2V …… (1)
Consider potential drop across 2Ω2Ω resistor to be V2V2 .
Again, using Ohm’s law, the value of V2V2 can be obtained as:
V2=2×2=4VV2=2×2=4V …… (2)
Consider potential drop across 3Ω3Ω resistor to be V3V3 .
V3=2×3=6VV3=2×3=6V …… (3)
Clearly, the potential drops across 1Ω1Ω , 2Ω2Ω and 3Ω3Ω resistors are 2V2V , 4V4V and 6V6V respectively.
4.
a) Three resistors 2Ω2Ω , 4Ω4Ω and 5Ω5Ω are combined in parallel. What is the total resistance of the combination?
Ans: In the above question it is given that there are three resistors of resistances 2Ω2Ω , 4Ω4Ω and 5Ω5Ω connected in parallel as shown below:
(Image will be uploaded soon)
Let
R1=2Ω;R2=4Ω;R3=5ΩR1=2Ω;R2=4Ω;R3=5Ω
Now, using the rule for parallel combination of resistors, total resistance RR of the combination will be:
1R=1R1+1R2+1R31R=1R1+1R2+1R3
⇒1R=12+14+15=1920⇒1R=12+14+15=1920
⇒R=2019Ω⇒R=2019Ω
Clearly, total resistance of the given parallel combination is 2019Ω2019Ω .
b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Ans: In the above question, it is given that the parallel combination mentioned in (a) is connected to a battery and negligible internal resistance as shown:
(Image will be uploaded soon)
It is known that R1=2Ω;R2=4Ω;R3=5ΩR1=2Ω;R2=4Ω;R3=5Ω.
Also given that,
Emf of the battery, V=20VV=20V
Ohm’s law can be used to determine the individual current flowing through each resistor and hence, the total current in the circuit.
If current I1I1 is flowing through resistor R1R1, it is given by:
I1=VR1I1=VR1
⇒I1=202=10A⇒I1=202=10A
If current I2I2 is flowing through resistor R2R2, it is given by:
I2=VR2I2=VR2
⇒I2=204=5A⇒I2=204=5A
If current I3I3 is flowing through resistor R3R3, it is given by:
I3=VR3I3=VR3
⇒I3=205=4A⇒I3=205=4A
Now, the total current will be:
I=I1+I2+I3I=I1+I2+I3
⇒I=10+5+4=19A⇒I=10+5+4=19A
Clearly, the current through resistors 2Ω2Ω , 4Ω4Ω and 5Ω5Ω are 10A10A, 5A5A and 4A4A respectively while the total current in the circuit is 19A19A.
5. At room temperature 27.0∘C27.0∘C, the resistance of a heating element is 100Ω100Ω. What is the temperature of the element if the resistance is found to be 117Ω117Ω, given that the temperature coefficient of the material of the resistor is 1.70×10−4∘C−11.70×10−4∘C−1 ?
Ans: In the above question it is given that at room temperature (T=27.0∘C)(T=27.0∘C), the resistance of the heating element is 100Ω100Ω (say R).
Also, the heating element’s temperature coefficient is given to be α=1.70×10−4∘C−1α=1.70×10−4∘C−1.
Now, it is said that the resistance of the heating element at an increased temperature (say T1T1) is 117Ω117Ω (say R1R1). To compute this unknown increased temperature T1T1, the formula for temperature coefficient of a material can be used. It is known that temperature co-efficient of a material provides information on the nature of that material with respect to its change in resistance with temperature. Mathematically,
α=R1−RR(T1−T)α=R1−RR(T1−T)
⇒T1−T=R1−RRα⇒T1−T=R1−RRα
Substituting the given values,
⇒T1−27=117−100100×1.70×10−4⇒T1−27=117−100100×1.70×10−4
⇒T1−27=1000⇒T1−27=1000
⇒T1=1027∘C⇒T1=1027∘C
Clearly, it is at 1027∘C1027∘C when the resistance of the element is 117Ω117Ω.
6. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0×10−7m26.0×10−7m2 , and its resistance is measured to be 5.0Ω5.0Ω . What is the resistivity of the material at the temperature of the experiment?
Ans: In the above question it is given that:
Length of the wire, l=15ml=15m
Area of cross-section of the wire, a=6.0×10−7m2a=6.0×10−7m2
Resistance of the material of the wire, R=5.0ΩR=5.0Ω
Let resistivity of the material of the wire be ρρ
It is known that, resistance is related with the resistivity as:
R=ρlAR=ρlA
⇒ρ=RAl⇒ρ=RAl
⇒ρ=5×6.0×10−715⇒ρ=5×6.0×10−715
⇒ρ=2×10−7m2⇒ρ=2×10−7m2
Therefore, the resistivity of the material is 2×10−7m22×10−7m2 .
7. A silver wire has a resistance of
2.1Ω2.1Ω
at 27.5∘C27.5∘C , and a resistance of 2.7Ω2.7Ω at 100∘C100∘C. Determine the temperature coefficient of resistivity of silver.
Ans: In the above question it is given that:
Temperature, T1=27.5∘CT1=27.5∘C.
Resistance of the silver wire at T1T1 is R1=2.1ΩR1=2.1Ω .
Temperature, T2=100∘CT2=100∘C .
Resistance of the silver wire at T2T2 is R2=2.7ΩR2=2.7Ω .
Let the temperature coefficient of silver be αα . It is known that temperature co-efficient of a material provides information on the nature of that material with respect to its change in resistance with temperature. Mathematically, it is related with temperature and resistance by the formula:
α=R2−R1R1(T2−T1)α=R2−R1R1(T2−T1)
⇒α=2.7−2.12.1(100−27.5)=0.0039∘C−1⇒α=2.7−2.12.1(100−27.5)=0.0039∘C−1
Clearly, the temperature coefficient of silver is 0.0039∘C−10.0039∘C−1.
8. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27∘C27∘C ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70×10−4∘C−11.70×10−4∘C−1 .
Ans: In the above question it is given that:
Supply voltage is V=230VV=230V
Initial current drawn is I1=3.2AI1=3.2A.
Let the initial resistance be R1R1 .
Therefore, using Ohm’s law,
R1=VI1R1=VI1
⇒R1=2303.2=71.87Ω⇒R1=2303.2=71.87Ω
Steady state value of the current is I2=2.8AI2=2.8A.
Let the resistance of the steady state be R2R2 .
Therefore, using Ohm’s law.
R2=VI2R2=VI2
⇒R2=2302.8=82.14Ω⇒R2=2302.8=82.14Ω
Temperature co-efficient of nichrome is α=1.70×10−4∘C−1α=1.70×10−4∘C−1 .
Initial temperature of nichrome is T1=27∘CT1=27∘C.
Let steady state temperature reached by nichrome be T2T2 .
Now, it is known that temperature co-efficient of a material provides information on the nature of that material with respect to its change in resistance with temperature. Mathematically, it is given by
α=R2−R1R1(T2−T1)α=R2−R1R1(T2−T1)
⇒(T2−T1)=R2−R1R1α⇒(T2−T1)=R2−R1R1α
Substituting the given values,
⇒(T2−27)=82.14−71.8771.87×1.70×10−4⇒(T2−27)=82.14−71.8771.87×1.70×10−4
⇒T2−27=840.5⇒T2−27=840.5
⇒T2=867.5∘C⇒T2=867.5∘C
Clearly, the steady temperature of the heating element is 867.5∘C867.5∘C.
9. Determine the current in each branch of the network shown in figure:
(Image will be uploaded soon)
Ans: Current flowing through various branches of the circuit is represented in the given figure.
(Image will be uploaded soon)
Consider
I1=I1=Current flowing through the outer circuit
I2=I2=Current flowing through branch AB
I3=I3=Current flowing through branch AD
I2−I4=I2−I4=Current flowing through branch BC
I3+I4=I3+I4=Current flowing through branch CD
I4=I4=Current flowing through branch BD
For the closed circuit ABDA, potential is zero i.e.,
10I2+5I4−5I3=010I2+5I4−5I3=0
2I2+I4−I3=02I2+I4−I3=0
I3=2I2+I4I3=2I2+I4 …… (1)
For the closed circuit BCDB, potential is zero i.e.,
5(I2−I4)−10(I3+I4)−5I4=05(I2−I4)−10(I3+I4)−5I4=0
5I2+5I4−10I3−10I4−5I4=05I2+5I4−10I3−10I4−5I4=0
5I2−10I3−20I4=05I2−10I3−20I4=0
I2=2I3+4I4I2=2I3+4I4 …… (2)
For the closed circuit ABCFEA, potential is zero i.e.,
−10+10(I1)+10(I2)+5(I2−I4)=0−10+10(I1)+10(I2)+5(I2−I4)=0
10=15I2+10I1−5I410=15I2+10I1−5I4
3I3+2I1−I4=23I3+2I1−I4=2 …… (3)
From equations (1) and (2), we obtain
I3=2(2I3+4I4)+I4I3=2(2I3+4I4)+I4
I3=4I3+8I4+I4I3=4I3+8I4+I4
−3I3=9I4−3I3=9I4
−3I4=+I3−3I4=+I3 …… (4)
Putting equation (4) in equation (1), we obtain
I3=2I2+I4I3=2I2+I4
−4I4=2I2−4I4=2I2 …… (5)
It is evident from the given figure that,
I1=I3+I2I1=I3+I2 ……. (6)
Putting equation (6) in equation (1), we obtain
3I2+2(I3+I2)−I4=23I2+2(I3+I2)−I4=2
5I2+2I3−I4=25I2+2I3−I4=2 …… (7)
Putting equations (4) and (5) in equation (7), we obtain
5(−2I4)+2(−3I4)−I4=25(−2I4)+2(−3I4)−I4=2
−10I4−6I4−I4=2−10I4−6I4−I4=2
17I4=−217I4=−2
I4=−217AI4=−217A
Equation (4) reduces to
I3=−3(I4)I3=−3(I4)
I3=−3(−217)=617AI3=−3(−217)=617A
I2=−2(I4)I2=−2(I4)
I2=−2(−217)=417AI2=−2(−217)=417A
I2−I4=417−(−217)=617I2−I4=417−(−217)=617
I3+I4=617+(−217)=417AI3+I4=617+(−217)=417A
I1=I3+I2I1=I3+I2
∴I1=617+417=1017A∴I1=617+417=1017A
Therefore, current in branch AB =417A=417A
Current in branch BC =617A=617A
Current in branch CD =−417A=−417A
Current in branch AD =617A=617A
Current in branch BD =(−217)A=(−217)A
Total current =417+617+−417+617+−217=1017A=417+617+−417+617+−217=1017A .
10.
a) In a metre bridge the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5Ω12.5Ω . Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
Ans: A metre bridge with resistors XX and YY is represented in the given figure.
(Image will be uploaded soon)
A meter bridge works with the same principle as that of a Wheatstone bridge, which is an electrical circuit used to measure unknown resistances. It forms a bridge circuit, wherein the two legs of the circuit are balanced, out of which, one leg has the unknown resistance. Here, the concept of balance point is used, when there is no deflection in the galvanometer (no current flow condition).
In the above question it is given that:
Balance point from end A is at distance, l1=39.5cml1=39.5cm
Resistance of the resistor
Y=12.5ΩY=12.5Ω
Condition for the balance with respect to the given meter bridge is given by
XY=100−l1l1XY=100−l1l1
⇒X=100−39.539.5×12.5=8.2Ω⇒X=100−39.539.5×12.5=8.2Ω
Thus, the resistance of resistor XX is 8.2Ω8.2Ω.
The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips which helps to minimize the resistance. Hence it is not taken into consideration in the bridge formula.
b) Determine the balance point of the bridge above if X and Y are interchanged.
Ans: When XX and YYare interchanged, then l1l1 and 100−l1100−l1 also get interchanged.
Thus, the balance point of the bridge would be 100−l1100−l1 from AA.
∴100−l1=100−39.5=60.5cm∴100−l1=100−39.5=60.5cm
Clearly, the balance point is 60.5cm60.5cm from AA.
c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
Ans: When the galvanometer and cell are swapped at the balance point of the bridge, the galvanometer would show zero deflection. As there is null deflection, zero current would flow through the galvanometer.
11. A storage battery of emf 8.0 V and internal resistance 0.5Ω0.5Ω is being charged by a 120 V DC supply using a series resistor of 15.5Ω15.5Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Ans: In the above question it is given that:
Emf of the storage battery is E=0.8VE=0.8V.
Internal resistance of the battery is r=0.5Ωr=0.5Ω .
DC supply voltage is V=120VV=120V
Resistance of the resistor is R=15.5ΩR=15.5Ω.
Consider effective voltage in the circuit to be V′V′, which would be the difference in the supply voltage and the emf of the battery.
V′=V−EV′=V−E
⇒V′=120−8=112V⇒V′=120−8=112V
Now, current flowing in the circuit is II and the resistance RR is connected in series to the storage battery.
Therefore, using Ohm’s law,
I=V′R+rI=V′R+r
⇒I=11215.5+0.5=7A⇒I=11215.5+0.5=7A
Thus, voltage across resistor RRwould be:
IR=7×15.5=108.5VIR=7×15.5=108.5V
DC supply voltage = Terminal voltage of battery + Voltage drop across RR
Terminal voltage of battery =120−108.5=11.5V=120−108.5=11.5V
A series resistor in a charging circuit takes the responsibility for controlling the current drawn from the external source. Excluding this series resistor is dangerous as the current flow would be extremely high if so.
12. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Ans: A potentiometer arrangement facilitates adjustable voltage dividing. It can be used to compare the emf’s of two different cells with the help of balance points in each case.
Balance point or null point is the point when the galvanometer in the circuit shows no deflection. i.e., when there is no current flowing in the circuit.
In the above question it is given that when the cell has an emf E1=1.25VE1=1.25V, the balance point of the potentiometer is given to be at a distance, l1=35cml1=35cm.
Now, when the cell is replaced by another cell of emf E2E2, the balance point of the potentiometer is given to be at a distance l2=63cml2=63cm.
The balance condition to compare the emf’s of two cells using a potentiometer setup is given by the relation,
E1E2=l1l2E1E2=l1l2
⇒E2=E1×l1l2⇒E2=E1×l1l2
⇒E2=1.25×6335=2.25V⇒E2=1.25×6335=2.25V
Clearly, the emf of the second cell is 2.25V2.25V .
13. The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5×1028m−38.5×1028m−3 . How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0×10−6m22.0×10−6m2 and it is carrying a current of 3.0 A.
Ans: In the above question it is given that:
Number density of free electrons in a copper conductor is n=8.5×1028m−3n=8.5×1028m−3.
Length of the copper wire is l=3.0ml=3.0m.
Area of cross-section of the wire is A=2.0×10−6m2A=2.0×10−6m2.
Current carried by the wire is I=3.0AI=3.0A.
Now we know that:
I=nAeVdI=nAeVd
Where,
ee is the electric charge of magnitude 1.6×10−19C1.6×10−19C.
VdVd is the drift velocity and
Drift velocity=Length of the wire (l)Time taken to cover (t)Drift velocity=Length of the wire (l)Time taken to cover (t)
I=nAeltI=nAelt
⇒t=nAelI⇒t=nAelI
⇒t=3×8.5×1028×2×10−6×1.6×10−193.0⇒t=3×8.5×1028×2×10−6×1.6×10−193.0
∴t=2.7×104s∴t=2.7×104s .
Hence the time taken by an electron to drift from one end of the wire to the other is 2.7×104s2.7×104s.
14. The earth’s surface has a negative surface charge density of 10−9Cm−210−9Cm−2 . The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralize the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37×106m6.37×106m.)
Ans: In the above question it is given that:
Surface charge density of the earth is σ=10−9Cm−2σ=10−9Cm−2.
Current over the entire globe is I=1800AI=1800A.
Radius of the earth is r=6.37×106mr=6.37×106m.
Surface area of the earth is given by:
A=4πr2A=4πr2
⇒A=4π×(6.37×106)2⇒A=4π×(6.37×106)2
⇒A=5.09×1014m2⇒A=5.09×1014m2
Charge on the earth surface is given by:
q=σAq=σA
⇒q=10−9×5.09×1014⇒q=10−9×5.09×1014
⇒q=5.09×105C⇒q=5.09×105C
Now, when the time taken to neutralize the surface of the earth is taken to be tt, then,
Current, I=qtI=qt
⇒t=qI⇒t=qI
⇒t=5.09×1051800=282.77s⇒t=5.09×1051800=282.77s
Clearly, the time taken to neutralize the surface of the earth is 282.77s282.77s.
15.
a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015Ω0.015Ω are joined in series to provide a supply to a resistance of 8.5Ω8.5Ω . What is the current drawn from the supply and its terminal voltage?
Ans: In the above question, it is given that six lead-acid type of secondary cells are joined in series as shown below:
(Image will be uploaded soon)
Here, number of secondary cells is n=6n=6.
Emf of each secondary cell is E=2.0VE=2.0V.
Internal resistance of each cell is r=0.015Ωr=0.015Ω .
Series resistor is connected to the combination of cells.
Resistance of the resistor RR is 8.5Ω8.5Ω.
If the current drawn from the supply is II, then
I=nER+nrI=nER+nr
⇒I=6×28.5+6×0.015⇒I=6×28.5+6×0.015
⇒I=1.39A⇒I=1.39A
Now the terminal voltage would be
V=IR=1.39×8.5=11.87VV=IR=1.39×8.5=11.87V
Clearly, the current drawn from the supply is 1.39A1.39A and the terminal voltage is 11.87V11.87V.
b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380Ω380Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
Ans: It is given that after using a secondary cell for long, emf of the secondary cell is E=1.9VE=1.9V.
Also, the internal resistance of the cell is r=380Ωr=380Ω.
Here, using Ohm’s law, the maximum current =Er=Er
⇒Er=1.9380=0.005A⇒Er=1.9380=0.005A.
Clearly, the maximum current used from the cell is 0.005A0.005A. To start the motor of a car, a large amount of current is required. Thus, this cell which produces just 0.005A cannot be used for this purpose.
16. Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables.
ρAl=2.63×108Ωm;ρCu=1.72×10−8Ωm;ρAl=2.63×108Ωm;ρCu=1.72×10−8Ωm;Relative density of Al=2.7;Relative density of Cu=8.9Relative density of Al=2.7;Relative density of Cu=8.9
Ans: In the above question it is given that:
Resistivity of aluminium is ρAl=2.63×108ΩmρAl=2.63×108Ωm.
Relative density of aluminium is d1=2.7d1=2.7.
Consider l1l1 to be the length of aluminium wire, m1m1 as its mass, the resistance of the copper wire as R2R2, and area of cross-section of the copper wire as A2A2.
Therefore, using the relation between resistance and resistivity,
R1=ρ1l1A1R1=ρ1l1A1 …… (1)
And
R2=ρ2l2A2R2=ρ2l2A2 …… (2)
However, we have R1=R2R1=R2
⇒ρ1l1A1=ρ2l2A2⇒ρ1l1A1=ρ2l2A2
Also, we have l1=l2l1=l2.
⇒ρ1A1=ρ2A2⇒ρ1A1=ρ2A2
⇒A1A2=ρ1ρ2⇒A1A2=ρ1ρ2
⇒A1A2=2.63×1081.72×10−8=2.631.72⇒A1A2=2.63×1081.72×10−8=2.631.72
Now, mass of the aluminium wire is given by:
m1=Volume×Densitym1=Volume×Density
⇒m1=A1l1×d1=A1l1d1⇒m1=A1l1×d1=A1l1d1 …… (3)
Similarly mass of the copper wire is given by:
m2=Volume×Densitym2=Volume×Density
⇒m2=A2l2×d2=A2l2d2⇒m2=A2l2×d2=A2l2d2 …… (4)
Dividing equation (3) by equation (4), we get:
m1m2=A1l1d1A2l2d2m1m2=A1l1d1A2l2d2
As l1=l2l1=l2,
m1m2=A1d1A2d2m1m2=A1d1A2d2
As
A1A2=2.631.72A1A2=2.631.72
m1m2=2.631.72×2.78.9=0.46m1m2=2.631.72×2.78.9=0.46
It indicated that m1<m2m1<m2.
Clearly, aluminium is lighter than copper.
Since aluminium is lighter, it is prioritized for overhead power cables instead of copper.
17. What conclusion can you draw from the following observations on a resistor made of alloy manganin?
| Current (A) | Voltage (V) | Current (A) | Voltage (V) |
| 0.2 | 3.94 | 3.0 | 59.2 |
| 0.4 | 7.87 | 4.0 | 78.8 |
| 0.6 | 11.8 | 5.0 | 98.6 |
| 0.8 | 15.7 | 6.0 | 118.5 |
| 1.0 | 19.7 | 7.0 | 138.2 |
| 2.0 | 39.4 | 8.0 | 158.0 |
