Class 12 Physics Answer Key | Uttar-Pradesh Exam 2025-26 Board Term 1

Question 1

How can a galvanometer be converted into an ammeter and a voltmeter? Explain with the correct circuit.

Correct Answer: By connecting a low resistance shunt in parallel for an ammeter and a high resistance in series for a voltmeter.

Solution:

Step 1: Understanding the Concept

A galvanometer is an instrument used to detect very small electric currents. Since it is highly sensitive and has relatively high resistance, it cannot directly measure large currents or voltages unless it is modified.

Step 2: Key Formula or Approach

• Conversion into Ammeter: A low resistance called a shunt (S) is connected in parallel with the galvanometer.
  Formula: S = (Ig / (I − Ig)) × G
  Where I = total current, Ig = galvanometer current for full-scale deflection, and G = resistance of galvanometer.
• Conversion into Voltmeter: A high resistance R is connected in series with the galvanometer.
  Formula: R = (V / Ig) − G
  Where V = desired voltage range and Ig = galvanometer current.

Step 3: Detailed Explanation

(a) Conversion into Ammeter

To measure current, the instrument should have very low resistance so that it does not disturb the circuit current. A small resistance called a shunt (S) is connected in parallel with the galvanometer (G). Most of the current flows through the shunt while only a small portion passes through the galvanometer, protecting its delicate coil.

Circuit Representation:

Total current I enters a junction and divides into two parts: Ig flows through the galvanometer and (I − Ig) flows through the shunt.

(b) Conversion into Voltmeter

To measure potential difference, the instrument should have very high resistance so that it draws very little current from the circuit. A high resistance R is connected in series with the galvanometer.

This increases the total resistance and allows the galvanometer to measure larger voltages safely.

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(b) Conversion into Voltmeter

For voltage measurement, the instrument must have very high resistance so that it draws minimal current.

A large resistance (R) is connected in series with the galvanometer (G).

This ensures that the current through the galvanometer remains limited to Ig.

Circuit Layout:

Terminal A → Resistor R → Galvanometer G → Terminal B

All components are arranged in a single series path.

Step 4: Final Answer

• To convert a galvanometer into an ammeter, a low resistance shunt is connected in parallel.
• To convert a galvanometer into a voltmeter, a high resistance is connected in series.

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Question 2

Explain the construction and working theory of a transformer.

Correct Answer: A transformer works on the principle of mutual induction using a primary coil, secondary coil, and a soft iron core.

Solution:

Step 1: Understanding the Concept

A transformer is a stationary electrical device that transfers electrical energy from one circuit to another using electromagnetic induction.

It is commonly used to either raise the voltage (step-up transformer) or lower the voltage (step-down transformer) in an AC circuit.

Step 2: Key Formula or Approach

The relation for an ideal transformer is based on the turns ratio:

Vs / Vp = Ns / Np = K

Where:

• Vs = Secondary voltage
• Vp = Primary voltage
• Ns = Number of turns in secondary coil
• Np = Number of turns in primary coil
• K = Transformer ratio

Step 3: Detailed Explanation

Construction:

• Core: Made of laminated soft iron sheets to reduce energy losses such as eddy currents.
• Primary Coil: Connected to the input AC supply.
• Secondary Coil: Connected to the load where the transformed voltage is obtained.

Both coils are electrically insulated and wound on the same soft iron core.

Working Theory:

• When an alternating voltage is applied to the primary coil, an alternating current flows through it.
• This current produces a time-varying magnetic flux in the iron core.
• The secondary coil, wound on the same core, links with this changing magnetic flux.
• According to Faraday’s Law of Electromagnetic Induction, an alternating EMF is induced in the secondary coil.
• This process is known as mutual induction.

Step 4: Final Answer

A transformer consists of a laminated soft iron core with primary and secondary windings, and it works on the principle of mutual induction to transfer electrical energy between circuits.

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Question 3

What are electromagnetic waves? By drawing its propagation diagram, show the electric field and magnetic field components in it.

Correct Answer: Electromagnetic waves are waves consisting of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of propagation.

Solution:

Step 1: Understanding the Concept

Electromagnetic (EM) waves are waves formed due to the combined oscillations of electric and magnetic fields.

Unlike mechanical waves, electromagnetic waves do not require any material medium and can travel through vacuum.

Step 2: Definition

Electromagnetic waves are transverse waves produced by accelerating charges. In these waves:

• The electric field (E) oscillates in one direction.
• The magnetic field (B) oscillates in a direction perpendicular to the electric field.
• Both fields are perpendicular to the direction of wave propagation.

Step 3: Propagation Diagram (Description)

Consider a three-dimensional coordinate system (x, y, z):

• The electromagnetic wave travels along the x-axis.
• The electric field (E) oscillates along the y-axis.
• The magnetic field (B) oscillates along the z-axis.

Thus, the three directions are mutually perpendicular:

• Electric Field (E) ⟂ Magnetic Field (B)
• Electric Field (E) ⟂ Direction of Propagation
• Magnetic Field (B) ⟂ Direction of Propagation

Simple Representation:

        Electric Field (E)
              ↑
              |
              |
--------------→----------------  Direction of Propagation
              |
              |
             ⊙
        Magnetic Field (B)

Step 4: Final Answer

Electromagnetic waves are transverse waves consisting of oscillating electric and magnetic fields that are mutually perpendicular to each other and also perpendicular to the direction of propagation.

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Question 4

How is a p-n junction diode used as a half-wave rectifier? Explain its working with circuit diagram and waveforms.

Correct Answer: A p-n junction diode acts as a half-wave rectifier by allowing current to pass only during the positive half-cycle of AC and blocking the negative half-cycle.

Step 1: Understanding the Concept

Rectification is the process of converting alternating current (AC) into direct current (DC). A diode works as a one-way device allowing current only in forward bias.

Step 2: Key Formula

Average output voltage of a half-wave rectifier:

Vdc = Vm / π

Where Vm is the peak value of the input voltage.

Step 3: Circuit Construction

The circuit consists of:

• AC supply connected to a step-down transformer
• A diode connected in series
• A load resistor (RL)

 AC Supply
    ~
    |
 Transformer
    |
    |----|>|---- RL ----|
         Diode

Step 4: Working

Positive Half Cycle:

• The p-side becomes positive and n-side negative.
• The diode becomes forward biased.
• Current flows through RL.
• Output voltage appears across the load.

Negative Half Cycle:

• The polarity reverses.
• The diode becomes reverse biased.
• No current flows.
• Output voltage becomes zero.

Waveforms

Input AC Wave

      / \      / \      / \
     /   \    /   \    /   \
----/-----\--/-----\--/-----\----


Output Rectified Wave

      / \          / \
     /   \        /   \
----/-----\------/-----\---------

Final Answer

A half-wave rectifier uses a single diode to convert AC into pulsating DC by allowing only positive half cycles to pass through the load.

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Question 5

What is wavefront? Explain refraction of waves using Huygens' principle.

Correct Answer: A wavefront is the locus of points vibrating in the same phase. Refraction occurs because wavelets travel with different speeds in different media.

Step 1: Definition of Wavefront

A wavefront is a surface joining all points of a wave that are vibrating in the same phase.

• Point source → Spherical wavefront
• Distant source → Plane wavefront

Step 2: Huygens Principle

According to Huygens:

• Every point on a wavefront acts as a source of secondary wavelets.
• The new wavefront is the envelope of these secondary wavelets.

Step 3: Refraction Explanation

1. Consider a plane wavefront AB incident on a boundary between two media.
2. The speed of waves in medium 1 = v1.
3. The speed in medium 2 = v2.
4. When the wave enters the second medium, its speed changes.
5. Because of this speed difference, the direction of the wavefront bends.
6. This bending of light is called refraction.

Snell's Law obtained:

sin i / sin r = v1 / v2 = μ

Final Answer

A wavefront is a surface of constant phase. Refraction occurs when wavelets travel at different speeds in different media causing the wavefront to change direction.

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Question 6

Derive the condition for equilibrium of Wheatstone bridge using Kirchhoff's laws.

Correct Answer:

P / Q = R / S

Step 1: Concept

A Wheatstone bridge is used to measure unknown resistance. It consists of four resistors arranged in a diamond shape with a galvanometer connected between two junctions.

The bridge is balanced when no current flows through the galvanometer.

Step 2: Circuit Representation

        P
   A -------- B
   |          |
   |          |
   R          Q
   |          |
   D -------- C
        S

Battery is connected across A and C and a galvanometer between B and D.

Step 3: Applying Kirchhoff’s Laws

When bridge is balanced:

Ig = 0

Therefore currents in the arms satisfy:

I1P = I2R
I1Q = I2S

Dividing both equations:

(I1P)/(I1Q) = (I2R)/(I2S)

Cancelling currents gives:

P / Q = R / S

Final Answer

The Wheatstone bridge is balanced when the ratio of resistances of one pair of opposite arms equals the ratio of the other pair.

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Last Updated on : March 26, 2026