JEE Challenger
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PhyMedAdv 2021 • Paper 2

Internal Resistance Measurement Using Meter Bridge Setup

To find the internal resistance r1r_1 of a battery having emf EE, a meter bridge having wire resistance R0=50 ΩR_0 = 50\ \Omega, an auxiliary resistance R0/2R_0/2, a secondary cell with emf E/2E/2 (having internal resistance rr), and a galvanometer G are connected as illustrated in the circuit diagram. If the balance point is detected at l=72 cml = 72\ \text{cm}, evaluate the value of r1r_1 in Ω\Omega.

Question Diagram 1
Official Numerical Answer3

Topics & Concepts

Current ElectricityMeasuring InstrumentsOhm's Law

Step-by-Step Solution

To find the internal resistance r1r_1 of the cell, we analyze the circuit under the condition of zero current through the galvanometer (null point).

1. Primary Circuit Analysis

The primary circuit forms a single closed loop consisting of:

  • A cell of emf EE with internal resistance r1r_1
  • An auxiliary resistance of R02\frac{R_0}{2}
  • The meter bridge wire of resistance R0=50 ΩR_0 = 50\ \Omega and length L=100 cmL = 100\ \text{cm}

The total resistance of the primary circuit loop is: Rtotal=r1+R02+R0=r1+32R0R_{\text{total}} = r_1 + \frac{R_0}{2} + R_0 = r_1 + \frac{3}{2} R_0

The current II flowing through the primary circuit loop is: I=Er1+32R0I = \frac{E}{r_1 + \frac{3}{2} R_0}

2. Resistance of the Wire Segment

The resistance of the wire segment of length l=72 cml = 72\ \text{cm} from the left end AA to the jockey position JJ is: RAJ=R0(l100)=50×72100=36 ΩR_{AJ} = R_0 \left( \frac{l}{100} \right) = 50 \times \frac{72}{100} = 36\ \Omega

3. Potential Difference and Balance Condition

Let VAV_A be the electric potential at the left terminal AA of the meter bridge wire.

  • The potential at the balance point JJ along the wire is: VJ=VA−IRAJV_J = V_A - I R_{AJ}

  • Let CC be the node between internal resistance r1r_1 and auxiliary resistance R0/2R_0/2. The potential at node CC is related to VAV_A through the upper left branch: VA−VC=E−Ir1  ⟹  VC=VA−E+Ir1V_A - V_C = E - I r_1 \implies V_C = V_A - E + I r_1

  • The potential difference between point JJ and node CC is: VJ−VC=(VA−IRAJ)−(VA−E+Ir1)=E−I(r1+RAJ)V_J - V_C = (V_A - I R_{AJ}) - (V_A - E + I r_1) = E - I (r_1 + R_{AJ})

At the balance point, no current flows through the galvanometer branch. Therefore, the potential difference VJ−VCV_J - V_C across this branch must equal the emf of the secondary cell E2\frac{E}{2}: VJ−VC=E2V_J - V_C = \frac{E}{2}

Equating both expressions for VJ−VCV_J - V_C: E−I(r1+RAJ)=E2E - I (r_1 + R_{AJ}) = \frac{E}{2} I(r1+RAJ)=E2I (r_1 + R_{AJ}) = \frac{E}{2}

4. Calculation of r1r_1

Substitute I=Er1+32R0I = \frac{E}{r_1 + \frac{3}{2} R_0} into the balance equation: Er1+32R0(r1+RAJ)=E2\frac{E}{r_1 + \frac{3}{2} R_0} (r_1 + R_{AJ}) = \frac{E}{2}

Canceling EE from both sides: 2(r1+RAJ)=r1+32R02 (r_1 + R_{AJ}) = r_1 + \frac{3}{2} R_0 2r1+2RAJ=r1+32R02 r_1 + 2 R_{AJ} = r_1 + \frac{3}{2} R_0 r1=32R0−2RAJr_1 = \frac{3}{2} R_0 - 2 R_{AJ}

Substitute R0=50 ΩR_0 = 50\ \Omega and RAJ=36 ΩR_{AJ} = 36\ \Omega: r1=32(50)−2(36)=75−72=3 Ωr_1 = \frac{3}{2}(50) - 2(36) = 75 - 72 = 3\ \Omega

Final Answer: The value of r1r_1 is 33.

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