Question

In order to measure the internal resistance r₁ of a cell of emf E, a meter bridge of wire resistance R₀ = 50 Ω, a resistance \(\frac{R_0}{2}\), another cell of emf \(\frac{E}{2}\) (internal resistance r) and a galvanometer G are used in a circuit, as shown in the figure. If the null point is found at l = 72 cm, then the value of r ₁ = ____

Answer 1

Correct Answer: 3

Step 1: Expression for current in the main circuit
The total resistance in the circuit is given as the sum of the internal resistance \( r_1 \) and the effective resistance of the external circuit, which is \( \frac{3R_0}{2} \).
So, the total current in the circuit is:
\( I = \frac{E}{r_1 + \frac{3R_0}{2}} \)

Step 2: Apply Kirchhoff’s loop rule
For a loop containing the emf source and the resistors, the KVL (Kirchhoff’s Voltage Law) equation is:
\( +E - I \cdot R_0 \cdot 0.72 - I \cdot r_1 - \frac{E}{2} = 0 \)
Here,
- The potential drop across a portion of \( R_0 \) is \( I \cdot R_0 \cdot 0.72 \)
- \( I \cdot r_1 \) is the internal resistance drop
- \( \frac{E}{2} \) is a given opposing emf term

Step 3: Substitute current expression into KVL equation
Replace \( I \) with \( \frac{E}{r_1 + \frac{3R_0}{2}} \) in the KVL equation:
\( \frac{E}{2} = \frac{2E}{2r_1 + 3R_0} \cdot (0.72R_0 + r_1) \)

Step 4: Simplify the equation
Multiply both sides by 2:
\( 2r_1 + 3R_0 = 4(0.72R_0 + r_1) \)
Expand the right-hand side:
\( 2r_1 + 3R_0 = 2.88R_0 + 4r_1 \)

Step 5: Rearranging the terms
Bring like terms together:
\( 3R_0 - 2.88R_0 = 4r_1 - 2r_1 \)
\( 0.12R_0 = 2r_1 \)

Step 6: Solve for internal resistance \( r_1 \)
Divide both sides by 2:
\( r_1 = \frac{0.12R_0}{2} = 0.06R_0 \)
If \( R_0 = 50Ω \), then:
\( r_1 = 0.06 × 50 = 3Ω \)

Final Answer:
The internal resistance \( r_1 = \mathbf{3 \, \Omega} \)