Question

For the given circuit, if R = 12\(\Omega\), balancing length is 180 cm. When value of R is 4 \(\Omega\), then balancing length is 120 cm. Find internal resistance of cell E. 

• A. 2 \(\Omega\)
• B. 5 \(\Omega\)
• C. 4 \(\Omega\)
• D. 1 \(\Omega\)

Hint:

Using the ratio \(\frac{l_1}{l_2} = \frac{R_1(R_2+r)}{R_2(R_1+r)}\) saves time in potentiometer problems involving variable resistors.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The internal resistance (\(r\)) of a cell can be determined using a potentiometer. The terminal voltage is balanced against the potential gradient of the wire.
Step 2: Key Formula or Approach:
Terminal voltage \(V = \frac{ER}{R+r} = k l\).
Ratio: \(\frac{R_1}{R_1+r} \cdot \frac{R_2+r}{R_2} = \frac{l_1}{l_2}\).
Step 3: Detailed Explanation:
Case 1: \(R_1 = 12 \Omega, l_1 = 180 \text{ cm}\).
\[ \frac{12E}{12+r} = k(180) \quad \dots(1) \]
Case 2: \(R_2 = 4 \Omega, l_2 = 120 \text{ cm}\).
\[ \frac{4E}{4+r} = k(120) \quad \dots(2) \]
Divide equation (1) by equation (2):
\[ \frac{12(4+r)}{4(12+r)} = \frac{180}{120} \]
\[ \frac{3(4+r)}{12+r} = \frac{3}{2} \implies \frac{4+r}{12+r} = \frac{1}{2} \]
\[ 8 + 2r = 12 + r \]
\[ r = 4 \Omega \]
Step 4: Final Answer:
The internal resistance of the cell is 4 \(\Omega\).