The circuit diagram of potentiometer used to measure the internal resistance of a cell (E) is shown in figure. The key 'K' is kept closed so as to send constant current through potentiometer wire. When key 'K₁' is kept open the null point is found to be at 120 cm on the potentiometer wire. When the key 'K₁' is closed the null point is shifted at 80 cm at the potentiometer wire. The internal resistance of the given cell is _____________ Ω.

Solution

Physics sir · 4 steps

Open key gives you E, closed key gives you the terminal voltage; their ratio is (R + r)/R.

1. K1 open. No current flows through the cell at the null point, so the wire balances the full emf: E = k l₁ with l₁ = 120 cm. Here k is the potential drop per cm of the wire, fixed by the main circuit.
2. K1 closed. Now the cell drives a current I = E/(R + r) through R = 4 Ω. The wire balances only the terminal voltage V = IR = k l₂ with l₂ = 80 cm.
3. Take the ratio. Dividing, E/V = l₁/l₂, and E/V = (R + r)/R. Solving for r gives the standard result.

   r = R(l₁/l₂ - 1) = 4(120/80 - 1) = 4 × 0.5

4. So. The internal resistance of the cell is r = 2 Ω.

The common slip: 6. Writing r = Rl₁/l₂ and forgetting to subtract 1 gives 6, which is actually R + r, not r.

The idea to keep. In a potentiometer, balancing length is proportional to voltage. Open circuit reads the emf, loaded circuit reads the terminal voltage, and the drop between them is caused by r.

Answer from the NTA final answer key. Solution by PrepWiser.