Question

A \(10 \mathrm{~m}\) wire potentiometer is connected to an accumulator of steady voltage. A \(7.8 \mathrm{~m}\) length of it balances the emf of a cell on 'open- circuit'. When cell delivers current through a conductor of resistance $10 \Omega\( it is balanced against \)7.0 \mathrm{~m}$ of the same potentiometer. What is the internal resistance of the cell ? (A) \(1.24 \Omega\) (B) \(1.36 \Omega\) (C) \(1.14 \Omega\) (D) \(1 \Omega\)

Short answer

The internal resistance of the cell is approximately \(1.14\Omega\), which corresponds to option (C).

Step-by-step solution

Open-circuit condition

In the open-circuit condition, the length of the potentiometer wire that balances the emf of the cell is \(7.8 \mathrm{~m}\). We have total length of wire as \(10 \mathrm{~m}\). We can calculate the potential gradient across the potentiometer:^(V_{gradient})^: \(V_{gradient} = \dfrac{V}{L}\) Where V is the voltage across the full length (10m) of the potentiometer and L is the length of the potentiometer (10m). Therefore, the voltage across the 7.8m part of the potentiometer (V1) can be found: \(V_1 = V_{gradient} \times 7.8 \) Since the potentiometer balances the emf, this voltage is equal to the emf of the cell (\(E\)): \(E = V_1\)

Closed-circuit condition

In the closed-circuit scenario, the cell delivers current through a conductor of resistance \(10\Omega\). The length of the potentiometer that balances this situation is \(7.0\mathrm{~m}\). We can find the voltage across the 7.0m part of the potentiometer (V2): \(V_2 = V_{gradient} \times 7.0\) Now, we can use Ohm's Law with the values we have. Let the current through the conductor be I and the internal resistance of the cell be r: \(V_2 = I (10 + r)\)

Solve for internal resistance (r)

We have the two equations E = V1 and V2 = I(10 + r). We can substitute the values of V1 and V2 from steps 1 and 2 into these equations: \(E = V_{gradient} \times 7.8\) \(V_{gradient} \times 7.0 = I (10 + r)\) Now, use the first equation to isolate V_gradient and substitute that into the second equation to solve for r: \(\dfrac{7.8}{7.0}E = I(10 + r)\) As we know E = IR + Ir, where R is \(10\Omega\): \(\dfrac{7.8}{7.0}(I(R + r)) = I(10 + r)\) Now, the I on both sides of the equation cancels out: \(\dfrac{7.8}{7.0}(10 + r) = 10 + r\)

Calculate the internal resistance (r)

Now we can solve for r: \(\dfrac{7.8}{7.0}(10 + r) - (10 + r) = 0\) \(\dfrac{0.8}{7.0}(10 + r) =r\) \(\dfrac{0.8}{7.0}(10 + r) - r = 0\) \(\dfrac{0.8}{7.0} \times 10 = r - \dfrac{0.8}{7.0}r\) \(r = \dfrac{0.8}{7.0} \times 10 \times \dfrac{7}{7-0.8}\) \(r \approx 1.14 \Omega\) The internal resistance of the cell is approximately \(1.14\Omega\), which corresponds to option (C).