First slide

potentiometer

Question

A resistance of $4 Ω$ and a wire of length 5 metres and resistance $5 Ω$ are joined in series and connected to a cell of e.m.f. 10 V and internal resistance $1 Ω$ . A parallel combination of two identical cells is balanced across 300 cm of the wire. The e.m.f. E of each cell is

Moderate

A

1.5 V
B

3.0 V
C

0.67 V
D

1.33 V

Solution

$\begin{array}{l} E = xl = \frac{V}{l} = \frac{iR}{L} \times l \Rightarrow E = \frac{e}{(R + R_{h} + r)} \times \frac{R}{L} \times l \\ \Rightarrow E = \frac{10}{(5 + 4 + 1)} \times \frac{5}{5} \times 3 = 3 V \end{array}$

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