A potentiometer wire has a gradient of 2.5 mV/cm…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A potentiometer wire has a gradient of $2.5\ \text{mV/cm}$ . A cell balances at $360\ \text{cm}$ when open. A $10\ \Omega$ resistor is then joined across the cell, and the balance point moves to $300\ \text{cm}$ . Find the emf and the internal resistance of the cell, and the current through the resistor.

Need a reminder? The key ideaShow

The gradient is $k = \dfrac{IR_w}{L}$ , and the potential at a distance $l$ is $V = kl$ . At balance $E = kl$ , so $\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}$ . A balance with the cell open and with $R$ across it gives $r = R\,\dfrac{l_1 - l_2}{l_2}$ . A voltmeter of resistance $R_v$ across a cell reads $V = \dfrac{ER_v}{R_v + r}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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