A cell of emf $E$ and internal resistance $r$ is connected to two external resistance $R_{1}$ and $R_{2}$ and a perfect ammeter. The current in the circuit is measured in four different situations:
(i) without any external resistance in the circuit
(ii) with resistance $R_{1}$ only
(iii) with $R_{1}$ and $R_{2}$ in series combination
(iv) with $R_{1}$ and $R_{2}$ in parallel combination
The currents measured in the four cases are $0.42 A, 1.05 A, 1.4 A$ and $4.2 A$ , but not necessarily in the order. Identify the currents corresponding to the four cases mentioned above.

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Solution

Using Ohm's law, $V = I R$

for (i) : $V = E, I = I_{1}, R = r$ so $I_{1} = \frac{E}{r}$

for (ii): $V = E, I = I_{2}, R = r + R_{1}$ so $I_{2} = \frac{E}{r + R_{1}}$

for (iii): $V = E, I = I_{3}, R = r + (R_{1} + R_{2})$ so $I_{3} = \frac{E}{r + R_{1} + R_{2}}$

for (iv): $V = E, I = I_{4}, R = r + \frac{R_{1} R_{2}}{R_{1} + R_{2}}$ so $I_{4} = \frac{E}{r + \frac{R_{1} R_{2}}{R_{1} + R_{2}}}$

Thus , $I_{1} > I_{4} > I_{2} > I_{3}$ (The equivalent resistance for parallel combination is less than of each resistance )

Hence, 4.2 A is for (i), 1.4 A for (iv), 1.05 A for (ii) and 0.42 A for (iii).

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