When two resistors are connected in series with a cell of emf $2 V$ and negligible internal resistance, a current of $2 / 5 A$ flows in the circuit. When the resistors are connected in parallel the main current is $\frac{5}{3} A$ . Calculate the resistances.

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Solution

Let $R_{1}$ and $R_{2}$ be the two resistance.
When the resistances are connected in series $R_{s} = R_{1} + R_{2}$
Current in the circuit is
$I = \frac{ϵ}{R + r}$
i.e., $\frac{2}{5} = \frac{2}{R_{s} + 0}$
i.e., $\Rightarrow R_{s} = R_{1} + R_{2} = 5 . . . . (1)$
When two resistances are connected in parallel
$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{R_{1} + R_{2}}{R_{1} R_{2}} = \frac{5}{R_{1} R_{2}} (∵ f r o m (1))$
$R_{p} = \frac{R_{1} R_{2}}{5}$
Current in the circuit is
$I = \frac{ϵ}{R_{p} + r}$
$\frac{5}{3} = \frac{2}{(\frac{R_{1} R_{2}}{5}) + 0} = \frac{10}{R_{1} R_{2}}$
$R_{1} R_{2} = 6 . . . . (2)$
Using the relation $(R_{1} - R_{2})^{2} = (R_{1} + R_{2})^{2} - 4 R_{1} R_{2}$
$(R_{1} - R_{2})^{2} = (5)^{2} - 4 (6) = 25 - 24 = 1$
$∴ R_{1} - R_{2} = 1 . . . . (3)$
Adding equations $(1)$ and $(3)$ , we get
$2 R_{1} = 6$ or $R_{1} = 3 Ω$
$R_{2} = 5 - R_{2} = 5 - 3 = 2 Ω$ .

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When two resistors are connected in series with a cell of emf $2 V$ and negligible internal resistance, a current of $2 / 5 A$ flows in the circuit. When the resistors are connected in parallel the main current is $\frac{5}{3} A$ . Calculate the resistances.