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Two inductors $L_{1}$ (inductance 1mH, internal resistance 3 $Ω$ ) and $L_{2}$ (inductance 2mH, internal resistance 4 $Ω$ ), and a resistor R (resistance 12 $Ω$ ) are all connected in parallel across a 5 V battery. The circuit is switched on a time t=0 The ratio of the maximum to the minimum current $(I_{m a x} / I_{m i n})$ drawn from the battery is

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Detailed Solution

$At t = 0, current will flow only in 12 Ω resistance$

$∴ I_{\min} = \frac{5}{12} A$

$At t \to \infty {both L}_{1} and L {}_{2}{behave as conducting wires}$

$\begin{array}{l} ∴ R_{e f f} = \frac{3}{2} Ω, \\ I_{\max} = \frac{V}{R_{e f f}} = \frac{5}{\frac{3}{2}} = \frac{10}{3} A \\ ∴ \frac{I_{\max}}{I_{\min}} = 8 \end{array}$

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