Two inductors L i (inductance 1 mH , internal resistance 3 Ω ) and L 2 (inductance 2 mH , internal resistance 4 Ω ), and a resistor R (resistance 12 Ω ) are all connected in parallel across a 5 V batt

Question

Two inductors L i (inductance 1 mH , internal resistance 3 Ω ) and L 2 (inductance 2 mH , 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 at time t = 0 . The ratio of the maximum current I m a x I m i n drawn from the battery is:

TestHub · Accepted Answer

Initally, the inductances will offer infinite resistance and the current will be maximum.

$I_{m a x} = \frac{ε}{R} = \frac{5}{12} A$ (Initially at t = 0)

Upon reaching steady state, the inductances will act as pure resistors and hence the current will be minimum.
$I_{m i n} = \frac{ε}{R_{e q}} = ε (\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{R})$ (Finally in steady state)
$= 5 (\frac{1}{3} + \frac{1}{4} + \frac{1}{12})$
$= \frac{10}{3} A$
$\frac{I_{m a x}}{I_{m i n}} = 8$

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