JEE Advanced 2016 · Paper 1 · Physics Q18

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PhysicsInteger+3Question 18 of 54

Two inductors $L_1$ (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 to the minimum current $(I_{\max}/I_{\min})$ drawn from the battery is

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Answer and worked solution

Official answer: 8

At t = 0. The inductors carry no current, so only $R$ conducts:
$I_{min}=\frac5{12}\text{ A}.$

Long after. The inductors act as their internal resistances, all in parallel:
$I_{max}=5\left(\frac1{12}+\frac13+\frac14\right)=5\times\frac8{12}\text{ A}.$

Ratio. $I_{max}/I_{min}=8$ .

Answer: 8

Topics:LR circuits

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JEE Advanced 2016 · Paper 1, Physics question 18