SATHEE: Electromagnetic Induction and Alternating Current 4 Question 16

Electromagnetic Induction and Alternating Current 4 Question 16

####16. Two inductors $L_{1}$ (inductance $1 m H$ , internal resistance $3 Ω$ ) and $L_{2}$ (inductance $2 m H$ , 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

(2016 Adv.)

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Answer:

Correct Answer: 16. $(8)$

Solution:

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

$\begin{aligned} I_{min} & = \frac{ε}{R_{e q}} = ε \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{R} (finally in steady state) \\ = 5 \frac{1}{3} + \frac{1}{4} + \frac{1}{12} = \frac{10}{3} A \end{aligned}$

$\frac{I_{max}}{I_{min}} = 8$

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Electromagnetic Induction and Alternating Current 4 Question 16

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