Electromagnetic Induction And Alternating Current Ques 52
52. Two inductors $L_{1}$ (inductance $1 \mathrm{mH}$, internal resistance $3 \Omega$ ) and $L_{2}$ (inductance $2 \mathrm{mH}$, internal resistance $4 \Omega$ ), and a resistor $R$ (resistance $12 \Omega$ ) are all connected in parallel across a $5 \mathrm{~V}$ battery. The circuit is switched on at time $t=0$. The ratio of the maximum to the minimum current $\left(I_{\max } / I_{\min }\right)$ drawn from the battery is
(2016 Adv.)
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Answer:
Correct Answer: 52.$(8)$
Solution:
Formula:
Growth Of Current in Series R-L Circuit:
$I_{\max }=\frac{\varepsilon}{R}=\frac{5}{12} \mathrm{~A}($ Initially at $t=0)$
$$ \begin{aligned} I_{\min } & =\frac{\varepsilon}{R_{\mathrm{eq}}}=\varepsilon \frac{1}{r_{1}+r_{2}+R} \quad \text { (finally in steady state) } \\ & =5 \frac{1}{3}+\frac{1}{4}+\frac{1}{12}=\frac{10}{3} \mathrm{~A} \end{aligned} $$
$$ \frac{I_{\max }}{I_{\min }}=8 $$
BETA
