Electromagnetic Induction and Alternating Current 4 Question 16
####16. Two inductors $L _1$ (inductance $1 mH$, internal resistance $3 \Omega$ ) and $L _2$ (inductance $2 mH$, internal resistance $4 \Omega$ ), and a resistor $R$ (resistance $12 \Omega$ ) 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 $\left(I _{\max } / I _{\min }\right)$ drawn from the battery is
(2016 Adv.)
Show Answer
Answer:
Correct Answer: 16. $(8)$
Solution:
$I _{\max }=\frac{\varepsilon}{R}=\frac{5}{12} A($ Initially at $t=0)$
$$ \begin{aligned} I _{\min } & =\frac{\varepsilon}{R _{eq}}=\varepsilon \frac{1}{r _1}+\frac{1}{r _2}+\frac{1}{R} \quad \text { (finally in steady state) } \\ & =5 \frac{1}{3}+\frac{1}{4}+\frac{1}{12}=\frac{10}{3} A \end{aligned} $$
$$ \frac{I _{\max }}{I _{\min }}=8 $$
