Electromagnetic Induction And Alternating Current Ques 69
- At time $t=0$, terminal $A$ in the circuit shown in the figure is connected to $B$ by a key and an alternating current $I(t)=I _0 \cos (\omega t)$, with $I _0=1 A$ and $\omega=500 rad s^{-1}$ starts flowing in it with the initial direction shown in the figure. At $t=7 \pi / 6 \omega$, the key is switched from $B$ to $D$. Now onwards only $A$ and $D$ are connected. A total charge $Q$ flows from the battery to charge the capacitor fully. If $C=20 \mu F, R=10 \Omega$ and the battery is ideal with emf of $50 V$, identify the correct statement(s).
(2014 Adv.)
(a) Magnitude of the maximum charge on the capacitor before $t=\frac{7 \pi}{6 \omega}$ is $1 \times 10^{-3} C$
(b) The current in the left part of the circuit just before $t=\frac{7 \pi}{6 \omega}$ is clockwise
(c) Immediately after $A$ is connected to $D$, the current in $R$ is $10 A$
(d) $Q=2 \times 10^{-3} C$
Show Answer
Answer:
Correct Answer: 69.(c, d)
Solution:
Formula:
- $\frac{d Q}{d t}=I$
$$ \begin{aligned} & \Rightarrow \quad Q=\int I d t=\int\left(I _0 \cos \omega t\right) d t \\ & \therefore \quad Q _{\max }=\frac{I _0}{\omega}=\frac{1}{500}=2 \times 10^{-3} C \end{aligned} $$
Just after switching
In steady state
At $t=\frac{7 \pi}{6 \omega}$ or $\omega t=\frac{7 \pi}{6}$
Current comes out to be negative from the given expression. So, current is anti-clockwise. Charge supplied by source from $t=0$ to $t=\frac{7 \pi}{6 \omega}$
Apply Kirchhoff’s loop law, just after changing the switch to position $D$
$$ 50+\frac{Q _1}{C}-I R=0 $$
Substituting the values of $Q _1, C$ and $R$, we get
$$ I=10 A $$
In steady state $Q _2=C V=1 mC$
$\therefore \quad$ Net charge flown from battery $=2 mC$
BETA
