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Q. 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 $\omega=500 \,rad / s$ 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).Physics Question Image

Alternating Current

Solution:

If $q$ represents the charge on capacitor's upper plate:
$I(t)=I_{0} \cos (\omega t)=\frac{d q}{d t}$
$ \Rightarrow q(t)=\frac{I_{0}}{\omega} \sin (\omega t) $
Max charge $=\frac{1 A }{500 rad s ^{-1}}=2 \times 10^{-3} C$
$I(t)=I_{0} \cos (\omega t)=\frac{d q}{d t}$
$ \Rightarrow q(t)=\frac{I_{0}}{\omega} \sin (\omega t)$
Max charge $=\frac{1 A }{500 rads ^{-1}}=2 \times 10^{-3} C$
Charge on upper plate at $t=\frac{7 \pi}{6 \omega}$
$=\frac{I_{0}}{\omega} \sin \left(\frac{7 \pi}{6}\right)=-\frac{I_{0}}{2 \omega}$
When capacitor is fully charged, charge on upper plate
$=50 V \times 20 \mu F =1 \times 10^{-3} C$
$\therefore Q=1 \times 10^{-2} C -\left(-\frac{I_{0}}{2 \omega}\right)=2 \times 10^{-3} C$
When capacitor is fully charged, charge on upper plate
$=50 \,V \times 20 \,\mu F =1 \times 10^{-3} C $
$\therefore Q=1 \times 10^{-2} C -\left(-\frac{I_{0}}{2 \omega}\right)=2 \times 10^{-3} C$
Voltage across capacitor when $A$ and $D$ are connected
$=\frac{1 \times 10^{-3} C }{20 \times 10^{-6} \mu F }=50\, V$
Total voltage across resistor $=100\, V$
$\Rightarrow $ current $=\frac{100\,V }{10 \Omega}=10 \,A$
Current is anticlockwise.