Question

An uncharged capacitor is connected in circuit as shown in figure. Power ratings of bulbs are given in diagram. At $t=0$, switch is closed. Select correct alternative(s) :

• A. At t = 0 power consumption in circuit is $\frac{3P_0}{2}$.
• B. After a long time power consumption in circuit is$\frac{P_0}{2}$.
• C. Brightness of $B_1$ decreases with time.
• D. Initially Brightness of bulb $B_2$ is less than brightness of $B_1$ , but later on $B_2$ will be brighter.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A At t = 0 power consumption in circuit is $\frac{3P_0}{2}$.
B After a long time power consumption in circuit is$\frac{P_0}{2}$.
C Brightness of $B_1$ decreases with time.
D Initially Brightness of bulb $B_2$ is less than brightness of $B_1$ , but later on $B_2$ will be brighter.

At $t=0$ capacitor is uncharged, so the simplified circuit is shown in figure.
Equivalent resistance for this circuit is $R_{eq}=\frac{2R\times R}{2R+R}=\frac{2R}{3}$
Let, $P_0=\frac{V_0^2}{R}$
$P_{consume}=\frac{V_0^2}{R_{eq}}=\frac{V_0^2}{\frac{2R}{3}}=\frac{3P_0}{2}$
After a long time interval, simplified circuit is shown in figure
$R_{eq}=2R$
$P_{consume}=\frac{V_0^2}{2R}=\frac{P_0}{2}$
Since current is decreasing in $B_1$ with time so brightness will also decrease.
Initially $B_1$ is brighter than $B_2$ but later $B_2$ will be brighter .