Question

Figure shows a part of complete circuit at $t=0$. Charge on capacitor as a function of time is given by $q=3(1–e^{–t})$ coulombs. At $t=0$, mark the correct statement $\left(s\right)$.

• A. Current through branch containing capacitor is $3\text{A}$
• B. Current through $R$ is $7\text{A}$
• C. The potential of $P$ is $14\text{V}$
• D. The value of $R$ is $2\Omega$

Answer 1

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

The value of $R$ is $2\Omega$


Given, charge on capacitor,
$q=3(1-e^{-t})$
$\therefore$ In branch containing capacitor, assume current $=I_1$
$\therefore I_1=\frac{dq}{dt}=3e^{-t}$
$\therefore$ At $t=0,I_1=3\text{A}$
$\to$ For potential at point $P$, taking same branch
$20-V_P=I_{1}R$
$\Rightarrow V_P=20-I_{1}R=20-3\times 2=14\text{V}$
$\to$ Now in branch with resistor and battery, current is $I_2$
$20-2-I_2.1=+14$
$\Rightarrow I_2=20-16=4\text{A}$
$\to$ Now, current through resistor $R$ is,
$I=I_1+I_2$
$I=3+4=7\text{A}$
$\to$ branch with resistor $R$ ;
$V_P-0=IR$
$\Rightarrow 14=7R$
$\Rightarrow R=\frac{14}{7}=2\Omega$