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 $t=0$, which of the following statement is/are correct.

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

Answer 1

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

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

Current passing through the branch containing $1\Omega$ resistor at $t=0\text{sec}$ is given by

$20-2-(1\times I_1)=V_P\Rightarrow I_1=18-V_P......\left(1\right)$

Current passing through the branch containing $2\Omega$ resistor at $t=0\text{sec}$ is given by

$20-\frac{q}{C}-(2\times I_2)=V_P\Rightarrow I_2=\frac{17-V_P}{2}.......\left(2\right)$

Current passing through $R$ at $t=0\text{sec}$ is given by

$I=\frac{0-V_P}{R}......\left(3\right)$

At $t=0$, the capacitor acts as short circuit $($i.e $C\to \infty )$

From $\left(2\right)$ we can then say that,

$I_2=\frac{20-V_P}{2}.......\left(4\right)$

But according to the question, $q=3(1-e^{-t})$

Differentiating the above equation with respect to time on both sides at $t=0$ gives, $I_2=3\text{A}$

Substituting this in $\left(4\right)$ we get, $V_P=14\text{V}$

From $\left(1\right)$, $I_1=18-14=4\text{A}$

So the net current passing through resistance $R$ is $I=I_1+I_2=4+3=7\text{A}$

Potential difference across resistance $R$ is $\Delta V=14\text{V}$

Therefore, Resistance $R=\frac{\Delta V}{I}=\frac{14}{7}=2\Omega$

Hence, all the options are the correct answers.