Question

In the figure shown, $V_{AB}$ ​at $t=1\text{s}$ is :

• A. $30\text{V}$
• B. $-30\text{V}$
• C. $20\text{V}$
• D. $20\text{V}$

Answer 1

The correct option is B $-30\text{V}$


Given:

$q=4t^2$

At $t=1\text{s}$, $q=4\text{C}$

Now, $i=\frac{dq}{dt}=8t$

At $t=1\text{s}$, $i=8\text{A}$

Also, $\frac{di}{dt}=8$

The charge on the capacitor is increasing. So, charge on positive plate is also increasing. Hence, direction of current is towards left.

Now, from KVL,

$V_A+8\times 2-4+2\times 8+\frac{4}{2}=V_B$

$\Rightarrow V_A-V_B=-30\text{V}$