Question

In the figure shown, the charge on the capacitor is $q=\left(4t^2\right)\text{C}$ then $V_{ab}$ 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 $-30\text{V}$

$i=\frac{dq}{dt}=\left(8t\right)A$
$\frac{di}{dt}=8\text{A/s}$
At $t=1\text{s},q=4\text{C},i=8\text{A}$
and $\frac{di}{dt}=8\text{A/s}$
Charge on capacitor is increasing. So, charge on positive plate is also increasing. Hence, direction of current is towards left.

Now, $V_a+2\times 8-4+2\times 8+\frac{4}{2}=V_b$
$\therefore V_a-V_b=-30\text{V}$