Question

In the figure shown $V_{ab}$ at $t=1s$ is :

• A. $30V$
• B. $-30V$
• C. $20V$
• D. $-20V$

Answer 1

Answer: (A)

$30V$

$i=\frac{dq}{dt}=\left(8t\right)A$
$\frac{di}{dt}=8A/s$
At $t=1s,q=4C,i=8A$ and $\frac{di}{dt}=8A/s$
Charge on capacitor is increasing. So, charge on positive plate is also increasing. Hence direction of current is towards left.
Now $V_a+2\times 8-4+2\times 8+\frac{4}{2}=V_b$
$\therefore$ $V_a-V_b=-30V$

$\therefore V_{ab}=V_b-V_a=30V$