Question

In the given branch $AB$ of a circuit, a current, $I=(10t+5)\text{A}$ is flowing, where $t$ is time in second. At $t=0$, the potential difference between points $A$ and $B$, $V_A-V_B$ is -

• A. $15\text{V}$
• B. $-5\text{V}$
• C. $-15\text{V}$
• D. $5\text{V}$

Answer 1

The correct option is A $15\text{V}$



Given:

$I=(10t+5)\text{A}$

At $t=0$, $I=5\text{A}$

Also, $\frac{dI}{dt}=10\text{A/s}$

Now, from KVL,

$V_B-10+1\times 10+5\times 3=V_A$

$\Rightarrow V_A-V_B=15\text{V}$