Question

Find the voltage $V_{ab}$ in the circuit shown in the figure.

(A) $+3\text{V}$
(B) $-3\text{V}$
(C) $+6\text{V}$
(D) $\text{ - 6V}$

Answer 1

Hint
To solve this question, start from the initial potential point and reach to the final potential point through any path. While travelling through a path, write all the potential drops and gains which come in between the path.

Complete step by step answer
As in the first mesh, the current source of $2\text{A}$ is present, so the current in the whole mesh is $2\text{A}$. Now, we need to find the current in the second mesh using KVL.
We assume a current of $I$ in the second mesh, as shown in the below diagram.

Applying KVL in the second mesh, we get
$\Rightarrow -30+6I+4I=0$
$\Rightarrow 10I=30$
Dividing both the sides by $10$, we get
$\Rightarrow I=3\text{A}$
The current in the branch containing the resistance $10\mathrm{\Omega }$ is zero, since it does not form any closed path. So, the potential drop across the $10\mathrm{\Omega }$ resistance is zero, and hence it can be discarded out. So, the circuit can be redrawn as


 

Now, for finding $V_{ab}$, we start from the point a, and travel along the path acdb to reach the final point b.
$\Rightarrow V_a+5(2)+5-4(3)=V_b$
On rearranging the terms, we get
$\Rightarrow V_a-V_b=-3\text{V}$
Or, $V_{ab}=-3V$
Thus, the voltage $V_{ab}$ is equal to $-3V$
Hence, the correct answer is option B.

Note
Do not apply KVL along the path, in between where a current source is there. This is because the potential difference across a current source is unknown to us. Do not assume it to be zero. It can only be found out analytically. So, applying KVL along the path containing a current source will not be possible.