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 $10\Omega$ is zero, since it does not form any closed path. So, the potential drop across the $10\Omega$ 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.