Question

Potential of the points $P$ and $Q$ in the circuit shown in the figure are

• A. $6\text{V}$ and $-2\text{V}$ respectively
• B. $6\text{V}$ and $+2\text{V}$ respectively
• C. $-6\text{V}$ and $-2\text{V}$ respectively
• D. $-6\text{V}$ and $+2\text{V}$ respectively

Answer 1

The correct option is A $6\text{V}$ and $-2\text{V}$ respectively


Let $i$ be current in the circuit
Applying KVL
$10-i\left(1\right)-i\left(3\right)-i\left(1\right)=0$
$i=\frac{10}{5}=2\text{A}$
As point D is earthed potential of D $\left(V_D\right)$ is zero

Potential difference across $3\Omega$ resistor is $V_P-V_D=i\left(3\right)=6\Rightarrow V_P=6\text{V}$

Potential difference across $1\Omega$ resistor is $V_D-V_Q=i\left(1\right)=2\Rightarrow V_Q=-2\text{V}$