At some instant, the current in the circuit shown below is 1 A, and it is decreasing at a rate of $10^2 A/s$. The value of the potential difference $(V_P - V_Q)$ at that instant is:

33 V

The correct answer is Option (1) → 33 V

Given: $i = 1 \, \text{A}$ and $\frac{di}{dt} = -10^2 \, \text{A/s}$, $L = 50 \, \text{mH}$, $R = 2 \, \Omega$, battery $= 30 \, \text{V}$.

Induced emf (magnitude):

$E_L = L \frac{di}{dt} = 50 \times 10^{-3} \times (-100) = -5 \, \text{V}$

So the inductor provides $+5 \, \text{V}$ when traversed from $P$ to the right (opposes the decrease in leftward current).

Resistor drop:

$V_R = I R = 1 \times 2 = 2 \, \text{V}$ (drop, so $-2 \, \text{V}$ contribution).

Battery (left to right): $+30 \, \text{V}$

Therefore,

$V_P - V_Q = (+5) + (+30) + (-2)$

$V_P - V_Q = 33 \, \text{V}$

Answer: $ \frac{V_P - V_Q}{ } = 33 \, \text{V}$