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:
Answer & explanation
Correct answer: option 1
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}$