At some instant, the current in the circuit shown below is 1 A, and it is decreasing at a rate of \( 10^{2}\,\mathrm{A/s} \). 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
The potential difference between the ends of a branch is found by adding the changes of potential across each element. Across an inductor the potential falls by \( L\,dI/dt \) in the direction of the current.
\( I = 1\,\mathrm{A}\;(\text{from Q to P}) \)
\( R = 2\,\Omega,\; E = 30\,\mathrm{V},\; L = 50\,\mathrm{mH} = 0.05\,\mathrm{H} \)
\( dI/dt = -10^{2}\,\mathrm{A\,s^{-1}}\;(\text{decreasing}) \)
Go from Q to P along the current:
across the resistor the potential falls by \( IR = 1 \times 2 = 2\,\mathrm{V} \);
across the cell, from the short (negative) plate to the long (positive) plate, it rises by 30 V;
across the inductor it changes by \( -L\,dI/dt = -(0.05)(-100) = +5\,\mathrm{V} \), a rise, because the falling current induces an emf that tries to keep it flowing.
\( \displaystyle V_P = V_Q - IR + E - L\frac{dI}{dt} \)
\( \displaystyle V_P - V_Q = -2 + 30 - (0.05)(-100) = -2 + 30 + 5 = 33\,\mathrm{V} \)