As per the given figure, if dI/dt=-1 A / s then the value of V_(AB) at this instant will be ____________ V.

Solution

Physics sir · 4 steps

Walk from A to B along the current and add up every drop, keeping the minus sign on dI/dt.

1. Resistor. The current of 2 A flows from A towards the inductor, so going with it through R the potential falls by IR = 2 × 12 = 24 V.
2. Inductor. Going with the current, the drop across the inductor is LdI/dt = 6 × (-1) = -6 V. A negative drop means the potential actually rises by 6 V. The current is decreasing, so the inductor pushes to keep it flowing, like a small battery aiding it.
3. Battery. The long thin plate is the positive terminal and it faces the inductor side. The short thick plate faces B. Moving from + to −, the potential falls by 12 V.

   V_A - IR - LdI/dt - E = V_B ⇒ V_(AB) = 24 + (-6) + 12

4. So. V_(AB) = V_A - V_B = 30 V, so the answer is 30.

The common slip: 42. Taking dI/dt as +1 A/s makes the inductor a 6 V drop, giving 24 + 6 + 12 = 42 instead of 30.

The idea to keep. Treat an inductor as a drop of LdI/dt in the direction of current. Put in the sign of dI/dt as given: a falling current turns that drop into a rise.

Answer from the NTA final answer key. Solution by PrepWiser.