In a purely inductive circuit with a DC input after a long time, the inductor behaves as which?

• A. A resistor.
• B. A capacitor.
• C. An open circuit.
• D. A short circuit.

Answer: (D)

This tests how an inductor behaves under a constant (DC) input after all transients have died out. The relation v = L di/dt means the voltage across an inductor depends on how quickly the current is changing. When the input is DC and we wait long enough, the current becomes steady, so di/dt = 0. That makes the voltage across the inductor zero. With zero voltage drop and no resistance in an ideal, purely inductive element, the inductor effectively becomes a short circuit—it behaves like a wire that carries the constant current without dropping any voltage. This is why, in DC steady state, an ideal inductor looks like a short. If you think about contrast, a capacitor would have acted as an open circuit at DC, since it stops DC current once it’s charged, while a real resistor would always drop some voltage. But for an ideal inductor with DC after a long time, zero voltage and a direct path mean a short.