A coil and a 10 Ω resistor are connected in series to a 10 V DC source, and a 100 Ω resistor is connected in parallel with the coil. After a long time, a switch disconnects the source and the 10 Ω resistor. What is the magnitude of the voltage across the 100 Ω resistor immediately after disconnection? (Ignore the coil's winding resistance.)

Answer: C. 100 V

In steady state the coil's voltage is 0 (like a short circuit), so no current flows in the 100 Ω resistor and the coil carries 10 V/10 Ω = 1 A. The coil current cannot change suddenly, so 1 A keeps flowing right after disconnection, and its only path is the 100 Ω resistor. The resistor therefore develops 1 A × 100 Ω = 100 V, ten times the supply voltage. The current then decays with time constant L/(100 Ω). This is why switching off a relay coil produces a high voltage, which is limited by giving the current a path through a diode.

• A. 10 V is the supply voltage; it ignores that the coil current is maintained.
• B. 0 V is the voltage across the 100 Ω before disconnection.
• C. Correct. 1 A keeps flowing through 100 Ω, giving 1 A × 100 Ω = 100 V.