EMC Question of the Week: September 28, 2026
Each of the four circuits in the figure has a switch that closes at t=0. Which circuit will be the first to develop 9 volts across the capacitor?
Answer
The correct answer is (b.) In all four circuits, the voltage across the capacitor rises exponentially to its steady-state value with a time constant equal to RC. R is the resistance as viewed from the capacitor terminals with the switch closed. Circuits (a.) and (d.) never reach 9 volts. The steady-state voltage in (a.) is about 8.3 volts and the steady-state voltage in (d.) is 2 volts.
The steady-state voltage in Option (b.) is 9.8 volts. The RC time constant is approximately 2 Ω times 1 μF, which is 2 μs. It will take approximately 2.5 time constants (or 5 μs) to reach 9 volts.
The steady-state voltage in Option (c.) is 10 volts. The RC time constant is 10 Ω times 1 μF, which is 10 μs. It will take approximately 2.3 time constants (or 23 μs) to reach 9 volts.
Note that the 2-Ω resistor in Option (c.) is in series with a current source. It has no effect on the steady-state voltage across the capacitor or the time constant.
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