An RC circuit has R = 1 kΩ and C = 10 μF. What is the time constant τ, and what is the meaning?

An RC circuit has R = 1 kΩ and C = 10 μF. What is the time constant τ, and what is the meaning?

• A. τ = 0.1 s; time to reach about 63% of final value
• B. τ = 0.01 s; time constant is time to reach about 63% of final value
• C. τ = 0.01 s; time to reach about 100% of final value
• D. τ = 0.001 s; time to reach about 50% of final value

Answer: (B)

The main idea here is the time constant, which for an RC circuit is RC. It tells you how quickly the capacitor voltage moves toward its final value. With R = 1 kΩ and C = 10 μF, the time constant is RC = (1000 Ω)(10 × 10^-6 F) = 0.01 s. What this time constant means: after a time equal to RC, the capacitor voltage has reached about 63% of the way from its starting value to its final value. If you’re charging, the voltage follows Vc(t) = V_final(1 − e^−t/RC); at t = RC, Vc = V_final(1 − e^−1) ≈ 0.632 V_final. If you’re discharging, it follows Vc(t) = V_initial e^−t/RC; at t = RC, Vc ≈ 0.368 V_initial, which is a 63% drop from the starting value. So the correct option is the one that gives a time constant of 0.01 s and describes the 63% progress toward the final value.

The main idea here is the time constant, which for an RC circuit is RC. It tells you how quickly the capacitor voltage moves toward its final value. With R = 1 kΩ and C = 10 μF, the time constant is RC = (1000 Ω)(10 × 10^-6 F) = 0.01 s.

What this time constant means: after a time equal to RC, the capacitor voltage has reached about 63% of the way from its starting value to its final value. If you’re charging, the voltage follows Vc(t) = V_final(1 − e^−t/RC); at t = RC, Vc = V_final(1 − e^−1) ≈ 0.632 V_final. If you’re discharging, it follows Vc(t) = V_initial e^−t/RC; at t = RC, Vc ≈ 0.368 V_initial, which is a 63% drop from the starting value.

So the correct option is the one that gives a time constant of 0.01 s and describes the 63% progress toward the final value.