What is the time constant of a circuit consisting of two 100-microfarad capacitors and two 470-kilohm resistors all in series?

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Multiple Choice

What is the time constant of a circuit consisting of two 100-microfarad capacitors and two 470-kilohm resistors all in series?

When charging a circuit with both resistors in series and capacitors in series, the relevant time constant comes from the total resistance times the equivalent capacitance seen by that resistance. For capacitors in series, the combined capacitance is lower: C_eq = (C1 × C2) / (C1 + C2). For resistors in series, the combined resistance is simply the sum: R_eq = R1 + R2. So the time constant τ = R_eq × C_eq. Here, C1 and C2 are both 100 μF, so C_eq = (100 μF × 100 μF) / (100 μF + 100 μF) = 50 μF. The resistors add: R_eq = 470 kΩ + 470 kΩ = 940 kΩ. Convert to consistent units: R_eq = 940,000 Ω and C_eq = 50 × 10^-6 F. Then τ = 940,000 × 50 × 10^-6 = 47 seconds. So the time constant is 47 seconds.

When charging a circuit with both resistors in series and capacitors in series, the relevant time constant comes from the total resistance times the equivalent capacitance seen by that resistance. For capacitors in series, the combined capacitance is lower: C_eq = (C1 × C2) / (C1 + C2). For resistors in series, the combined resistance is simply the sum: R_eq = R1 + R2. So the time constant τ = R_eq × C_eq.

Here, C1 and C2 are both 100 μF, so C_eq = (100 μF × 100 μF) / (100 μF + 100 μF) = 50 μF. The resistors add: R_eq = 470 kΩ + 470 kΩ = 940 kΩ. Convert to consistent units: R_eq = 940,000 Ω and C_eq = 50 × 10^-6 F. Then τ = 940,000 × 50 × 10^-6 = 47 seconds.

So the time constant is 47 seconds.

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