Time constant of a circuit with two 100 µF capacitors and two 470 kΩ resistors in series is.

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Multiple Choice

Time constant of a circuit with two 100 µF capacitors and two 470 kΩ resistors in series is.

Time constant is determined by the resistance that the capacitor network “sees” multiplied by the equivalent capacitance of the network. When components are in series, you add resistances and you combine capacitors using the series formula. Two 470 kΩ resistors in series give a total resistance of 940 kΩ. Two 100 µF capacitors in series give an equivalent capacitance of (C1*C2)/(C1+C2) = (100 µF × 100 µF)/(100 µF + 100 µF) = 50 µF. Convert to consistent units and multiply: 940,000 Ω × 50 × 10^-6 F = 940,000 × 0.00005 = 47 seconds. Therefore, the time constant is 47 seconds. The other numbers would come from using different equivalent values or mixing up series vs parallel combinations, which isn’t the case here.

Time constant is determined by the resistance that the capacitor network “sees” multiplied by the equivalent capacitance of the network. When components are in series, you add resistances and you combine capacitors using the series formula.

Two 470 kΩ resistors in series give a total resistance of 940 kΩ. Two 100 µF capacitors in series give an equivalent capacitance of (C1*C2)/(C1+C2) = (100 µF × 100 µF)/(100 µF + 100 µF) = 50 µF.

Convert to consistent units and multiply: 940,000 Ω × 50 × 10^-6 F = 940,000 × 0.00005 = 47 seconds.

Therefore, the time constant is 47 seconds. The other numbers would come from using different equivalent values or mixing up series vs parallel combinations, which isn’t the case here.

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