A series RC circuit contains two resistors and two capacitors. The resistors are 39 Ω and 68 Ω. The capacitors have capacitive reactances of 60 Ω and 50 Ω. The applied voltage is 240 V. What is the voltage drop on the capacitor that has 60 Ω of reactance?

Vc = 94 V

Explanation:

In a series circuit, powered by a sinusoidal source, the current is the same across all circuit elements, and can be calculated based on the impedance concept, that takes into account that the voltage and the current in a given circuit element have a phase angle, which means that voltage and current don't reach to a maximum at the same time.

We define the impedance as the complex sum of the resistive part of the circuit (in which current and voltage be in phase) and the reactive part (in which current and voltage are 90º out of phase), as follows:

Z = R + j (XL - XC)

In order to obtain the magnitude of Z, we can do the following:

Z = √R² + (Xl-Xc) ²

The value of the current can be obtained just dividing the applied voltage between the magnitude of the impedance.

For our circuit, we can get the magnitude of Z as follows:

Z = √ (39+68) ² + (60+50) ² = 154 Ω

⇒ I = V/Z = 240 V / 154 Ω = 1.56 A

In order to get the magnitude of the voltage drop across any circuit element, we just need to apply the generalized version of Ohm's Law.

For a capacitor, the voltage drop can be calculated as follows:

Vc = I * Xc

⇒ Vc = 1.56 A * 60Ω = 94 V