An RLC series circuit has an applied voltage of 240 volts. R = 48 Ω, XL = 100 Ω, XC = 36 Ω, and Z = 80 Ω. What is the voltage drop across the capacitor?

108V

Explanation:

Voltage drop across the capacitor (Vc) = IXc where

I is the current in the circuit

Xc is the capacitive reactance

Given Xc = 36Ω

To get the current I, we will use the formula for calculating the total voltage across the circuit which gives;

V = IZ

Where V is the applied voltage = 240Ω

I is the current flowing in the circuit.

Z is the impedance = 80Ω

I = V/Z

I = 240/80

I = 3A

Voltage drop across the capacitor Vc = 3*36 = 108V

The current 3A was used since the elements are connected in series and same current flows in the elements of a series connected circuit.

I44volts

Explanation:

Given that,

Vc = IXc

I = V/Z

Where V = 240 volts, Z = 80 ohms

Hence, I = V/Z = 240:80

I = 4A

Xc=36 ohms

Vc = IXc = 4*36

Vc = 144volts