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20 February, 14:34

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?

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Answers (2)
  1. 20 February, 16:37
    0
    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
  2. 20 February, 18:22
    0
    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.
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