A 2μF capacitor holds a charge of 24μC. The capacitor is discharged through a 3Ω resistor. What is the voltage across the capacitor at the t

Question

A 2μF capacitor holds a charge of 24μC. The capacitor is discharged through a 3Ω resistor. What is the voltage across the capacitor at the time that the current through the resistor is 3A?

a. VC = 9V
b. VC = 8V
c. VC = 12V
d. VC = 3V

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Trung Dũng 7 months 2021-07-29T20:26:43+00:00 2 Answers 3 views 0

Answers ( )

  1. thienthi
    0
    2021-07-29T20:27:50+00:00
    Reply

    Answer:

    The voltage drop across the capacitor is 3V

    (d) VC = 3V

    Explanation:

    Given;

    capacitance of the capacitor, C =  2μF

    charge of the capacitor, Q =  24μC

    resistance of the resistor, R = 3Ω

    current through the resistor, I = 3 A

    Initial voltage across the capacitor, V₀:

    VC₀ = Q/C = 24μ / 2μ = 12 V

    voltage across the resistor: V_R

    V_R = IR = 3 x 3 = 9 V

    Voltage drop across the capacitor at the time that the current through the resistor is 3A:

    VC = VC₀  – V_R

    VC = 12V – 9V

    VC = 3V

    Therefore, the voltage drop across the capacitor is 3V

    (d) VC = 3V

  2. Orla
    0
    2021-07-29T20:28:05+00:00
    Reply

    Answer:

    d. VC = 3V

    Explanation:

    If a 2μF capacitor holds a charge of 24μC, the voltage across the capacitor initially Vc is expressed as;

    Vc = Q/C {since Q = CV}

    Vc = 24μC/2μF

    Vc = 12Volts

    The voltage across the resistor is derived according to ohm’s law;

    VR = IR

    Given R = 3Ω and I = 3A

    VR = 3×3

    VR = 9V

    If the capacitor is discharged through the resistor, the voltage across the capacitor at the time that the current through the resistor is 3A is expressed as;

    VC = Vc – VR

    Vc is the voltage across the capacitor during discharging.

    VC = 12-9

    VC = 3V

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