An emf source with emf of 200 V, a resistor with R =50.0 Ω, and a capacitor with C = 4.0 mF are connected in series. As the capacitor charge

Question

An emf source with emf of 200 V, a resistor with R =50.0 Ω, and a capacitor with C = 4.0 mF are connected in series. As the capacitor charges, when the current in the resistor is 1.00 A, what is the magnitude of the charge on each plate of the capacitor?

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Farah 8 hours 2021-07-22T04:30:32+00:00 1 Answers 0 views 0

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    2021-07-22T04:32:30+00:00

    Answer:

    0.6 C

    Explanation:

    From the question,

    Since the Resistor and the capacitor are connected in series.

    The voltage across the resistor + Voltage across the capacitor = Total voltage.

    Vt = Vr+Vc…………… Equation 1

    Where Vt = Total voltage, Vr = Voltage across the resistor, Vc = Voltage across the capacitor

    make Vc the subject of the equation

    Vc = Vt-Vr………….. Equation 2

    From ohm’s law,

    Vr = IR………………. equation 3

    Where I = current, R = Resistance.

    Given: I = 1 A, R = 50 Ω

    Substitute into equation 3

    Vr = 1(50)

    Vr = 50 V

    Also given: Vt = 200 V

    Substitute into equation 2

    Vc = 200-50

    Vc = 150 V.

    Applying,

    Q = CVc………………… Equation 4

    Where Q = charge on each plate of the capacitor, C = Capacitance of the capacitor

    Given: C = 4.0 mF = 4.0×10⁻³ F, Vc = 150 V

    Substitute into equation 4

    Q = 4.0×10⁻³(150)

    Q = 0.6 C

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