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

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