An RC circuit, hooked up to a battery as shown in the figure, starts with an unc

Question

An RC circuit, hooked up to a battery as shown in the figure, starts with an uncharged capacitor. The resistance in the circuit is R = 680.0 ? the capacitor has capacitance of C = 16.0 ?F and the battery maintains the emf of ? = 19.0 V. The switch is closed at time t = 0.0 s and the capacitor begins to charge.

A. What is the time constant for this circuit?

B. What is the charge on the capacitor after the switch has been closed for t = 4.68×10-3 s?

C. What is the current through the circuit after the switch has been closed for t = 4.68×10-3 s?

D. What is the voltage across the capacitor after the switch has been closed for t = 4.68×10-3 s?

Explanation / Answer

A) time constant, T = RC

T = 680 x 16 x 10^-6 = 0.0109 s

B) PD across capacitor, Vc = E [ 1 - e^(-t/T) ]

Vc = 19 [ 1 -e^-(0.00468 / 0.0109) ]

Vc = 6.64 Volt

Q = CV = 16uF x 6.64 =106.27 uC

C) I = (e - Vc) / R = (19 - 6.64) / 680 = 0.018 A

D) Vc = 6.64 volt

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.