The circuit below has two capacitors, three resistors, a battery and two switche

Question

The circuit below has two capacitors, three resistors, a battery and two switches. The switches, S1 and S2, have been open for

a long time and the capacitors are uncharged. At time t = 0 both switches are closed. 1 ?F = 10-6 F. a. Immediately after both switches are closed, what is the current I2?

b. After the switches have been closed for a very long time, what is the current I1?

c. After the switches have been closed for a very long time, what is the total energy, U, stored in the capacitors C1 and C2?

d. After the switches have been closed for a very long time, at time T both switches are opened. Explain what happens next.

Explanation / Answer

a) at t=0 both capacitors will act as short circuit.

therfore I2 = e/R2

I2 = 18/50 = 0.36 A (answer)

b)after very long time capacitors act as open circuit

therefore I1 = 0

c) after long time no current flows through capacitors , and potential difference across them is V= e=18 V

energy stored in C1 = E1 = (C1V^2/2) = 1.62*10^(-8) Joules (answer

energy stored in C2 = E2 = (C2V^2/2) =3.24*10^(-8) Joules (answer)

d) capacitor C2 starts discharging

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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