At t = 0, the switch is flipped to position 2 which connects V s to the RC circu

Question

At t = 0, the switch is flipped to position 2 which connects Vs to the RC circuit. The capacitor charges up, and its voltage increases exponentially towards Vs, with a time constant i (= RC).

After one millisecond, the switch is put back into position 1. The capacitor now discharges, and its voltage decreases exponentially towards 0V.

The switch is toggled at this rate between the two pole positions.

For the charging interval, at t=i (assuming at t=0, vC(0) = 0V), vC(i) = 0.632*Vs = 3.16V (for Vs = 5V).

For the discharging interval, at t=i (assuming at t=0, vC(0) = Vs), vC(i) = 0.368*Vs = 1.84V (for Vs = 5V).

For Figure 1 find its time constant and complete the table below

Explanation / Answer

the time constant = RC = 10000 (15 x 10^-9) = 1.5 x 10^-4 s

at t = 0 - :

since no current flows => i(0-) = 0 and Vc(0-) = VR = 0 V

and dVc/dt (0-) = 0

during charging Vc(t) is given by :

Vc(t) = Vs (1 - e^-(t/) ) = 5 (1 - e^-( t /0.15ms) )

so at t = 0 u get :

Vc(0+) = 0 V

iC(t) = C dVc/dt = 15x10^-9 x 5 (1/0.15ms) e^-( t /0.15ms)

=> iC(t) = 5x10^-4 e^-( t /0.15ms) A

so iC(0+) = 5x10^-4 A

hence dVc/dt (0+) = iC(0+) / C = 5x10^-4 / (15x10^-9) = 3.33 x 10^4 V/s

and VR(0+) = iC(0+) x R = 5x10^-4 x 10^4 = 5 V

at t = 0.3 ms u get :

iC(0.3ms) = 5x10^-4 e^-( 0.3ms /0.15ms)

=> iC(0.3ms) = 6.77 x 10^-5 A

and Vc(0.3ms) = 5(1 - e^-2) = 4.32 V

so VR(0.3ms) = 5 - 4.32 = 0.68 V

and dVc/dt (0.3ms) = iC(0.3ms) / C

=  6.77 x 10^-5 / (15 x 10^-9)

= 4.51 x 10^3 V/s

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