Solution Topic 5

Solution for topic 5 τ = R th C where R th is the Thevenin equivalent at the capacitor terminals. R th = 120 || 80 + 12 = 60 Ω τ = 60 × 0.5 × 10 -3 = 30 ms Problem

τ = RTh C = 20 x 0.3 = 6s

(b) RTh = 20 //(5 + 25) + 8 = 20Ω, Example

The circuit in Fig. (a) is equivalent to the one shown in Fig. (b). 8Ω

io +

+ 12 Ω

6Ω

vx

vo

−

1/3 F

−

+ vc −

(a) R eq = 8 + 12 || 6 = 12 Ω τ = R eq C = (12)(1 / 3) = 4 s v c = v c (0) e - t τ = 30 e - t 4 = 30 e -0.25 t V vx =

4 v = 10 e -0.25t V 4+8 c

vx = vo + vc io =

 →

vo = - 2.5 e -0.25t A 8

v o = v x − v c = -20 e -0.25t V

+ Req

v − (b)

1/3 F

Example When t < 0, the switch is closed as shown in Fig. (a). 6Ω + + −

24 V

vc(0)

12 Ω

4Ω

− (a) R eq = 4 || 12 = 3 Ω

v c (0) =

3 (24) = 8 V 3+ 6

When t > 0, the switch is open as shown in Fig. (b). 6Ω

24 V

+ −

t=0

1/6 F

(b) τ = R eq C = (3)(1 / 6) = 1 / 2 s v( t ) = v c (0) e - t τ = 8 e -2 t V w c (0) =

1 2 1 1 Cv c (0) = × × 64 = 5.333J 2 2 6

Problem v( t ) = v(4) e -(t -4) τ τ = RC = (20)(0.1) = 2 where v(4) = 24 , -(t - 4) 2 v( t ) = 24 e v(10) = 24 e -6 2 = 1.195 V

3Ω

Tutorial 1.

τ = R eq C

v( t ) = v(0) e- t τ ,

R eq = 2 + 8 || 8 + 6 || 3 = 2 + 4 + 2 = 8 Ω τ = R eq C = (0.25)(8) = 2 v( t ) = 20 e -t 2 V 2.

3. (a)

Before t = 0, v = 12 V . After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v(∞) = 4 , v(0) = 12 , τ = RC = (2)(3) = 6 -t 6 v( t ) = 4 + (12 − 4) e v( t ) = 4 + 8 e - t 6 V

(b)

Before t = 0, v = 12 V . After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ After transforming the current source, the circuit is shown below. t=0 2Ω

12 V

v(0) = 12 , v = 12 V

+ −

v(∞) = 12 ,

4Ω

5F

τ = RC = (2)(5) = 10