Chapter 7, Solution 1. Applying KVL to Fig. . 1 t ∫ i dt + Ri = 0 C -∞ Taking the derivative of each term, i di +R =0 C dt di dt or =− i RC Integrating, i( t ) - t = ln I 0 RC i( t ) = I 0 e - t RC v( t ) = Ri( t ) = RI 0 e - t RC or v(t ) = V0e- t RC Chapter 7, Solution 2. τ = R th C where R th is the Thevenin equivalent at the capacitor terminals. R. | Chapter 7 Solution 1. Applying KVL to Fig. . 1 F C Lidt Ri 0 Taking the derivative of each term i di R 0 C dt or di dt i RC Integrating i li t I Inl----I I Io J -t RC i t Ioe-t RC v t Ri t RI0e-t RC or v t V0e IR Chapter 7 Solution 2. t RthC where Rth is the Thevenin equivalent at the capacitor terminals. Rth 120 80 12 60 Q t 60 x x 10-3 30 ms Chapter 7 Solution 3. a RTh 10 10 5kQ t RnC 5x103 x2x10-6 10 ms b RTh 20 5 25 8 20Q t R C 6s In --- Chapter 7 Solution 4. T ReqCeq C1C2 where Ceq T R R R1R2C1C2 R1 R2 C1 C2 Chapter 7 Solution 5. v t v 4 e- t-4 T where v 4 24 t RC 20 2 v t 24e- t-4 2 v 10 24e-62 V Chapter 7 Solution 6. Vo v 0 24 4 V 0 10 2 v t voe-t T t RC 40x10-6x2x103 0 25 v t 4e 12-5tV Chapter 7 Solution 7. v t v 0 e T t R C where Rth is the Thevenin resistance across the capacitor. To determine Rth we insert a 1-V voltage source in place of the capacitor as shown below. 1 1 - 1 i L ------------- 1 10 2 8 16 1 13 i i1 i2 0 16 80 1 80 Rth i 13 80 8 t RthC - x 0T - v t 20 e 43t 8 V Chapter 7 Solution 8. a 1 t RC - 4 dv -i C dt C 10 -4 e-4t --- C 5 mF b c d R 50 Q 4C --------- 1 t RC - s 4 -------- wC 0 cV 2 5 X 10-3 100 250 mJ w 1X 1CV2 CV2 1-e-2t0 T wr 2 X 2cV0 2 cV0 1 e 1 - e-8t0 ---- e-8t0 1 2 or e8t0 2 1 t0 In 2 ms 8 --------- Chapter 7 Solution 9. v t v 0 e- t R C Req 2 8 8 6 3 2 4 2 8 Q t ReqC 8 2 v t 20e 1 2
