Bài giải phần giải mạch P7

Chapter 7, Solution 1. Applying KVL to Fig. 7.1. 1 t ∫ i dt + Ri = 0 C -∞ Taking the derivative of each term, i di +R =0 C dt di dt or =− i RCIntegrating,  i( t )  - t = ln  I 0  RCi( t ) = I 0 e - t RC v( t ) = Ri( t ) = RI 0 e - t RC or v(t ) = V0e- t RCChapter 7, Solution 2.τ = R th C where R th is the Thevenin equivalent at the capacitor terminals.R...
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Bài giải phần giải mạch P7Chapter 7, Solution 1. Applying KVL to Fig. 7.1. 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 RCChapter 7, Solution 2. 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 msChapter 7, Solution 3. (a) RTh = 10 // 10 = 5kΩ, τ = RTh C = 5 x10 3 x 2 x10 −6 = 10 ms (b) RTh = 20 //(5 + 25) + 8 = 20Ω, τ = RTh C = 20 x0.3 = 6sChapter 7, Solution 4. τ = R eq C eq C1C 2 R 1R 2 where C eq = , R eq = C1 + C 2 R1 + R 2 R 1 R 2 C1C 2 τ= ( R 1 + R 2 )(C 1 + C 2 )Chapter 7, Solution 5. v( t ) = v(4) e -(t -4) τ where v(4) = 24 , τ = RC = (20)(0.1) = 2 v( t ) = 24 e -(t - 4) 2 v(10) = 24 e -6 2 = 1.195 VChapter 7, Solution 6. 2 v o = v ( 0) = (24) = 4 V 10 + 2 2 v( t ) = voe − t / τ , τ = RC = 40 x10−6 x 2 x103 = 25 v( t ) = 4e −12.5t VChapter 7, Solution 7.v( t ) = v(0) e - t τ , τ = R th Cwhere R th is the Thevenin resistance across the capacitor. To determine R th , we insert a1-V voltage source in place of the capacitor as shown below. 8Ω i2 i i1 + + + 0.5 V − 10 Ω v=1 1V − − 1 1 − 0.5 1 i1 = = 0.1 , i2 = = 10 8 16 1 13 i = i1 + i 2 = 0.1 + = 16 80 1 80 R th = = i 13 80 8 τ = R th C = × 0.1 = 13 13 v( t ) = 20 e -13t 8 VChapter 7, Solution 8. 1 (a) τ = RC = 4 dv -i = C dt - 0.2 e = C (10)(-4) e-4t -4t  → C = 5 mF 1 R= = 50 Ω 4C 1 (b) τ = RC = = 0.25 s 4 1 1 (c) w C (0) = CV02 = (5 × 10 -3 )(100) = 250 mJ 2 2 1 1 1 (d) w R = × CV02 = CV02 (1 − e -2t 0 τ ) 2 2 2 1 0.5 = 1 − e -8t 0  → e -8t 0 = 2 or e =2 8t 0 1 t 0 = ln (2) = 86.6 ms 8Chapter 7, Solution 9. v( t ) = v(0) e- t τ , τ = R eq C 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 VChapter 7, Solution 10. io 15 Ω i 10 Ω iT + 10 mF v 4Ω − (10)(3) 15 i o = 10 i  → i o = =2A 15 i.e. if i(0) = 3 A , then i o (0) = 2 A i T (0) = ...