chap08

chap08 - Chapter 8 Solution 1(a At t = 0 the circuit has...

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Chapter 8, Solution 1. (a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a). + v L 6 10 H + v (a) 6 + 6 V S 10 µ F (b) i(0-) = 12/6 = 2A, v(0-) = 12V At t = 0+, i(0+) = i(0-) = 2A , v(0+) = v(0-) = 12V (b) For t > 0, we have the equivalent circuit shown in Figure (b). v L = Ldi/dt or di/dt = v L /L Applying KVL at t = 0+, we obtain, v L (0+) – v(0+) + 10i(0+) = 0 v L (0+) – 12 + 20 = 0, or v L (0+) = -8 Hence, di(0+)/dt = -8/2 = -4 A/s Similarly, i C = Cdv/dt, or dv/dt = i C /C i C (0+) = -i(0+) = -2 dv(0+)/dt = -2/0.4 = -5 V/s (c) As t approaches infinity, the circuit reaches steady state. i( ) = 0 A , v( ) = 0 V
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Chapter 8, Solution 2. (a) At t = 0-, the equivalent circuit is shown in Figure (a). 25 k 20 k i R + + v i L 60 k 80V (a) 25 k 20 k i R + i L 80V (b) 60||20 = 15 kohms, i R (0-) = 80/(25 + 15) = 2mA. By the current division principle, i L (0-) = 60(2mA)/(60 + 20) = 1.5 mA v C (0-) = 0 At t = 0+, v C (0+) = v C (0-) = 0 i L (0+) = i L (0-) = 1.5 mA 80 = i R (0+)(25 + 20) + v C (0-) i R (0+) = 80/45k = 1.778 mA But, i R = i C + i L 1.778 = i C (0+) + 1.5 or i C (0+) = 0.278 mA
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(b) v L (0+) = v C (0+) = 0 But, v L = Ldi L /dt and di L (0+)/dt = v L (0+)/L = 0 di L (0+)/dt = 0 Again, 80 = 45i R + v C 0 = 45di R /dt + dv C /dt But, dv C (0+)/dt = i C (0+)/C = 0.278 mohms/1 µ F = 278 V/s Hence, di R (0+)/dt = (-1/45)dv C (0+)/dt = -278/45 di R (0+)/dt = -6.1778 A/s Also, i R = i C + i L di R (0+)/dt = di C (0+)/dt + di L (0+)/dt -6.1788 = di C (0+)/dt + 0, or di C (0+)/dt = -6.1788 A/s (c) As t approaches infinity, we have the equivalent circuit in Figure (b). i R ( ) = i L ( ) = 80/45k = 1.778 mA i C ( ) = Cdv( )/dt = 0 . Chapter 8, Solution 3. At t = 0 - , u(t) = 0. Consider the circuit shown in Figure (a). i L (0 - ) = 0, and v R (0 - ) = 0. But, -v R (0 - ) + v C (0 - ) + 10 = 0, or v C (0 - ) = -10V. (a) At t = 0 + , since the inductor current and capacitor voltage cannot change abruptly, the inductor current must still be equal to 0A , the capacitor has a voltage equal to –10V . Since it is in series with the +10V source, together they represent a direct short at t = 0 + . This means that the entire 2A from the current source flows through the capacitor and not the resistor. Therefore, v R (0 + ) = 0 V . (b) At t = 0 + , v L (0+) = 0, therefore Ldi L (0+)/dt = v L (0 + ) = 0, thus, di L /dt = 0A/s , i C (0 + ) = 2 A, this means that dv C (0 + )/dt = 2/C = 8 V/s . Now for the value of dv R (0 + )/dt. Since v R = v C + 10, then dv R (0 + )/dt = dv C (0 + )/dt + 0 = 8 V/s .
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40 40 + 10V + v C 10 2A i L + v R + v R + 10V + v C 10 (b) (a) (c) As t approaches infinity, we end up with the equivalent circuit shown in Figure (b). i L ( ) = 10(2)/(40 + 10) = 400 mA v C ( ) = 2[10||40] –10 = 16 – 10 = 6V v R ( ) = 2[10||40] = 16 V Chapter 8, Solution 4. (a) At t = 0 - , u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in Figure (a). i(0 - ) = 40/(3 + 5) = 5A, and v(0 - ) = 5i(0 - ) = 25V. Hence, i(0 + ) = i(0 - ) = 5A v(0 + ) = v(0 - ) = 25V 3 5 i + v + 40V (a)
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0.25 H 3 i R i C + + v L i 5 0.1F 4 A 40V (b) (b) i C = Cdv/dt or dv(0 + )/dt = i C (0 + )/C For t = 0 + , 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b). Since i and v cannot change abruptly, i R = v/5 = 25/5 = 5A, i(0 + ) + 4 =i C (0 + ) + i R (0 + ) 5 + 4 = i C (0 + ) + 5 which leads to i C (0 + ) = 4 dv(0 + )/dt = 4/0.1 = 40 V/s Chapter 8, Solution 5.
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