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HW2S

# HW2S - EE202L Homework#2 Solutions 7.13 10 K 12 V t=0 A 20...

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EE202L Homework #2 Solutions 7.13 Find Vc(t) for t 0. Identify the forced & natural response. Vc(0 - ) = V 3 12 ) 10 20 10 ( 10 = ! + + Vc(0 + ) = V(0 - ) = 3 V Vc( ) = V 6 12 10 10 10 = ! + R th =10k || 10k = 5k τ = R th · C = 5k (1 × 10 -6 ) = 5 × 10 -3 Vc(t) = Vc( ) + [Vc(0) – Vc( )]· ! t e " = 6 - 3· 200t e ! 7.18 Find i L (t) for t > 0 and sketch the waveform i L (0 - ) = 5/5k = 1mA i L (0 + ) = i L (0 - ) = 1mA + 12 V - A B t = 0 20 K Ω 10 K Ω 10 K Ω 1 μ F + Vc(t) - + 12 V - 20 K Ω 10 K Ω 10 K Ω + Vc(t) - t < 0 + 12 V - 10 K Ω 10 K Ω + Vc(t) - t = Forced response Natural response + 5 V - t = 0 100 mH 1 K Ω i L (t) 5 K Ω 5 V i L (t) 5 K Ω t < 0 + -

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i L ( ) = 5/1k = 5mA R th = 1 k Ω τ = L/R th = 100 mH/1 k Ω = 10 -4 = 0.0001 S i L (t) = i L ( ) + [i L (0)-i L ( )] e -t/ τ = 5 + (1-5) e -10000t = 5 – 4e -10000t mA 7.21 Find Vc(t) for t 2ms t < 0. Vc(0 - ) = 0 V = Vc(0 + ) If we assume that the switch is closed at t = 0, and leave it there for a long time. t Vc( ) = V 18 225 15 3 15 15 = ! + R th = 3 k || 15 k = 2.5 k Ω τ = Rth · C = 2.5 × 10 -3 · (10 -6 ) = 0.0025 S Vc(t) = Vc( ) + (Vc(0) – Vc( )) · e -t/ τ = 225/18 + (0-225/18) · e -t/0.0025 Thus, Vc(2ms - ) = (225/18) (1-e -36/45 ) = 6.9 V Vc(2ms + ) = 6.9 V = Vc(2ms - ) After the switch goes back to open at t = 2ms.
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