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Unformatted text preview: MAE140 Linear Circuits Fall 2006 Final exam solutions. Tuesday December 5. Question 1: LRC Circuit Analysis Part (i): This is the circuit transformed into the s-domain with initial condition current sources, as explained in the text p. 450. Part (ii): KCL (summing currents leaving) at Node B yields sC ( V o- V A )- Cv C (0) = , V A ( s ) = V o ( s )- 1 s v C (0) . Now using KCL at Node A , we have 1 R ( V A- V i ) + 1 sL V A + sC ( V A- V o ) + Cv C (0) + i L (0) s = , parenleftbigg 1 R + 1 sL + sC parenrightbigg V A ( s )- 1 R V i ( s )- sCV o ( s ) + Cv C (0) + i L (0) s = . bracketleftbigg 1 R + 1 sL + sC bracketrightbiggbracketleftbigg V ( s )- 1 s v C (0) bracketrightbigg- sCV o ( s ) = 1 R V i ( s )- Cv C (0)- i L (0) s , bracketleftbigg 1 R + 1 sL bracketrightbigg V o ( s )- bracketleftbigg 1 sR + 1 s 2 L bracketrightbigg v c (0) = 1 R V i ( s )- i L (0) s , V ( s ) = 1 /R 1 /R + 1 /sL V i ( S ) + 1 s v C (0)- i L (0) s [1 /R + 1 /sL ] . = s s + R/L V i ( s ) + 1 s v C (0)- R s + R/L i L (0) . Part (iii): Steady-state is determined by setting the inductor voltage to zero and the capacitor current to zero. If the inductor voltage is zero, then the current through the resistor R is i R = V S R = i L , because the capacitor current is zero. The capacitor current is zero and the A terminal voltage is zero. Since the output port is open-circuit, the capacitor can theoretically hold any voltage without losing it. In practice, it would leak away to zero due to an internal resistance in the capacitor. Note that this formulawould leak away to zero due to an internal resistance in the capacitor....
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- Spring '08