BME211HW8 - Homework #8 Solutions P8.3-1 Here is the...

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Homework #8 Solutions P8.3-1 Here is the circuit before t = 0, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the initial capacitor voltage, v (0). By voltage division ( 29 ( 29 6 0 12 4 V 6 6 6 v = = + + Next, consider the circuit after the switch closes. The closed switch is modeled as a short circuit. We need to find the Thevenin equivalent of the part of the circuit connected to the capacitor. Here’s the circuit used to calculate the open circuit voltage, V oc . ( 29 oc 6 12 6 V 6 6 V = = + Here is the circuit that is used to determine R t . A short circuit has replaced the closed switch. Independent sources are set to zero when calculating R t , so the voltage source has been replaced by a short circuit. ( 29 ( 29 t 6 6 3 6 6 R = = + Then ( 29 t 3 0.25 0.75 s R C τ = = = Finally, ( 29 ( 29 ( 29 oc oc 1.33 / 0 6 2 V for > 0 t t v t V v V e e t - - = + - = -
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P8.3-6 Before the switch opens, ( 29 ( 29 o o 3 5 0.25 mA 0 0.25 mA 20 10 i t i = = = P . After the switch opens the part of the circuit connected to the inductor can be replaced by it's Norton equivalent circuit to get: Therefore 3 5 0.25 ms 20 10 τ = = × . Next, 4000 L L ( ) ( (0) ) 0.5 0.25 mA for 0 t t sc sc i t i i i e e t - - = + - = - Finally , ( 29 ( 29 4000 5 5 V for 0 t o L d v t i t e t dt - = = P8.3-7 At = 0 (steady-state) t - Since the input to this circuit is constant, the inductor will act like a short circuit when the circuit is at steady-state: for t > 0 ( 29 ( 29 ( ) 20 0 6 A R L t t L L i t i e e - - = =
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P8.3-8 Before the switch opens, the circuit will be at steady state. Because the only input to this circuit is the constant voltage of the voltage source, all of the element currents and
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BME211HW8 - Homework #8 Solutions P8.3-1 Here is the...

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