Test8Solution

Test8Solution - Solution of ECE 300 Test 8 F09 1 In the space provided graph this voltage versus time Put a scale on the vertical axis so that

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Solution of ECE 300 Test 8 F09 1. In the space provided graph this voltage versus time. Put a scale on the vertical axis so that actual voltages could be read from the graph. v t ( ) = 4u t ( ) 6u 2 t ( ) + 5u t 4 ( ) 2. Find the numerical values of the capacitor current and voltage and the switch current at the times indicated. v C 0 ( ) = ____________V v C 0 + ( ) = ____________V v C ( ) = ____________V i C 0 ( ) = ____________mA i C 0 + ( ) = ____________mA i C ( ) = ____________mA i s 0 ( ) = ____________mA i s 0 + ( ) = ____________mA i s ( ) = ____________mA What is the numerical time constant of this circuit after the switch is closed. τ = ____________ s I 1 = 2mA , R 2 = 5k Ω , R 3 = 10k Ω , V 4 = 15V , C 5 = 4 μ F The circuit has been in the configuration with the switch open for a very long time and all currents and voltages are constant. For the current source only: The current source current must all flow through resistor R 2 developing a voltage of 10V (positive on top). Then the capacitor voltage would also be 10V. The switch current is zero. For the voltage source only: No current flows anywhere in the circuit so the capacitor voltage must be -15V. Overall: The capacitor voltage is -5V and the capacitor current is zero. At time t = 0 + : 10 10 -10 -10 v( t ) t I 1 R 2 t = 0 C 5 V 4 R 3 v C i C i s
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The capacitor voltage remains at -5V. All the current source current now flows through the switch. The voltage across R 3 is 10V (positive on the left). Therefore the current through the capacitor, voltage source and R 3 is 1mA (counter-
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This note was uploaded on 05/01/2010 for the course ECE 300 taught by Professor Watson during the Spring '09 term at Carson-Newman.

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Test8Solution - Solution of ECE 300 Test 8 F09 1 In the space provided graph this voltage versus time Put a scale on the vertical axis so that

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