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Unformatted text preview: Solution of ECE 300 Test 6 F09 1. Find the numerical energy stored in the capacitor and in the inductor. V 1 = 24V , R 2 = 18 Ω , C 3 = 1 μ F , L 4 = 100mH The source voltage is constant. Therefore all currents and voltages are constant. Therefore the capacitor is an open circuit and the inductor is a short circuit. Then the current through the resistor and inductor is 24V/18 Ω = 1.333A . The inductor stored energy is E L = 1/ 2 ( ) 100mH ( ) 1.333A ( ) 2 = 88.89mJ . The voltage across the capacitor is zero. Therefore its stored energy is zero. 2. Find the numerical values of the inductor current and capacitor voltage. I 1 = 2A , R 2 = 50 Ω , L 3 = 10mH , R 4 = 20 Ω , C 5 = 10 μ F , V 6 = 50V The source voltage and source current are constant. Therefore all currents and voltages are constant. Therefore the capacitor is an open circuit and the inductor is a short circuit. Using superposition, the current through the inductor is I L = I 1 + = I 1 = 2A . The voltage across the capacitor is...
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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.
- Spring '09