This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
This note was uploaded on 05/01/2010 for the course ECE 300 taught by Professor Watson during the Spring '09 term at CarsonNewman.
 Spring '09
 Watson
 Volt

Click to edit the document details