240_quiz1a

# 240_quiz1a - Thus writing KVL around the outer loop 4 6 –...

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Name __________________________ ECE 240 – Quiz #1 January 28, 2010 For the circuit shown above, find: a. i b. v c. the power delivered by the box “A” d. The power delivered by the 6V source e. The power delivered to the 6A source a. KCL at the node to the right of Box A: i+6+4 = 0 i = -10 b. the voltage across the right 1 Ω resistor is 4(1) = 4V with the positive on the bottom.
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Unformatted text preview: Thus writing KVL around the outer loop: 4 + 6 – v = 0 v = 10 c. –iv = -(-10)(10) = 100W d. the left 3 Ω resistor has 2A going down, thus the current going down in the 6V source is 10 – 2 = 8A power delivered by the 6V is -8*6 = -48W e. 4(6) = 24W (the current enter the plus side of the voltage) +-!" # ! !\$ % ! &\$ ’ ( * + \$...
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## This note was uploaded on 02/27/2010 for the course ECE 240 taught by Professor Drsharlenekatz during the Spring '10 term at CSU Northridge.

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