ECE
lect8.pdf

# Kvl 1 10 5i 1 3i 1 i 2 2i 1 i 3 0 kvl sm 5 4i 3 2i 3

• 14

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KVL-1: 10 – 5i 1 – 3(i 1 - i 2 ) – 2(i 1 -i 3 ) = 0 KVL-SM: –5 – 4i 3 – 2(i 3 -i 1 ) – 3(i 2 -i 1 ) – 1i 2 = 0 I source: i 3 – i 2 = 3A 3 equations and 3 unknowns -- solve for i 1 , i 2 , i 3 Once the currents are known, the circuit behavior is known. _ + 5 W 1 W 3 W 2 W 4 W _ + 5V 3A 10V i 1 i 2 i 3 Supermesh

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EE201 Lecture 8 P. 13 Solution : Use KVL to obtain equations. KVL-1: 10 – 5i 1 – 3(i 1 +i 2 ) – 2(i 1 -i 3 ) = 0 KVL-SM: 5 – 1i 2 – 3(i 2 +i 1 ) – 2(i 1 -i 3 ) – 4(-i 3 ) = 0 I source: i 3 + i 2 = 3A Exact problem as on P. 12, but notice difference in grouping of terms. _ + 5 W 1 W 3 W 2 W 4 W _ + 5V 3A 10V i 1 i 2 i 3 Supermesh
EE201 Lecture 8 P. 14 Exercise : Write equations for mesh currents above. KVL-1: 10 – 5i 1 – 3(i 1 +i 2 ) – 2(i 1 +i 3 ) = 0 KVL-SM: 5 – 1i 2 – 3(i 2 +i 1 ) – 2(i 1 +i 3 ) – 4(i 3 ) = 0 I source: i 2 – i 3 = 3A _ + 5 W 1 W 3 W 2 W 4 W _ + 5V 3A 10V i 1 i 2 i 3 Supermesh
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• Fall '08
• ALL
• Mesh Analysis, Voltage drop, EE201

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