Engineering Board Exam Reviewer · Branch, Loop and Node Analyses · medium
In the two-loop circuit shown, V1=24 V, V2=15 V, R1=6, , R2=8, , and R3=3, . (The battery V2 has its positive terminal at the top.) Using mesh analysis with the clockwise mesh currents I1 (left loop) and I2 (right loop), find I1.
- 1.6 A
- 4.27 A
- 1.71 A
- -0.2 A
Answer
A. 1.6 A
Solution
Mesh 1: (R1+R2)I1-R2I2=V1 14, ,I1-8, ,I2=24 V. Mesh 2: -R2I1+(R2+R3)I2=-V2 -8, ,I1+11, ,I2=-15 V. Solving, I1=1.6 A and I2=-200 mA. (A negative sign would mean the current flows opposite to the assumed direction.)
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