Engineering Board Exam Reviewer · Branch, Loop and Node Analyses · medium
In the two-loop circuit shown, V1=30 V, V2=24 V, R1=15, , R2=2, , and R3=10, . (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.
- 2.04 A
- 1.56 A
- 1.76 A
- -1.74 A
Answer
B. 1.56 A
Solution
Mesh 1: (R1+R2)I1-R2I2=V1 17, ,I1-2, ,I2=30 V. Mesh 2: -R2I1+(R2+R3)I2=-V2 -2, ,I1+12, ,I2=-24 V. Solving, I1=1.56 A and I2=-1.74 A. (A negative sign would mean the current flows opposite to the assumed direction.)
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