Engineering Board Exam Reviewer · Branch, Loop and Node Analyses · medium
In the two-loop circuit shown, V1=30 V, V2=6 V, R1=10, , R2=5, , and R3=2, . (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.
- 3 A
- 2 A
- 2.25 A
- 0.75 A
Answer
C. 2.25 A
Solution
Mesh 1: (R1+R2)I1-R2I2=V1 15, ,I1-5, ,I2=30 V. Mesh 2: -R2I1+(R2+R3)I2=-V2 -5, ,I1+7, ,I2=-6 V. Solving, I1=2.25 A and I2=750 mA. (A negative sign would mean the current flows opposite to the assumed direction.)
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