Engineering Board Exam Reviewer · Branch, Loop and Node Analyses · medium
In the two-loop circuit shown, V1=24 V, V2=6 V, R1=6, , R2=15, , 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 I2.
- -240 mA
- 390 mA
- 585 mA
- 780 mA
Answer
D. 780 mA
Solution
Mesh 1: (R1+R2)I1-R2I2=V1 21, ,I1-15, ,I2=24 V. Mesh 2: -R2I1+(R2+R3)I2=-V2 -15, ,I1+25, ,I2=-6 V. Solving, I1=1.7 A and I2=780 mA.
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