Engineering Board Exam Reviewer · Branch, Loop and Node Analyses · hard
In the two-loop circuit shown, V1=9 V, V2=18 V, R1=15, , R2=10, , and R3=3, . (The battery V2 has its positive terminal at the top.) Find the current through R2 (taken as positive when it flows downward).
- -1.88 A
- -0.28 A
- 1.32 A
- -1.6 A
Answer
C. 1.32 A
Solution
Mesh 1: (R1+R2)I1-R2I2=V1 25, ,I1-10, ,I2=9 V. Mesh 2: -R2I1+(R2+R3)I2=-V2 -10, ,I1+13, ,I2=-18 V. I1=-280 mA and I2=-1.6 A. In R2, I1 flows downward and I2 flows upward, so IR2=I1-I2=1.32 A (downward if positive).
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