Engineering Board Exam Reviewer · Branch, Loop and Node Analyses · medium

In the two-loop circuit shown, V1=18 V, V2=20 V, R1=4, , R2=12, , 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 I2.

  1. -1.3 A
  2. 6.7 A
  3. 0.15 A
  4. -1.43 A

Answer

A. -1.3 A

Solution

Mesh 1: (R1+R2)I1-R2I2=V1 16, ,I1-12, ,I2=18 V. Mesh 2: -R2I1+(R2+R3)I2=-V2 -12, ,I1+14, ,I2=-20 V. Solving, I1=150 mA and I2=-1.3 A.

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