Engineering Board Exam Reviewer · Circuit Theorems and Conversions · medium
In the circuit shown, V1=24 V, V2=6 V, R1=15, , R2=10, , and R3=3, . Use superposition to find the current through R2 (taken positive downward) due to V2 acting alone (V1 replaced by a short circuit).
- 461.54 mA
- 400 mA
- 666.67 mA
- 600 mA
Answer
B. 400 mA
Solution
With V1 shorted, R1 R2=6, , so the current from V2 is V2/R3+R1 R2= 6 V9, =666.67 mA. It divides between R1 and R2: |IR2|=(666.67 mA)15, /15, +10, =400 mA (magnitude).
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