Discussion
Using superposition theorem, find the current through the 5 Ω resistor in the network shown below :
Answer: A
The current through the 5 Ω resistor is 3.875 A.
Take each source alone. With only the 16 V source (current source open), the 3 Ω and 5 Ω resistors are in series. With only the 5 A source (voltage source shorted), the current divides between the 5 Ω resistor and the 3 Ω resistor:
\[ \begin{aligned} I_{1} &= \frac{16}{3 + 5} = 2\ \text{A} \\ I_{2} &= 5 \times \frac{3}{3 + 5} = 1.875\ \text{A} \\ I &= I_{1} + I_{2} = 3.875\ \text{A} \end{aligned} \]
Both contributions flow down through the 5 Ω resistor, so they add. 2.00 A is the voltage-source part alone, and the other values come from wrong current-divider ratios.
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