Network Analysis Questions and Answers

Answer: A

A current source \(I\) with a resistance \(R\) in parallel is converted into a voltage source \(V = IR\) with the same resistance \(R\) in series (source transformation).

\[ V = 15 \times 2 = 30\ \text{V} \]

The series resistance stays at 2 ohms. The values 15 V and 7.5 V come from dividing instead of multiplying, and 5 V has no basis. The answer is 30 V, option A.

Answer: B

The power dissipated in a resistor is \(P = I^2 R\). The current of 20 mA is \(0.02\) A and the resistance is \(1000\ \Omega\).

\[ \begin{aligned} P = (0.02)^2 \times 1000 \\ &= 4 \times 10^{-4} \times 10^3 = 0.4\ \text{W} \end{aligned} \]

Check with \(V = IR = 20\) V, then \(P = VI = 20 \times 0.02 = 0.4\) W. The value 4 W would come from taking the current as 63 mA, and 0.04 W or 0.004 W from slips of a power of ten when converting milliamperes. The answer is 0.4 W, option B.

Answer: C

For two resistors in parallel the equivalent resistance is the product over the sum.

\[ \begin{aligned} R_{eq} = \frac{40 \times 60}{40 + 60} \\ &= \frac{2400}{100} = 24\ \Omega \end{aligned} \]

The result is smaller than the smaller resistor, 40 ohms, as it must be for a parallel combination, which rules out 50 ohms and 30 ohms only as being too close to the average. The value 12 ohms is smaller than needed. The answer is 24 ohms, option C.