Mesh and Nodal Analysis Questions and Answers

Answer: A

The 10 ohm resistor is directly across the 30 V source. The other branch is the 11 ohm resistor in series with the 4 ohm resistor, which is 15 ohms, also directly across the source.

\[ \begin{aligned} I_{10} = \frac{30}{10} \\ &= 3\ \text{A},\qquad I_{15} = \frac{30}{15} \\ &= 2\ \text{A} \end{aligned} \]

\[ I_{source} = 3 + 2 = 5\ \text{A} \]

Equivalently the parallel combination of 10 and 15 ohms is 6 ohms and 30/6 = 5 A. The values 6 A, 10 A and 12 A come from ignoring the series resistor or adding the wrong branches. The answer is 5 A, option A.

Answer: A

Nodal analysis is based on Kirchhoff's current law.

In nodal analysis one node is chosen as reference and the voltages of the other nodes are the unknowns. At each node, the currents in the connected branches, written in terms of the node voltages using Ohm's law, are made to sum to zero, which is KCL. The resulting equations are solved for the node voltages.

Mesh (loop) analysis is the method built on Kirchhoff's voltage law. Ohm's law is used within both methods but is not their basis, and Maxwell's name is attached to loop currents, not to nodal analysis.

Answer: D

The voltage needed is 80 V.

With c-d open, the 30 Ω and 10 Ω resistors form one path from a to b (a to c through 30 Ω, then c to b through 10 Ω), and the 5 Ω resistor forms a second path from a to d, which is joined directly to b. The two paths are in parallel across a-b.

For 20 V across the 10 Ω resistor, the current in the 30 Ω and 10 Ω path is

\[ \begin{aligned} I &= \frac{20}{10} = 2\ \text{A} \\ V_{ab} &= 2 \times (30 + 10) = 80\ \text{V} \end{aligned} \]

The 5 Ω path draws its own current but does not change the voltage across the other path. 20 V is the drop across the 10 Ω resistor alone, and 60 V is the drop across the 30 Ω resistor alone.