Network Theorems Questions and Answers

Answer: B

The key gives \(V_{TH}\) = 8.57 V.

The key takes the 24 Ω resistor as the load across A-B and removes it. The open-circuit voltage at A-B is then the voltage across the 12 Ω resistor, which forms a divider with the 2 Ω series resistor:

\[ \begin{aligned} V_{TH} &= 10 \times \frac{12}{2 + 12} \\ &= \frac{120}{14} \approx 8.57\ \text{V} \end{aligned} \]

If the 24 Ω resistor were left in place, the parallel combination (8 Ω) would give 8 V, which is not an option, so the question must be read with the 24 Ω resistor as the load. None of 7.92 V, 1.71 V or 8.35 V is the open-circuit voltage of this divider.

Answer: B

The Norton current source is connected in parallel with the load.

Norton's theorem replaces a linear network, as seen from two terminals, by a current source \(I_{N}\) (the short-circuit current at those terminals) in parallel with a resistance \(R_{N}\) (the resistance seen with sources set to zero). The load is then connected across those same two terminals, so it sits in parallel with both the source and \(R_{N}\), and the source current divides between them.

It is the Thevenin voltage source that appears in series with the load. 'Series with the source' and 'parallel with the source' describe the Norton resistance's relation to the source, not the source's connection to the load.

Answer: A

The key's answer is simplify parallel circuits.

Norton's theorem replaces any linear two-terminal network, however complex, by a single current source \(I_{N}\) in parallel with a single resistance \(R_{N}\). Because the equivalent is a parallel combination, the load can simply be added in parallel and the load current found with the current-divider rule, which makes parallel circuits and parallel loads easy to analyse.

Calculating the equivalent resistance is only one step in finding the Norton equivalent, not its purpose. Determining the open-circuit voltage is the corresponding step of Thevenin's theorem. The theorem applies to DC and AC linear circuits alike, so 'analyse AC circuits' does not describe its purpose.

Answer: A

The Norton current source is in parallel with a resistor.

The Norton equivalent of a linear network is a current source \(I_{N}\), equal to the short-circuit current at the terminals, in parallel with the Norton resistance \(R_{N}\), the resistance seen at the terminals with all independent sources turned off. \(R_{N}\) is the same as the Thevenin resistance. For AC networks it becomes an impedance, but in the standard statement it is a resistance.

A capacitor or an inductor alone is not the Norton element. A voltage source in series with a resistance is the Thevenin equivalent; putting a voltage source in parallel with the current source would not describe the network at all.

Answer: B

To solve for the load current by Thevenin's theorem the steps are taken in this order. First the load is removed, step (ii). Next the Thevenin voltage and Thevenin resistance of the remaining network are found, step (i). Then the equivalent circuit is formed by putting the Thevenin voltage and resistance in series, step (iii). Last the load is connected in series with it, and the current is found, step (iv). So the sequence is (ii), (i), (iii), (iv). The other sequences start with finding the Thevenin values before the load is removed, or connect the load before the equivalent circuit exists. The answer is (ii), (i), (iii), (iv).

Answer: D

Thevenin's theorem replaces a complicated linear network, as seen from the load terminals, by an ideal voltage source (the open-circuit voltage) in series with a single resistance (the resistance looking back into the network with the sources deactivated). That is option D. The parallel combination of an ideal current source and a resistor is the Norton equivalent, which is why options A and B, with a parallel resistor, and option C, with a current source, are wrong for Thevenin's theorem.