FE Electrical Circuit Analysis Practice Problems

Answer: A) 3 V

By Kirchhoff's voltage law on p. 363 of the Handbook the rises around the loop equal the drops, so the three resistor voltages add up to the 9 V of the battery. The same current flows through all three, the defining property of a series connection on p. 363, and equal resistances carrying equal current have equal drops by Ohm's law:

$$3V_R = 9 \quad\Rightarrow\quad V_R = 3\text{ V}$$

The result does not depend on the resistance value: 3 Ω or 3 kΩ, each takes one third. This is the voltage divider in its simplest form, \(V_R = V_s\,R/(R_S)\) with \(R_S = 3R\), and the general rule is that series elements share the source voltage in proportion to their resistances. The current is \(9/(3R)\) and the power dissipated in each resistor is \(V_R^2/R = 9/R\) watts; with unequal resistors the drops would differ but would still sum to 9 V, which is the check to apply to any series calculation.

9 V puts the full battery voltage across each resistor, which is the parallel connection, not the series one.

4.5 V divides by two instead of three.

27 V multiplies the source by the number of resistors, creating voltage that no source supplies.

Answer: A) 12 V

Nothing is connected across A and B, so no current leaves the terminals and the two resistors carry the same current. The source therefore divides between them in proportion to their resistances:

$$V_{Th} = V_{oc} = 30 \times \dfrac{4}{6+4} = 12\text{ V}$$

A quick way to see whether the answer is even plausible: the 4 \(\Omega\) is the smaller of the two resistors, so it must take the smaller share, which means less than half of 30 V. That single observation eliminates every other option here.

18 V is the drop across the 6 \(\Omega\) resistor. It is the other half of the same divider, and it is what you get by putting the wrong resistance in the numerator. The two add to 30 V, as they must.

20 V divides by the wrong total, \(30 \times 4/6\), using one resistance instead of the sum of both. A divider fraction can never exceed one, so the denominator has to be the whole series resistance.

30 V is the source itself, which would appear at the terminals only if the 6 \(\Omega\) resistor were absent.

Answer: C) 6.25 A

Real and reactive power combine in quadrature into the apparent power, \(|S| = \sqrt{P^2 + Q^2}\) from the Complex Power triangle on p. 368 of the Handbook, and \(|S| = V_{rms}I_{rms}\):

$$|S| = \sqrt{1200^2 + 900^2} = \sqrt{2.25 \times 10^6} = 1500\text{ VA}$$

$$I_{rms} = \dfrac{|S|}{V_{rms}} = \dfrac{1500}{240} = 6.25\text{ A}$$

The power factor is \(P/|S| = 1200/1500 = 0.8\), and the current lags or leads by \(\arccos 0.8 = 36.87^\circ\) according to the sign of \(Q\); positive var by the Handbook's convention means an inductive load and a lagging current. The same 6.25 A follows from \(I = P/(V\cos\theta) = 1200/(240 \times 0.8)\). The load's equivalent impedance is \(240/6.25 = 38.4\) Ω at 36.87°, or \(30.7 + j23.0\) Ω. Only 5 A of the 6.25 A do work; the remaining component, 3.75 A in quadrature, circulates reactive energy, and a capacitor supplying 900 var would remove it and reduce the supply current to 5 A for the same 1200 W. The wiring and the transformer must nevertheless be rated for the full 6.25 A.

5 A is \(P/V\), the in-phase component of the current only.

3.75 A is \(Q/V\), the quadrature component only.

8.75 A adds watts and vars arithmetically, \(2100/240\), instead of in quadrature.

Answer: C) 35.7 V

Kirchhoff's current law (p. 363 of the Handbook) at each node, currents leaving through the resistors:

$$\text{node 1:}\quad \dfrac{V_1}{10} + \dfrac{V_1 - V_2}{5} = 5$$

$$\text{node 2:}\quad \dfrac{V_2 - V_1}{5} + \dfrac{V_2}{20} = 0 \quad\Rightarrow\quad V_2 = 0.8V_1$$

Substituting into the node-1 equation, \(0.1V_1 + 0.2(V_1 - 0.8V_1) = 0.1V_1 + 0.04V_1 = 0.14V_1 = 5\), so

$$V_1 = \dfrac{5}{0.14} = 35.7\text{ V}, \qquad V_2 = 28.6\text{ V}$$

The reduction check: the 5 Ω and 20 Ω resistors are in series from node 1 to ground, 25 Ω, in parallel with the 10 Ω, \(10 \times 25/35 = 7.14\) Ω, and \(5 \times 7.14 = 35.7\) V. The source current splits into 3.57 A through the 10 Ω and 1.43 A through the 25 Ω path, adding to 5 A, and the 1.43 A drops 7.14 V across the 5 Ω, which is \(V_1 - V_2\). Node analysis needs no reduction and handles circuits where none is possible; here both routes agree, which is the point of doing the second.

50 V is \(5 \times 10\), the 10 Ω resistor alone with the other path ignored.

25 V puts 10 Ω in parallel with only the 5 Ω, forgetting the 20 Ω in series with it.

12.5 V puts all three resistors in parallel to ground, \(5 \times 2.5\).

Answer: B) 11 \(\Omega\)

Every delta side uses the same numerator - the sum of the three pairwise products of the wye arms - and is divided by the arm going to the terminal it does not touch. The side AB does not touch terminal C, so it is divided by \(R_c\):

$$R_aR_b + R_bR_c + R_cR_a = 18 + 54 + 27 = 99$$

$$R_{AB} = \dfrac{99}{R_c} = \dfrac{99}{9} = 11\;\Omega$$

The other two sides follow from the same numerator: \(R_{BC} = 99/3 = 33\) \(\Omega\) and \(R_{CA} = 99/6 = 16.5\) \(\Omega\). A delta side always comes out larger than either wye arm it spans, which is a useful check.

The Handbook does not tabulate this general conversion; the only wye-delta relation it carries is \(Z_\Delta = 3Z_Y\) for a balanced three-phase load, the special case of the formula above when all three arms are equal. That is why the formula is given to you in the question.

33 \(\Omega\) and 16.5 \(\Omega\) are the other two delta sides, \(R_{BC}\) and \(R_{CA}\), produced by dividing by \(R_a\) or \(R_b\) instead of \(R_c\). The rule that sorts them out is that each delta side is divided by the opposite arm, the one whose terminal the side does not touch.

9 \(\Omega\) is simply \(R_c\) copied across. The conversion is not a relabelling: a wye of 3, 6 and 9 \(\Omega\) becomes a delta of 33, 11 and 16.5 \(\Omega\), all of them larger than the arms they replace.