FE Electrical Circuit Analysis Practice Problems

Answer: B) 4 Ω

The Handbook's rule for two resistors in parallel on p. 358 is \(R_P = R_1R_2/(R_1 + R_2)\):

$$R_P = \dfrac{6 \times 12}{6 + 12} = \dfrac{72}{18} = 4\ \Omega$$

The general form, \(1/R_P = 1/6 + 1/12 = 3/12\), gives the same 4 Ω. A parallel combination is always smaller than its smallest member, because each added branch opens another path for current: the conductances add, \(1/6 + 1/12\) siemens, and the 12 Ω branch adds half as much conductance as the 6 Ω branch. With a source of \(V\) across the pair, the 6 Ω branch carries \(V/6\) and the 12 Ω branch \(V/12\), twice as much current through the smaller resistor, and the total \(V/4\) is what the equivalent 4 Ω would draw. The special case of two equal resistors gives half of either; here the result is two thirds of the smaller one.

18 Ω is the series sum, the two resistors carrying the same current instead of the same voltage.

9 Ω is the average of the two, which no connection produces.

0.25 Ω is the sum of reciprocals, \(1/6 + 1/12\), left uninverted; it is the conductance in siemens, not the resistance.

Answer: A) 5 Ω

The Handbook's Source Equivalents (p. 358, continuing on p. 359) describe the same network two ways: a Thévenin source \(V_{oc}\) in series with \(R_{eq}\), and a Norton source \(I_{sc}\) in parallel with the same \(R_{eq}\). Shorting the Thévenin model's terminals makes the whole \(V_{oc}\) appear across \(R_{eq}\), so the short-circuit current is \(V_{oc}/R_{eq}\), and the resistance follows from the two measurements:

$$R_{Th} = \dfrac{V_{oc}}{I_{sc}} = \dfrac{15}{3} = 5\ \Omega$$

This is the practical way to find the equivalent of a black box, with no knowledge of what is inside: two meter readings at the terminals fix both parameters. A third, equivalent method is to kill the internal sources and measure the resistance directly, or to apply a test source, but that requires access to the sources. Any load \(R_L\) across the terminals then draws \(I = 15/(5 + R_L)\); with a 5 Ω load, matched to \(R_{Th}\), the current is 1.5 A and the load takes 11.25 W, the most it can, as the Maximum Power-Transfer Theorem of p. 362 states.

15 Ω reports the open-circuit voltage as a resistance.

45 Ω multiplies the two readings instead of dividing them.

18 Ω adds them.

Answer: C) 2.4 A

Kirchhoff's voltage law on p. 358 of the Handbook, rises equal to drops around mesh 1, with the shared resistor carrying the net current \(I_1 - I_2\) because the two mesh currents oppose each other in it:

$$10 = 3I_1 + 2(I_1 - I_2) = 5I_1 - 2I_2$$

$$10 = 5I_1 - 2(1) \quad\Rightarrow\quad I_1 = \dfrac{12}{5} = 2.4\text{ A}$$

The total resistance around mesh 1 is indeed 5 Ω, but the shared resistor's drop is not \(2I_1\): the neighbouring mesh pushes 1 A the other way through it, which lowers the drop by 2 V and lets the 10 V source drive more current. That coupling term, \(-2I_2\), is the whole content of mesh analysis, and forgetting it is the same as pretending the second mesh does not exist. The net current in the shared resistor is \(2.4 - 1 = 1.4\) A in the direction of \(I_1\), dropping 2.8 V, and the 3 Ω resistor drops 7.2 V; the two add to 10 V, which closes the loop check. Had the mesh-2 current been defined so as to reinforce \(I_1\) in the shared branch, the equation would read \(10 = 5I_1 + 2I_2\) and give 1.6 A, which is why the direction convention is stated.

2.0 A is \(10/5\), the loop solved as if it stood alone; the \(2I_2\) term is dropped.

1.6 A takes the neighbouring current as aiding rather than opposing \(I_1\) in the shared resistor, \(10 = 5I_1 + 2\).

4.0 A leaves the shared resistor out of the loop, \(12/3\).

Answer: D) 2.73 A

Solve the pair by substitution. The second equation gives \(I_2 = \frac{2}{3}I_1\); in the first,

$$5I_1 - 2\left(\tfrac{2}{3}I_1\right) = 10 \quad\Rightarrow\quad \tfrac{11}{3}I_1 = 10 \quad\Rightarrow\quad I_1 = \dfrac{30}{11} = 2.73\text{ A}$$

and then \(I_2 = 20/11 = 1.82\) A. Both equations check: \(5(2.727) - 2(1.818) = 13.64 - 3.64 = 10\) and \(-2(2.727) + 3(1.818) = -5.45 + 5.45 = 0\). The equations describe a 10 V source in mesh 1, a shared 2 Ω resistor, 3 Ω in the mesh-1 loop besides the shared one, and 1 Ω in mesh 2 with no source, which is the physical content of the coefficients: the diagonal terms are the total mesh resistances and the off-diagonal term is the shared resistance with a negative sign, as Kirchhoff's voltage law (p. 358) produces for clockwise mesh currents. The shared resistor carries \(I_1 - I_2 = 10/11 = 0.91\) A and drops 1.82 V, which is also the voltage across the 1 Ω resistor of mesh 2, \(1 \times 1.82\).

2.0 A is \(10/5\), the first equation with the coupling term \(-2I_2\) dropped.

1.82 A is \(I_2\).

0.91 A is \(I_1 - I_2\), the current in the shared resistor.

Answer: B) 0.018 S

Admittances in parallel combine additively (p. 362 of the Handbook), and the table on that page gives the susceptances: a resistor has conductance \(1/R\), an inductor susceptance \(-1/(\omega L) = -1/X_L\), and a capacitor susceptance \(+\omega C = 1/X_C\):

$$Y = \dfrac{1}{100} - j\dfrac{1}{50} + j\dfrac{1}{200} = 0.01 - j0.02 + j0.005 = 0.01 - j0.015\text{ S}$$

$$|Y| = \sqrt{0.01^2 + 0.015^2} = \sqrt{1 \times 10^{-4} + 2.25 \times 10^{-4}} = 0.018\text{ S}$$

The net susceptance is negative, so the parallel combination is inductive: the inductor's larger susceptance outweighs the capacitor's, and the total current lags the voltage by \(\arctan(0.015/0.01) = 56.3^\circ\). The equivalent impedance is \(1/Y = 55.5\angle 56.3^\circ = 30.8 + j46.2\) Ω. With 100 V across the bank the resistor takes 1 A, the inductor 2 A lagging by 90° and the capacitor 0.5 A leading by 90°; the two reactive currents partly cancel to 1.5 A, which combines in quadrature with the 1 A to give 1.8 A total, the same \(|Y|\) times 100 V. Working in admittance is what makes parallel ac circuits as easy as series ones.

0.01 S is the conductance alone, the reactive branches ignored.

0.035 S adds all three magnitudes, treating the susceptances as if they were in phase with the conductance and with each other.

0.025 S combines the susceptances with the correct signs but adds the result arithmetically to the conductance, \(0.01 + 0.015\), instead of in quadrature.