FE Electrical and Computer Practice Exam

In a sinusoidal steady state the circuit is analysed with phasors and impedances. The inductive reactance depends on the frequency:

The series impedance is $Z = R + jX_L = 30 + j37.7\,\Omega$. Resistance and reactance are at right angles in the complex plane, so the magnitude of the impedance is found with the Pythagorean theorem, not by simple addition:

Ohm's law for phasors gives the rms current magnitude:

The current lags the voltage by $\theta = \arctan(37.7/30) = 51.5°$, as expected for an inductive circuit, but only the magnitude is asked for.

I ≈ 2.49 A (option B)

The most tempting wrong answer is A (1.77 A), which adds the resistance and reactance arithmetically ($30+37.7 = 67.7\,\Omega$). Because the resistor voltage and the inductor voltage are 90° apart, their impedances add as perpendicular vectors, and the arithmetic sum overstates the impedance. Option D (4.00 A) ignores the inductor entirely, and option C (3.18 A) ignores the resistor.

For a balanced three-phase load, the total apparent power is related to the line quantities by $S = \sqrt{3}\,V_L I_L$, regardless of whether the load is wye- or delta-connected.

First find the apparent power from the real power and the power factor, because the current is set by the apparent power, not the real power alone:

Then solve for the line current:

The same result follows on a per-phase basis. For a wye load the phase voltage is $V_p = 480/\sqrt{3} = 277$ V and each phase carries one third of the power, 5,000 W, so $I_p = 5{,}000/(277 \times 0.80) = 22.6$ A; in a wye connection the line current equals the phase current.

The load also draws reactive power $Q = S\sin\theta = 18{,}750(0.60) = 11{,}250$ var, which is why the current is 25% higher than it would be at unity power factor.

$I_L$ ≈ 22.6 A

(a) Closed-loop transfer function. For unity feedback, $T(s) = G(s)/[1 + G(s)]$. Multiply numerator and denominator by $s(s+4)$ to clear the fraction:

(b) Natural frequency and damping ratio. Match the denominator to the standard second-order form $s^2 + 2\zeta\omega_n s + \omega_n^2$:

Since $0<\zeta<1$ the response is underdamped; the closed-loop poles are at $s = -2 \pm j3.46$.

(c) Percent overshoot. For an underdamped second-order system the overshoot depends only on $\zeta$:

(d) Steady-state ramp error. The open-loop function has one pole at the origin, so the system is type 1: zero error to a step, finite error to a ramp. The velocity error constant is

and the steady-state error to a unit ramp is $e_{ss} = 1/K_v$:

Raising $K$ would reduce this error but also reduce $\zeta$ and increase the overshoot, which is the classic trade-off in this system.

T(s) = 16/(s² + 4s + 16); $\omega_n$ = 4 rad/s, ζ = 0.5; %OS ≈ 16.3%; $e_{ss}$ = 0.25