Placement Test Practice — Electrical Engineering
Placement Test Practice — Electrical Engineering
These problems cover circuit analysis, AC phasors, transfer functions, poles and zeros, and Laplace transforms.
Practice Test — 25 Questions
Show Answer Key
1. $1/R = 1/10+1/15+1/30 = 6/30 = 1/5$; $R_{eq} = 5\,\Omega$
2. $I = 24/12 = 2$ A; $P = 2^2 \times 8 = 32$ W
3. $V_{th} = 30 \times 20/30 = 20$ V
4. $X_L = 2\pi(60)(0.2) = 75.4\,\Omega$; $|Z| = \sqrt{2500+5685} = 90.4\,\Omega$
5. $f_0 = 1/(2\pi\sqrt{0.1 \times 10^{-5}}) = 159.2$ Hz
6. $Z = R = 10\,\Omega$
7. $\mathbf{V} = (170/\sqrt{2})\angle 30° = 120.2\angle 30°$ V rms, or $170\angle 30°$ peak phasor
8. $V_{\text{rms}} = 100/\sqrt{2} = 70.7$ V
9. $P = 240 \times 5 \times 0.8 = 960$ W
10. $s = (-4 \pm \sqrt{16-52})/2 = -2 \pm 3j$
11. Real parts $= -2 < 0$ → stable
12. $s^2 - 4 = (s-2)(s+2) = 0$; zeros at $s = 2$ and $s = -2$
13. $\frac{5 \cdot 4}{(s+3)^2 + 16} = \frac{20}{s^2+6s+25}$
14. $2(1 - e^{-3t})$
15. $20/4 = 5$
16. $\tau = RC = 2000 \times 0.5 \times 10^{-6} = 10^{-3}$ s $= 1$ ms
17. $63.2\%$
18. $E = \frac{1}{2}CV^2 = 0.5 \times 10^{-4} \times 2500 = 0.125$ J
19. $E = \frac{1}{2}LI^2 = 0.5 \times 0.5 \times 9 = 2.25$ J
20. $I = 2 + (-0.5) = 1.5$ A
21. $I_N = V_{th}/R_{th} = 3$ A; $R_N = R_{th} = 4\,\Omega$
22. $|Z_C| = 1/(500 \times 20 \times 10^{-6}) = 100\,\Omega$
23. $P_{\max} = V_{th}^2/(4R_{th}) = 100/20 = 5$ W
24. $s = -\omega_n\zeta \pm \omega_n\sqrt{\zeta^2-1}$. Since $\zeta < 1$: $s = -7.07 \pm j7.07$
25. $H(s) = \frac{K}{(s+2)(s+5)}$; DC gain: $H(0) = K/10 = 1$ → $K=10$; $H(s) = \frac{10}{(s+2)(s+5)}$