Updated for the new July 2020 FE Exam Topics

Free FE Practice Test# PrepFE™

## Free FE Electrical and Computer Example Practice Problems

We've selected 10 diverse practice problems from our question bank that you can use to review for the Electrical and Computer engineering FE exam and give you an idea about some of the content we provide.

$i_L(0) = 0.25 \si A$ and $v_C(0) = 0$.

◯ A.

$e^{-\frac{1}{2}t} \text{ V}$

◯ B.

$e^{-\frac{1}{2}t}\sin\frac{1}{2}t + \cos\frac{1}{2}t \text{ V}$

◯ C.

$\sin\frac{1}{2}t \text{ V}$

◯ D.

$e^{-\frac{1}{2}t}\sin\frac{1}{2}t \text { V}$

The constant "m" is most nearly:

◯ A.

$-2$

◯ B.

$4$

◯ C.

$0$

◯ D.

$2$

◯ A.

$-67.7\si{W}$

◯ B.

$67.7\si{W}$

◯ C.

$1690\si{W}$

◯ D.

$-1690\si{W}$

◯ A.

perform parallel processing.

◯ B.

reduce memory access time.

◯ C.

increase clock speed.

◯ D.

increase the number of functional units.

◯ A.

$R_2 = 25\si {k\Omega}, R_4 = 25 \si {k\Omega}$

◯ B.

$R_2 = 10 \si {k\Omega}, R_4 = 15 \si {k\Omega}$

◯ C.

$R_2 = 5\si {k\Omega}, R_4 = 5 \si {k\Omega}$

◯ D.

$R_2 = 20 \si {k\Omega}, R_4 = 25 \si {k\Omega}$

◯ A.

$0.726 \si{s}$

◯ B.

$5.111 \si{s}$

◯ C.

$1.231 \si{s}$

◯ D.

$0.940 \si{s}$

◯ A.

$.105 \si{m}$

◯ B.

$.420 \si{m}$

◯ C.

$1.05 \si{m}$

◯ D.

$2.02 \si{m}$

The radio horizon is defined as the distance for which propagating electromagnetic waves are locally tangent to the Earth's surface.

For an antenna in nominal weather conditions this is, $$d_{\si{km}} = 4.1\sqrt{h_\si{m}}$$ Where $h_\si m$ is the antenna height and $d_{\si{km}}$ is the radio horizon.

What is the minimum propagation delay if the two towers are at the maximum separation from one another?

◯ A.

$273.33 \mu \si{s}$

◯ B.

$341.15 \mu \si{s}$

◯ C.

$120.55 \mu \si{s}$

◯ D.

$65.23\mu \si{s}$

◯ A.

$\bar A $

◯ B.

$\bar B$

◯ C.

$\bar A \bar B$

◯ D.

$\bar A + \bar B$

◯ A.

$120\angle 30^\circ \si{V_rms}$

◯ B.

$90\angle -90^\circ \si{V_rms}$

◯ C.

$120\angle 90^\circ \si{V_rms}$

◯ D.

$120\angle -90^\circ \si{V_rms}$

$i_L(0) = 0.25 \si A$ and $v_C(0) = 0$.

A.$e^{-\frac{1}{2}t} \text{ V}$

B.$e^{-\frac{1}{2}t}\sin\frac{1}{2}t + \cos\frac{1}{2}t \text{ V}$

C.$\sin\frac{1}{2}t \text{ V}$

D.$e^{-\frac{1}{2}t}\sin\frac{1}{2}t \text { V}$

The correct answer is D.

The homogenous solution is to be obtained. The homogenous solution is when the input function, the voltage source in this case, is set equal to 0. Therefore, we can write the loop equation for the RLC series circuit as, $$v_L + v_R + v_C = 0$$ Express $v_L$ as $v_L = L\frac{di_L}{dt}$, and express $v_R$ as $v_R = i_LR$.

Now, substitute $C\frac{dv_C}{dt}$ for $i_L$ to obtain the full homogenous equation in terms of $v_C$, $$LC\frac{d^2v_C}{dt^2}+RC\frac{dv_C}{dt} + v_C = 0$$ Divide through by LC, $$\frac{d^2v_C}{dt^2}+\frac{R}{L}\frac{dv_C}{dt} + \frac{1}{LC}v_C = 0$$ This is now a second order linear homogenous differential equation with constant coefficients in the form, $$r^2 + ar + b = 0$$ To determine the form of the solution for this second order linear homogenous equation with constant coefficients, compute if $a^2 \gt 4b$, or if $a^2 = 4b$, or if $a^2 \lt 4b$.

In our circuit, $a$ = $\frac{R}{L}$ = $\frac{(4\Omega)}{(4 \si H)} = 1$, and $b$ = $\frac{1}{LC} =\frac{1}{(4 \si H)(0.5 \si F)} = \frac{1}{2} $

Now determine $a^2$ and $4b$ and compare. $$a^2 = 1^2 =1$$ $$4b = 4(\frac{1}{2}) = 2$$ Because $a^2 \lt 4b$, the solution will take the form, $$y= e^{\alpha x} (C_1 \cos \beta x + C_2 \sin \beta x)$$ Where $\alpha$= – $\frac{a}{2}$ and $\beta$ = $\frac{\sqrt{4b-a^2}}{2}$

Compute $\alpha$, $$\alpha = -\frac{a}{2} = -\frac{1}{2}$$ Compute $\beta$, $$\beta = \frac{\sqrt{4b-a^2}}{2} = \frac{\sqrt{4(0.5)-1^2}}{2} = \frac{1}{2}$$ Thus the expression for $v_C(t)$ becomes, $$v_C(t) = e^{-\frac{1}{2}t}\left(C_1\cos\frac{1}{2}t+C_2\sin\frac{1}{2}t\right)$$ When t = 0, $v_C(0) = 0 = C_1$, then use $i_L$ to find $C_2$ as follows, $$\frac{i_L}{C} = \frac{dv_C}{dt} = -\frac{1}{2}e^{-\frac{1}{2}t}C_1\cos\frac{1}{2}t \\ - \frac{1}{2}e^{-\frac{1}{2}t}C_1\sin\frac{1}{2}t - \frac{1}{2}e^{-\frac{1}{2}t}C_2\sin\frac{1}{2}t+\frac{1}{2}e^{-\frac{1}{2}t}C_2\cos\frac{1}{2}t$$ Substitute the initial condition at $i_L(0)$, $$\frac{i_L(0)}{C} = \frac{dv_C(0)}{dt} = -\frac{1}{2}C_1 + \frac{1}{2}C_2$$ $$\frac{i_L(0)}{C} = \frac{0.25\si A}{0.5\si F} = \frac{1}{2} = -\frac{1}{2}C_1 + \frac{1}{2}C_2$$ Since $C_1$ is known to be 0, then $C_2$ = 1

Finally, the homogeneous solution is, $$v_C(t) = e^{-\frac{1}{2}t}\sin\frac{1}{2}t \text{ V}$$

The constant "m" is most nearly:

A.$-2$

B.$4$

C.$0$

D.$2$

The correct answer is D.

The divergence of a magnetic field is defined to be 0 by Maxwell's Equations. $$\nabla \bullet \mathbf B = 0$$ Therefore, calculate the divergence and set the expression equal to zero.

The divergence of a vector field is defined as, $$\nabla\bullet \mathbf V = \left (\frac {\partial}{\partial x}\mathbf i\ +\frac {\partial}{\partial y}\mathbf k +\frac {\partial}{\partial z}\mathbf j \right) \bullet (V_1 \mathbf i + V_2\mathbf j + V_3\mathbf k)$$ Thus, $$\nabla \bullet \mathbf B = \frac{d}{dx} 10y \mathbf i + \frac{d}{dy} 2y \mathbf j - \frac{d}{dz} mz \mathbf k = 0$$ $$\nabla \bullet \mathbf B = 0 + 2 - m = 0$$ $$m=2$$

A.$-67.7\si{W}$

B.$67.7\si{W}$

C.$1690\si{W}$

D.$-1690\si{W}$

The correct answer is D.

Begin by writing a KVL (loop) equation for the single loop. Assume the current flows in the clockwise direction. Note that it will ultimately not matter which direction the current is assumed to flow as shall be explained at the bottom. This yields, $$150\si{V} +0.8v_x - 20\si{V} = i(10\Omega + 50\Omega)$$ Note that the $v_x$ voltage is equal to $(i)(50)\Omega$ (by Ohm's law, the current flowing in the resistor is $i$ and the resistor voltage, $v_x$, is defined as the voltage drop across the $50\Omega$ resistor) so it may be substituted directly into the loop equation to yield, $$150\si{V} + 0.8(50i) - 20\si{V} = i(10\Omega + 50\Omega)$$ Solving for the unknown, $i$, produces, $$150\si{V} - 20\si{V} = i60\Omega - i40\Omega$$ $$130\si{V} = i20\Omega$$ $$i = \frac{130\si{V}}{20\Omega} = 6.5\si{A} $$ For a source to absorb power, then positive current must flow out of the negative terminal. For the dependent source, current flows out of the positive terminal. If the current had been assumed to be counterclockwise, then the current would have been negative and a simple sign reversal would yield the same result as below.

Thus, $$P = IV = -(0.8v_x)(6.5\si{A})$$ We know that $v_x = 50\Omega i = (50\Omega)(6.5 \si A)$. Therefore, $$P = -[(0.8)(50\Omega)(6.5\si{A})(6.5\si{A})] \\ = -1690\si{W} $$

A.perform parallel processing.

B.reduce memory access time.

C.increase clock speed.

D.increase the number of functional units.

The correct answer is A.

A.$R_2 = 25\si {k\Omega}, R_4 = 25 \si {k\Omega}$

B.$R_2 = 10 \si {k\Omega}, R_4 = 15 \si {k\Omega}$

C.$R_2 = 5\si {k\Omega}, R_4 = 5 \si {k\Omega}$

D.$R_2 = 20 \si {k\Omega}, R_4 = 25 \si {k\Omega}$

The correct answer is D.

First, determine the value of $R_2$. This can be found from considering the gain provided by the top op-amp. The output load is floating and the load seen by each op-amp is $\frac{R_L}{2}$. This can be accounted for in the gain equation by doubling the gain seen across the top op-amp. The input to the negative terminal of the op-amp across $R_1$ is $0$. Thus this is a non-inverting amplifier. The gain expression is, $$\frac{v_o}{v_1} = 2\left (1+\frac{R_2}{R_1} \right) = 10$$ If $R_1 = 5\si {k\Omega}$, then, $$2\left (1+ \frac{R_2}{5 \si {k\Omega}} \right ) = 10 $$ Solving for $R_2$, $$R_2 = 20 \si {k\Omega}$$ The bottom amplifier is connected as an inverting amplifier, whose gain is equal in magnitude and is given by, $$ 2K = \frac{R_4}{R_3} = 10 \\ \Rightarrow K = \frac{R_4}{R_3} = 5 $$ Substituting in the given $R_3 = 5 \si {k\Omega}$, $$ R_4 = 25 \si {k\Omega} $$

A.$0.726 \si{s}$

B.$5.111 \si{s}$

C.$1.231 \si{s}$

D.$0.940 \si{s}$

The correct answer is A.

The controller plant, $G(s)$, is $\frac{25}{s(s+5)}$.

This is a unity feedback controller. It can be deduced that this is a unity feedback controller from looking at the classical model of a negative feedback controller, which is shown in the FE Reference Handbook as depicted in the image and comparing it to our system (noting that $G_2(s) = 1$, $H(s) = 1$ and $L(s) = 0$).

A unity feedback model is also shown explicitly in the section on Control Systems.

Thus, $H(s) = 1$ and the closed loop transfer function, $T(s)$, is $\frac{Y(s)}{R(s)}$ which is, $$T(s) = \frac{G(s)}{1+G(s)}$$ Substituting in $G(s)$, $$T(s) = \frac{\frac{25}{s(s+5)}}{1+\frac{25}{s(s+5)}}$$ $$T(s) = \frac{25}{s^2 + 5s + 25}$$ This is a second-order control system of the form shown in the handbook, $$\frac{Y(s)}{R(s)} = \frac{K\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}$$ Where, $$\omega_n = \sqrt{25} = 5$$ $$2\zeta\omega_n = 5$$ Substituting $\omega_n$ and solving for $\zeta$ yields, $$\zeta = 0.5$$ Using the values for $\omega_n$ and $\zeta$, calculate the peak time from, $$t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}}$$ $$t_p = \frac{\pi}{5\sqrt{1-0.5^2}} = 0.726 \si{s}$$

A.$.105 \si{m}$

B.$.420 \si{m}$

C.$1.05 \si{m}$

D.$2.02 \si{m}$

The correct answer is A.

The wavelength of a sinusoid traveling on a transmission line is given as, $$\lambda = \frac{U}{f}$$ For an electromagnetic wave in free space, the velocity of propagation, $U$, is the speed of light, $$\lambda = \frac{c}{f}$$ The velocity factor is the ratio of the velocity of propagation in the medium to that of the wave in free space (note this relationship is not explicitly given in the FE Reference Handbook), $$k = \frac{v}{c}$$ Substituting $k$ into the first equation yields, $$\lambda = \frac{kc}{f}$$ Now substitute all known values to find the wavelength, $$\lambda = \frac{(0.7)(3 \times 10^8 {\frac{\si{m}}{\si{s}}})}{2 \times 10^9 \si{Hz}} = .105\si{m} $$

The radio horizon is defined as the distance for which propagating electromagnetic waves are locally tangent to the Earth's surface.

For an antenna in nominal weather conditions this is, $$d_{\si{km}} = 4.1\sqrt{h_\si{m}}$$ Where $h_\si m$ is the antenna height and $d_{\si{km}}$ is the radio horizon.

What is the minimum propagation delay if the two towers are at the maximum separation from one another?

A.$273.33 \mu \si{s}$

B.$341.15 \mu \si{s}$

C.$120.55 \mu \si{s}$

D.$65.23\mu \si{s}$

The correct answer is A.

Since the definition of the radio horizon is such when the propagating waves are tangential to the Earth's surface, then the maximum distance separation comes from summing the two antenna radio horizons together, as shown in the figure.

Now, calculate the maximum separation, $d$, $$d = 4.1\sqrt{100\si{m}} + 4.1\sqrt{100\si{m}}$$ $$d = 82\si{km}$$ The propagation delay is the time it takes for the electromagnetic wave to arrive at the receiver after being sent by the transmitter. Refer to the Delays in Computer Networks section in the

A.$\bar A $

B.$\bar B$

C.$\bar A \bar B$

D.$\bar A + \bar B$

The correct answer is A.

Begin by simplifying with the complement law on $(\bar B + B)$, $$\bar B + B = 1$$ The reduced expression is, $$(\overline{AB})(\bar A +B)$$ Apply De Morgan's Law to break $\overline{AB}$. We do this because addition does not distribute over multiplication in Boolean algebra. Consider that while, in general, $$a × [b + c] = [a × b] + [a × c]$$ The addition, however, will not be distributive over multiplication. That is, $$a + [b × c] \text{ does not, in general, equal } [a + b] × [a + c]$$ Returning to the problem, after applying DeMorgan's Law, $$(\bar A + \bar B)(\bar A + B)$$ Distribute the terms to obtain, $$(\bar A \bar A + \bar A B + \bar A \bar B + B\bar B)$$ Apply the complement law that a variable ANDed with the NOT of itself is 0 (i.e. $\bar B B = 0 )$ $$(\bar A \bar A + \bar A B + \bar A \bar B)$$ Apply the identity that $A \cdot A = A$, $$(\bar A + \bar A B + \bar A \bar B)$$ Factor out $\bar A$, $$\bar A (1 + B + \bar B)$$ Apply complement law one more time to finally obtain, $$\bar A $$

A.$120\angle 30^\circ \si{V_rms}$

B.$90\angle -90^\circ \si{V_rms}$

C.$120\angle 90^\circ \si{V_rms}$

D.$120\angle -90^\circ \si{V_rms}$

The correct answer is D.

Since the source is balanced, it follows that the magnitudes of all the phase voltages with respect to neutral are equal, $$|V_{an}| = |V_{bn}| = |V_{cn}|$$ Because it is given that this is a positive phase system, we know that $\vec V_{bn}$ lags $\vec V_{an}$ by $120^\circ$ and $\vec V_{cn}$ lags $\vec V_{bn}$ by $120^\circ$.

Since we know the lag between $\vec V_{bn}$ and $\vec V_{an}$ and that the magnitude of the voltages are equal, we can compute $\vec V_{bn}$ directly from the phasor expression, $$\vec V_{bn} = |V_{an}|\angle 30^\circ - 120^\circ = 120\angle -90^\circ \si{V_rms}$$