Fe Exam Review Electrical and Computer Engineering Sveum Pdf
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Free FE Electrical and Computer Example Exercise Problems
We've selected 10 diverse exercise problems from our question bank that you can use to review for the Electric and Calculator engineering Iron exam and give yous an thought about some of the content we provide.
ane) Obtain the homogenous solution to the differential equation that yields the voltage, $v_C(t)$ across the capacitor in the serial RLC circuit.
$i_L(0) = 0.25 \si A$ and $v_C(0) = 0$.
◯ A.
$e^{-\frac{one}{two}t} \text{ V}$
◯ B.
$e^{-\frac{1}{two}t}\sin\frac{1}{two}t + \cos\frac{1}{2}t \text{ 5}$
◯ C.
$\sin\frac{1}{two}t \text{ Five}$
◯ D.
$e^{-\frac{1}{two}t}\sin\frac{1}{2}t \text { V}$
two) A magnetic field has the vector field $\mathbf{\vec B} = 10y \mathbf i +2y \mathbf j - mz \mathbf m$.
The abiding "grand" is most almost:
3) In the circuit shown, find the power absorbed by the dependent source.
4) Pipelining improves CPU functioning by enabling the processor to:
◯ A.
perform parallel processing.
◯ B.
reduce retentivity access time.
◯ C.
increase clock speed.
◯ D.
increment the number of functional units.
five) For the given circuit, $R_1 = R_3 = 5 \si {k\Omega}$. Determine $R_2$ and $R_4$ to obtain a gain of $ten$.
◯ A.
$R_2 = 25\si {k\Omega}, R_4 = 25 \si {one thousand\Omega}$
◯ B.
$R_2 = 10 \si {k\Omega}, R_4 = xv \si {yard\Omega}$
◯ C.
$R_2 = 5\si {thousand\Omega}, R_4 = 5 \si {one thousand\Omega}$
◯ D.
$R_2 = 20 \si {k\Omega}, R_4 = 25 \si {g\Omega}$
vi) For the system shown in the figure, compute the peak value time:
7) A communication line uses a shield twisted pair (STP) cable that has a velocity factor of $0.7.$ What is about nigh the wavelength of a $2 \si {GHz}$ signal traveling on the STP cable?
8) A pair of line of sight communication antenna towers are to exist constructed, each at a meridian of $100\si{m}$.
The radio horizon is divers every bit the distance for which propagating electromagnetic waves are locally tangent to the Earth's surface.
For an antenna in nominal atmospheric condition conditions this is, $$d_{\si{km}} = four.1\sqrt{h_\si{one thousand}}$$ Where $h_\si m$ is the antenna elevation and $d_{\si{km}}$ is the radio horizon in $\si {km}$.
What is the minimum propagation delay if the two towers are at the maximum separation from i another?
◯ A.
$273.33 \mu \si{s}$
◯ B.
$341.15 \mu \si{southward}$
◯ C.
$120.55 \mu \si{southward}$
nine) Simplify the following expression: $$(\overline{AB})(\bar A +B)(\bar B + B)$$
10) A balanced three stage Y continued source has, using a positive phase sequence, $\vec V_{an} = 120\angle30^\circ \si{V_{rms}}$. Observe the $\vec V_{bn}$.
◯ A.
$120\angle 30^\circ \si{V_rms}$
◯ B.
$90\angle -xc^\circ \si{V_rms}$
◯ C.
$120\angle xc^\circ \si{V_rms}$
◯ D.
$120\angle -90^\circ \si{V_rms}$
Solutions
1) Obtain the homogenous solution to the differential equation that yields the voltage, $v_C(t)$ across the capacitor in the series RLC circuit.
$i_L(0) = 0.25 \si A$ and $v_C(0) = 0$.
A.$e^{-\frac{one}{2}t} \text{ V}$
B.$e^{-\frac{1}{2}t}\sin\frac{ane}{2}t + \cos\frac{1}{2}t \text{ Five}$
C.$\sin\frac{1}{two}t \text{ V}$
D.$e^{-\frac{1}{2}t}\sin\frac{1}{2}t \text { V}$
The correct answer is D.
Explanation:
Refer to FE Reference Handbook: 2nd-Guild Linear Homogeneous Differential Equations with Abiding Coefficients
The homogenous solution is to exist obtained. The homogenous solution is when the input part, the voltage source in this case, is set equal to 0. Therefore, nosotros can write the loop equation for the RLC series excursion as, $$v_L + v_R + v_C = 0$$ Express $v_L$ as $v_L = L\frac{di_L}{dt}$ (the voltage across an inductor equation), and express $v_R$ as $v_R = i_LR$ (Ohm's Law).
Now, substitute $C\frac{dv_C}{dt}$ for $i_L$ (the current through the capacitor is the same as the current through the inductor since the elements are in series) to obtain the full homogenous equation in terms of $v_C$, $$LC\frac{d^2v_C}{dt^2}+RC\frac{dv_C}{dt} + v_C = 0$$ Split through by LC, $$\frac{d^2v_C}{dt^2}+\frac{R}{50}\frac{dv_C}{dt} + \frac{1}{LC}v_C = 0$$ This is at present a 2d order linear homogenous differential equation with constant coefficients in the grade, $$r^2 + ar + b = 0$$ To decide the form of the solution for this second lodge linear homogenous equation with abiding coefficients, compute if $a^2 \gt 4b$, or if $a^2 = 4b$, or if $a^ii \lt 4b$.
In our circuit, $a$ = $\frac{R}{50}$ = $\frac{(4\Omega)}{(four \si H)} = 1$, and $b$ = $\frac{1}{LC} =\frac{1}{(4 \si H)(0.5 \si F)} = \frac{1}{ii} $
Now determine $a^2$ and $4b$ and compare. $$a^2 = 1^2 =1$$ $$4b = four(\frac{1}{ii}) = ii$$ Because $a^two \lt 4b$, the solution volition take the form, $$y= due east^{\alpha ten} (C_1 \cos \beta 10 + C_2 \sin \beta x)$$ Where $\alpha$= – $\frac{a}{2}$ and $\beta$ = $\frac{\sqrt{4b-a^2}}{2}$
Compute $\blastoff$, $$\blastoff = -\frac{a}{2} = -\frac{one}{2}$$ Compute $\beta$, $$\beta = \frac{\sqrt{4b-a^ii}}{2} = \frac{\sqrt{4(0.five)-one^2}}{2} = \frac{1}{two}$$ Thus the expression for $v_C(t)$ becomes, $$v_C(t) = e^{-\frac{one}{2}t}\left(C_1\cos\frac{1}{2}t+C_2\sin\frac{ane}{two}t\right)$$ When t = 0, $v_C(0) = 0 = C_1$, then use $i_L$ to find $C_2$ as follows, $$\frac{i_L}{C} = \frac{dv_C}{dt} = -\frac{1}{2}e^{-\frac{i}{2}t}C_1\cos\frac{1}{two}t \\ - \frac{1}{ii}e^{-\frac{ane}{2}t}C_1\sin\frac{1}{two}t - \frac{1}{2}e^{-\frac{one}{ii}t}C_2\sin\frac{i}{2}t+\frac{1}{two}e^{-\frac{1}{ii}t}C_2\cos\frac{1}{2}t$$ Substitute the initial condition at $i_L(0)$, $$\frac{i_L(0)}{C} = \frac{dv_C(0)}{dt} = -\frac{1}{2}C_1 + \frac{i}{2}C_2$$ $$\frac{i_L(0)}{C} = \frac{0.25\si A}{0.5\si F} = \frac{1}{two} = -\frac{1}{two}C_1 + \frac{one}{2}C_2$$ Since $C_1$ is known to be 0, and so $C_2$ = 1
Finally, the homogeneous solution is, $$v_C(t) = e^{-\frac{one}{2}t}\sin\frac{one}{two}t \text{ V}$$
ii) A magnetic field has the vector field $\mathbf{\vec B} = 10y \mathbf i +2y \mathbf j - mz \mathbf thousand$.
The constant "thou" is most nearly:
A.$-2$
B.$4$
C.$0$
D.$ii$
The correct respond is D.
Explanation:
Refer to the Electromagnetic Dynamic Fields section in the Electrical and Estimator Engineering chapter of the Fe Reference Handbook.
The departure of a magnetic field is defined to be 0 by Maxwell's Equations. $$\nabla \bullet \mathbf B = 0$$ Therefore, calculate the difference and set the expression equal to zero.
The divergence of a vector field is divers equally, $$\nabla\bullet \mathbf V = \left (\frac {\fractional}{\partial ten}\mathbf i\ +\frac {\partial}{\partial y}\mathbf yard +\frac {\fractional}{\partial z}\mathbf j \right) \bullet (V_1 \mathbf i + V_2\mathbf j + V_3\mathbf m)$$ Thus, $$\nabla \bullet \mathbf B = \frac{d}{dx} 10y \mathbf i + \frac{d}{dy} 2y \mathbf j - \frac{d}{dz} mz \mathbf one thousand = 0$$ $$\nabla \bullet \mathbf B = 0 + 2 - grand = 0$$ $$grand=2$$
3) In the circuit shown, observe the power absorbed by the dependent source.
A.$-67.7\si{Westward}$
B.$67.vii\si{Due west}$
C.$1690\si{W}$
D.$-1690\si{W}$
The correct answer is D.
Explanation:
Refer to the Power Absorbed by a Resistive Element & Kirchhoff's Laws sections in the Electrical and Figurer Engineering chapter of the Iron Reference Handbook.
Begin by writing a KVL (loop) equation for the single loop. Assume the current flows in the clockwise direction. Note that it will ultimately non matter which direction the current is causeless to period as shall be explained at the lesser. This yields, $$150\si{5} +0.8v_x - twenty\si{V} = i(10\Omega + fifty\Omega)$$ Note that the $v_x$ voltage is equal to $(i)(50)\Omega$ (by Ohm's constabulary, the current flowing in the resistor is $i$ and the resistor voltage, $v_x$, is defined as the voltage drop across the $l\Omega$ resistor) so it may exist substituted directly into the loop equation to yield, $$150\si{Five} + 0.8(50i) - 20\si{V} = i(x\Omega + 50\Omega)$$ Solving for the unknown, $i$, produces, $$150\si{V} - 20\si{V} = i60\Omega - i40\Omega$$ $$130\si{V} = i20\Omega$$ $$i = \frac{130\si{V}}{20\Omega} = 6.five\si{A} $$ 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 final. If the current had been assumed to be counterclockwise, then the electric current would have been negative and a elementary sign reversal would yield the same result equally below.
Thus, $$P = IV = -(0.8v_x)(vi.v\si{A})$$ We know that $v_x = fifty\Omega i = (fifty\Omega)(6.5 \si A)$. Therefore, $$P = -[(0.8)(50\Omega)(six.5\si{A})(6.v\si{A})] \\ = -1690\si{W} $$
four) Pipelining improves CPU operation past enabling the processor to:
A.perform parallel processing.
B.reduce memory access time.
C.increment clock speed.
D.increase the number of functional units.
The right reply is A.
Explanation:
Pipelining is an implementation technique where multiple instructions are overlapped in execution. The computer pipeline is divided in stages. Each stage completes a function of an educational activity in parallel. The stages are connected i to the next to form a pipe - instructions enter at one end, progress through the stages, and get out at the other end. Pipelining does not decrease the time for individual instruction execution. Instead, information technology increases instruction throughput. The throughput of the instruction pipeline is determined by how often an didactics exits the pipeline.
5) For the given circuit, $R_1 = R_3 = 5 \si {k\Omega}$. Make up one's mind $R_2$ and $R_4$ to obtain a gain of $10$.
A.$R_2 = 25\si {k\Omega}, R_4 = 25 \si {k\Omega}$
B.$R_2 = 10 \si {k\Omega}, R_4 = 15 \si {m\Omega}$
C.$R_2 = 5\si {k\Omega}, R_4 = 5 \si {k\Omega}$
D.$R_2 = 20 \si {k\Omega}, R_4 = 25 \si {m\Omega}$
The correct answer is D.
Caption:
Refer to Operational Amplifiers in the FE Reference Handbook.
First, determine the value of $R_2$. This tin be establish from considering the gain provided past the peak op-amp. The output load is floating and the load seen past each op-amp is $\frac{R_L}{ii}$. This can be accounted for in the gain equation by doubling the gain seen across the tiptop op-amp. The input to the negative final of the op-amp across $R_1$ is $0$. Thus this is a non-inverting amplifier. The gain expression is, $$\frac{v_o}{v_1} = 2\left (1+\frac{R_2}{R_1} \right) = ten$$ If $R_1 = five\si {one thousand\Omega}$, then, $$2\left (1+ \frac{R_2}{5 \si {k\Omega}} \correct ) = x $$ Solving for $R_2$, $$R_2 = 20 \si {grand\Omega}$$ The lesser amplifier is continued as an inverting amplifier, whose proceeds is equal in magnitude and is given by, $$ 2K = \frac{R_4}{R_3} = ten \\ \Rightarrow K = \frac{R_4}{R_3} = 5 $$ Substituting in the given $R_3 = 5 \si {k\Omega}$, $$ R_4 = 25 \si {k\Omega} $$
6) For the system shown in the effigy, compute the peak value time:
A.$0.726 \si{s}$
B.$5.111 \si{s}$
C.$one.231 \si{s}$
D.$0.940 \si{s}$
The correct answer is A.
Caption:
Refer to Control Systems in the Fe Reference Handbook.
The controller establish, $Thou(s)$, is $\frac{25}{south(due south+5)}$.
This is a unity feedback controller. Information technology can exist 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 equally depicted in the image and comparing it to our system (noting that $G_2(s) = 1$, $H(s) = 1$ and $50(southward) = 0$).
A unity feedback model is likewise shown explicitly in the section on Control Systems.
Thus, $H(due south) = 1$ and the closed loop transfer function, $T(s)$, is $\frac{Y(s)}{R(s)}$ (output over the input) which is, $$T(s) = \frac{G(s)}{1+1000(s)}$$ Substituting in $G(s)$, $$T(s) = \frac{\frac{25}{s(s+5)}}{1+\frac{25}{southward(south+5)}}$$ $$T(south) = \frac{25}{due south^2 + 5s + 25}$$ This is a second-order command system of the form shown in the handbook, $$\frac{Y(s)}{R(s)} = \frac{G\omega_n^2}{s^2 + two \zeta \omega_n s + \omega_n^2}$$ Where, $$\omega_n = \sqrt{25} = v$$ $$2\zeta\omega_n = five$$ Substituting $\omega_n$ and solving for $\zeta$ yields, $$\zeta = 0.5$$ Using the values for $\omega_n$ and $\zeta$, calculate the peak time from, $$t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}}$$ $$t_p = \frac{\pi}{5\sqrt{1-0.5^2}} = 0.726 \si{southward}$$
7) A communication line uses a shield twisted pair (STP) cable that has a velocity cistron of $0.7.$ What is almost nigh the wavelength of a $2 \si {GHz}$ signal traveling on the STP cable?
A.$.105 \si{thou}$
B.$.420 \si{m}$
C.$1.05 \si{m}$
D.$2.02 \si{yard}$
The correct answer is A.
Caption:
Refer to FE Reference Handbook: Lossless Manual Lines
The wavelength of a sinusoid traveling on a manual line is given equally, $$\lambda = \frac{U}{f}$$ For an electromagnetic wave in free space, the velocity of propagation, $U$, is the speed of low-cal, $$\lambda = \frac{c}{f}$$ The velocity factor is the ratio of the velocity of propagation in the medium to that of the wave in gratuitous infinite (annotation this relationship is not explicitly given in the FE Reference Handbook), $$k = \frac{v}{c}$$ Substituting $thou$ into the first equation yields, $$\lambda = \frac{kc}{f}$$ Now substitute all known values to find the wavelength, $$\lambda = \frac{(0.seven)(three \times 10^viii {\frac{\si{yard}}{\si{s}}})}{2 \times ten^ix \si{Hz}} = .105\si{m} $$
8) A pair of line of sight communication antenna towers are to exist constructed, each at a height of $100\si{m}$.
The radio horizon is divers equally the distance for which propagating electromagnetic waves are locally tangent to the Earth's surface.
For an antenna in nominal weather atmospheric condition this is, $$d_{\si{km}} = iv.ane\sqrt{h_\si{thousand}}$$ Where $h_\si chiliad$ is the antenna acme and $d_{\si{km}}$ is the radio horizon in $\si {km}$.
What is the minimum propagation filibuster if the two towers are at the maximum separation from one another?
A.$273.33 \mu \si{south}$
B.$341.15 \mu \si{s}$
C.$120.55 \mu \si{s}$
D.$65.23\mu \si{southward}$
The correct answer is A.
Explanation:
From the given definition of the radio horizon, and the fact these are antennas which must run into 1 another in order to communicate, we must account for the Earth'due south curvature.
Since the definition of the radio horizon is such when the propagating waves are tangential to the Globe's surface, then the maximum altitude separation comes from summing the ii antenna radio horizons together, every bit shown in the figure.
Now, summate the maximum separation, $d$, and remember that the formula given in the trouble argument is that $d$ is in $\si{km}$ if the antenna pinnacle is in $\si one thousand$, $$d = 4.1\sqrt{100} + 4.1\sqrt{100} \\ d = 41 \si {km} + 41\si {km}\\ d = 82\si{km}$$ The propagation filibuster is the fourth dimension it takes for the electromagnetic wave to arrive at the receiver after being sent by the transmitter. Refer to the Delays in Calculator Networks section in the Atomic number 26 Reference Manual. The propagation speed is the speed of low-cal, $c$, and the distance is the separation of the antenna towers. $$\text{distance} = \text{speed} \times \text{fourth dimension}$$ Thus the minimum propagation delay is that which merely accounts for the moving ridge travel time, ignoring all other electric or atmospheric effects. $$\frac{d}{c} = t_{\text{delay}}$$ $$t_{\text{filibuster}} = \frac{8.2 \times 10^four \si{m}}{3 \times x^8 \frac{\si{m}}{\si{s}}} = 0.00027333\si{s}$$ $$t_{\text{delay}} = 273.33 \times 10^{-half-dozen} = 273.33 \mu \si{s}$$
ix) Simplify the following expression: $$(\overline{AB})(\bar A +B)(\bar B + B)$$
A.$\bar A $
B.$\bar B$
C.$\bar A \bar B$
D.$\bar A + \bar B$
The correct respond is A.
Explanation:
Refer to Fe Reference Handbook: Logic Operations and Boolean Algebra
Begin by simplifying with the complement police force on $(\bar B + B)$, $$\bar B + B = 1$$ The reduced expression is, $$(\overline{AB})(\bar A +B)$$ Apply De Morgan'south Law to suspension $\overline{AB}$. We practice 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, even so, volition 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 police that a variable ANDed with the Not of itself is 0 (i.east. $\bar B B = 0 )$ $$(\bar A \bar A + \bar A B + \bar A \bar B)$$ Utilize the identity that $A \cdot A = A$, $$(\bar A + \bar A B + \bar A \bar B)$$ Factor out $\bar A$, $$\bar A (i + B + \bar B)$$ Apply complement law one more time to finally obtain, $$\bar A $$
10) A balanced three phase Y connected source has, using a positive phase sequence, $\vec V_{an} = 120\angle30^\circ \si{V_{rms}}$. Detect the $\vec V_{bn}$.
A.$120\angle 30^\circ \si{V_rms}$
B.$ninety\angle -xc^\circ \si{V_rms}$
C.$120\angle 90^\circ \si{V_rms}$
D.$120\angle -ninety^\circ \si{V_rms}$
The correct reply is D.
Explanation:
Refer to the Balanced Three-Stage (3-φ) Systems department in the Electric and Calculator Applied science affiliate of the FE Reference Handbook.
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, nosotros 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 betwixt $\vec V_{bn}$ and $\vec V_{an}$ and that the magnitude of the voltages are equal, we can compute $\vec V_{bn}$ directly from the phasor expression, $$\vec V_{bn} = |V_{an}|\angle 30^\circ - 120^\circ = 120\angle -90^\circ \si{V_rms}$$
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