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Ex. lesson 5
Informatics Engineering 2014-2015
1
F.F.I. — Degree on Informatics Engineering. Course 2014-2015. — Exercises for lesson 5
Alternating current voltage sources
1. An AC voltage source is made out of a rotating coil as shown in the figure. The coil have radius R, N turns,
~ shown is uniform and constant.
and rotates at constant angular velocity ω, measured in rad/s. The magnetic field B
~
If at t = 0 B is at an angle δ with the perpendicular to the coils, obtain: (a) the magnetic flux, Φ(t), through the
coil and the induced EMF ε(t); (b) the maximum and effective values ε0 and εe of ε(t) if N = 200, R = 1 cm,
B =0.5 T and the coil rotates at 50 revolutions per second.
Sol.: (a) Φ(t) = N BπR2 cos(ωt + δ) and ε(t) = N BπR2 ω sen(ωt + δ); (b) ε0 ' 9.87 and εe ' 6.98 V.
Phasors
2. Obtain the phasors associated with the following harmonic functions both in rectangular and polar forms:
8 cos(ωt), 6 cos(ωt +√π/3), 4 cos(ωt + 2π/3),
2 cos(ωt − 3π/4),
2√cos(ωt + π) and 8 sin(ωt + π/3). √
√
√
Sol.: 8 = 86 0o , 3 + 3 3j = 66 60o , −2 + 2 3j = 46 120o , − 2 −
86 − 30o .
2j = 26 − 135o , −2 = 26 180o and 4 3 − 4j =
3. If the angular frequency
if ω, find the
√ harmonic functions associated with the following phasors: 2, -7, −3j,
√
(1 + j), (−1 − j), ( 3 + j) and (2 − 12j).
√
√
Sol.: 2 cos(ωt), 7 cos(ωt + π), 3 cos(ωt − π/2),
2 cos(ωt + π/4),
2 cos(ωt − 3π/4), 2 cos(ωt + π/6) and 4 cos(ωt − π/3).
4. Consider the junction shown in the figure. Three currents are known: I1P
(t) = 2 cos(ωt) A, I2 (t) = 4 cos(ωt +
π/3) A and I3 (t) = 8 cos(ωt + 2π/3) A. Using Kirchhoff’s current law
Ii (t) = 0, and using phasors obtain
the current I4 (t).
√
√
√
√
Sol.: Ie1 = 2, Ie2 = (2 + 2 3j), Ie3 = (−4 + 4 3j) and Ie4 = −(Ie1 + Ie2 + Ie3 ) = −6 3j, then I4 (t) =6 3 cos(ωt − π/2) A.
Impedance
5. Obtain the impedance of a parallel association of an inductor with L = 12 mH and a capacitor with C = 250 µF
if ω = 103 rad/s. If the AC effective voltage across its terminals is Ve = 36 V, obtain the effective current Ie through
the association and the frequency f in Hz.
Sol.: Z = (12j|| − 4j) = −6j Ω; Ie = Ve /|Z| = 36/6 = 6 A; f = ω/(2π) =159.15 Hz.
√
6. The voltage drop across the terminals of an association in series of two elements is V (t) = 4 2 cos(105 t +
π/4) V, and the current through it is I(t) = 2 cos(105 t) mA. Obtain the impedance of the association, the type of
each element (resistor, inductor or capacitor) and their corresponding magnitudes L, C or R.
Sol.: Z = (2 + 2j) kΩ, then they are a resistor with R = 2 kΩ and an inductor with L = 20 mH.
7. Through the association in series of an inductor with L = 10 mH and a capacitor with C =12.5 µF flows a
current with effective value Ie =0.5 A. If the effective voltage across the capacitor is VC,e =10 V, find the frequency
f in Hz and the effective voltage across the inductor VL,e and across the whole association Ve . Check that Ve 6=
VC,e +VL,e and explain why.
Sol.: From VC,e = Ie /(ωC), ω = 4 × 103 rad/s and f =636.62 Hz ; VL,e = ωLIe = 20V ; Ve = |Z|Ie = |XL − XC |Ie =
|40 − 20|0.5=10 V 6= 10 + 20 V. Because the voltages are not in phase and have to be added as vectors or complex numbers
and not as scalars.
Ex. lesson 5
Informatics Engineering 2013-2014
2
Circuit analysis
8. The angular frequency of the circuit shown in the figure is ω = 104 /3 rad/s. (a) Applying Kirchhoff’s laws
obtain the three currents through the branches and draw the phasor diagram. (b) Obtain the total impedance as seen
from the source terminals and use it to calculate the current through the source, compare it with the result for the
previous question. (c) Obtain L and C. √
Sol.: (a) I1 (t) = 4 cos(104 t/3) A, I2 (t) = 2 5 cos(104 t/3 − 0.464) A, I3 (t) = 2 cos(104 t/3 + π/2) A. (b) Z =7.5 Ω,
I1 (t) = 4 cos(104 t/3) A which the same as in the previous result. (c) L =0.9 mH, C = 20 µF.
9. In the circuit shown with f =1 kHz (a) obtain the impedance seen from the source terminals A–B and the
current I(t) through the source √
(b) Obtain the impedance or capacitance of each element in the circuit
Sol.: (a) Z = 12 + 12j Ω; I(t) =
2 cos(2π × 103 t − π/4) A; (b) 1.91 mH, 0.796 mH, 13.26 µF and 5.31 µF.
10. The angular frequency in the circuit shown is ω = 103 rad/s. Obtain: (a) the current Ii (t) through each branch
and the phasor diagram; (b) the voltage drops VAB (t) and VBC (t); (c) the average or active power delivered by the
source and consumed
by the resistor, checking√that they are the equal.
√
√
Sol.: (a) I1 (t) = 2 cos(103 t − π/4) A, I2 (t) = 2 cos(103 t + 3π/4) A and I3 (t) = 2 2 cos(103 t − π/4) A;
√
(b) VAB (t) = 4 2 cos(103 t + π/4) V, VCB (t) = 8 cos(103 t + π) V; (c) Pε = PR = 4 W.
11. The circuit shown works with an angular frequency ω = 104 rad/s: (a) obtain the current Ii (t) through
each branch and draw the corresponding phasor diagram; (b) obtain VAB (t) and its effective value through three
different paths; √
(c) obtain the power delivered and consumed, checking the power
balance.
√
Sol.: (a) I1 (t) = 40 cos(104 t + arctan(3)) mA, I2 (t) = 2 cos(104 t) mA, I3 (t) = 52 cos(104 t + arctan(3/2)) mA;
(b) VAB (t) = 8 cos(104 t − π/2) V, VAB,e =5.66 V; (c) Active power delivered by the sources: P1 = 8 mW, P2 = 16 mW,
average power consumed by the resistors: P(R=1) = 20 mW, P(R=2) = 4 mW; Balance: (8 + 16) mW= (20 + 4) mW.
12. For the circuit shown ω = 400 rad/s. Obtain: (a) Z̃ab ; (b) I(t); (c) the active power delivered and the average
power consumed checking the balance; (d)(*) what element should be connected between the points a and b if we
want that the current and voltage at the source to be in phase. Obtain the new current and powers, checking that
they are the same as previously.
Sol.: (a) Zab = (5 + 5j) Ω; (b) I(t) = 44 cos(400t − π/4) A; (c) Pa = PR = 4840 W; (d) A capacitor with C 0 = 250 µF,
√
I 0 (t) = 22 2 cos(400t) A, the new powers are Pa0 = PR0 = 4840 W.
Resonance and changing frequency
13. A radio receiver is tuned to detect the signal emitted by a radio station. The tuning circuit can be considered
as a series RLC circuit– uses a capacitor with C =32.3 pF and a inductor with L=0.25 mH. Obtain the frequency
of the radio signal.
Sol.: f =1.77 MHz.
14. A coil with L =0.1 H is connected in series to a resistor with R = 10 Ω and a capacitor. The resonance
frequency is f =60 Hz. Obtain C and the voltage drop across each element if the source voltage has a maximum
value V0 =100 V (note: use phase 0o at the source).
Sol.: C = 70,4 µF; VR = 100 cos(120πt), VC (t) = 120π cos(120πt−π/2) and VL (t) = 120π cos(120πt+π/2) = −VC (t).
15. (a) Consider a series RC circuit connected to an AC source with fixed effective voltage Ve , and changing
angular frequency ω. Obtain the effective voltage across C as a function of ω VC,e (ω), and represent if graphically
(b) Repeat for a seriespRL circuit.
p
Sol.: (a)VC,e (ω) = Ve /
1 + (RCω)2 . (b)VL,e (ω) = Ve /
1 + (R/(Lω))2 .
Ex. lesson 5
Informatics Engineering 2013-2014
3
Figures for Lesson 5
I2(t)
Prob. 1
+
I3
-
-15jW
30 0o V
24
3jW
o
0 V
8 0o V
I1
4W
+
2 mH
A
I3(t)
6W
12jW
+
5jW
Prob. 9
I2 +e2= 16 0oV
I1
I2
I3
-30jW
-12jW
B
Prob. 8
C
Prob. 4
A
I2 6W
I1
I4(t)
I1(t)
B
250 mF
e 1= 8
Prob. 10 B
+
0V
1kW
-
I3
o
50nF
100 mH
2 kW
Prob. 11
a
I(t)
Ve =220 V
5W
+
0.01 H
125 mF
b
Prob. 12
A
400 mH
B
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