MBS3129 Electrical Principles

Tutorial 2A Solutions

1. Determine the period and frequency for the waveform of the below figure. How many cycles are
shown?
Ans:
1
∴f = =
T = 8s 8 μs 125 kHz
2 cycles are shown

2. For the waveform of the below figure, determine

a. period; b. frequency; c. peak-to-peak value
Ans:
1
a. 5 cycles in 525s. Thus, 5.25T = 525s and T = 100s
4
1 1
b. f = = =10 kHz
T 100 μs
c. Peak-to-peak = 2(75V) = 150V

3. Determine f, T, and amplitude for each of the following equation:

a. v = 75 sin 200πt V b. i = 8 sin 300t A
Ans:
a. ω = 200π rad/s b. ω = 300 rad/s
by 2πf = 200π by 2πf = 300
f =200π/2π =100 Hz f =300/2π =47.75 Hz
1 1 1 1
T= = =10ms T= = =21ms
f 100 f 47 . 75
Amplitude Vm = 75V Amplitude Im = 8A

4. Given v = 5 sin(ωt + 45º) V. If ω = 20π rad/s, what is v at t = 20 ms and 90 ms?

Ans:
180°
v(t =20ms) = 5 sin(20π x20x10-3x + 45º) = 5 sin(72º+ 45º) = 4.455V
π
180°
v(t =90ms) = 5 sin(20π x90x10-3x + 45º) = 5 sin(324º+ 45º) = 0.782V
π

5. Given i = 47 sin 8260t mA, determine current at t = 80s and 1200s.

Ans:
180°
i(t = 80s) = 47 sin(8260 x 80x10-6 x ) = 47 sin(37.86º) = 28.85 mA
π
180°
i(t = 1200s) = 47 sin(8260 x1200 x10-6 x ) = 47 sin(567.92º) = -22 mA
π

6. Write the equation as a function of time for the following waveforms. Express the phase angle in
degrees.

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(c)
Ans:
2π
T =10 μs ∴ ω= =200 π x 103 rad/s
(a) T
θ=1800 −40 0=140 0 . Thus,
v=80sin(200 π x 10 3 t−1400 )V
1 1
T= = =1200 μs
(b) f 833 .3
θ= (
160 μs
1200 μs )
(360 0 )=48 0 ;ω=2 πf =5236 rad/s
Thus, v = 100 sin(5236t + 48º)V

(c) θ= 90º − 54º = 36º

Φ = 360º − 36º = 324º
1620
t= x 36º = 180s
324
So period T = 1620s+180s = 1800s
2π 2π
ω= = =3490.66 rad/s
T 1800 μ
Thus, v = 100 sin(3490.66t + 36 º)V

7. Write the equation as a function of time for the following waveforms. Express the phase angle in
degrees.

Ans:
π 180
(a) i = 5 sin(1000t + × ) mA
5 π
i = 5 sin(1000t + 36º) mA

2π
(b) i = 10 sin( t
−3 + 180º − 60º) A
50× 10
i = 10 sin(125.66t + 120º) A
°
π 180
(c) v = 40 sin(2π900t − × )
4 π
v = 40 sin(5654.87t − 45º) V

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8. Sketch the following waveforms with the horizontal axis scaled in degrees and seconds:
o o
a. v =100 sin(232 .7 t +40 ) V b. i =20 sin(ωt −60 ) mA, f = 200Hz

Ans:

3.333

9. Sketch the following waveforms with the horizontal axis scaled in degrees:
a. v1 = 80 sin(ωt + 45º) V c. i1 = 10 cos ωt mA
b. v2 = 40 sin(ωt − 80º) V d. i2 = 5 cos(ωt − 20º) mA
Ans:

10. Consider the voltages of Question 9:

a. Sketch the phasors for v1 and v2.
b. What is the phase difference between v1 and v2?
c. Determine which voltage leads and which lags.
Ans:
a.

b. 125º
c. v1 leads

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11. With the aid of phasors, sketch the waveforms for each of the following pairs and determine the
phase difference and which waveform leads:
a. i = 40 sin(ωt + 80º), v = -30 sin(ωt − 70º)
b. v = 20 cos(ωt + 10º), i = 15 sin(ωt − 10º)
(Given: cos(t + ) = sin(t +  + 90) )
Ans:

12. For each of the following phasors, determine the equation for v or i as a function of time, and
sketch the waveform.

Ans:
(a) v = Vm sin(ωt + 70º)
v = 100 sin(ωt + 70º) V

(b) i = Im sin(ωt + 180º - 40º)

i = 20 sin(ωt + 140º) A (a) (b)

(c) i = Im sin(ωt - 35º)

i = 10 sin(ωt - 35º) mA

13. For the below waveforms, determine the phase differences. Which waveform leads? Draw the
phasors for the waveforms.

Ans:

(a) (b)

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 Phase difference = 90º A leads B Phase difference =150º A leads B

14. Determine the effective (rms) values of each of the following:

o
a. v =100 sin ωt V; b. i =8 sin 377 t A; c. v =40 sin (ωt +40 ) V; d. i =120 cos ωt mA

Ans:
a. 100Vx0.707 = 70.7 V b. 8A x0.707 = 5.66 A
c. 40V x0.707 = 28.3 V d. 120mAx0.707 = 84.8 mA

15. Compute the rms value for the below figure:

Ans:

I eff =
√ Area under squared curve
Base

√
(22 x 3 )+0+(22 x 1 )+(( -3 )2 x 1 )+(32 x 2)+0
= 10
= 2.07A

End

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MEC3129 Electrical Principles
Tutorial 2B Solutions:

Q1. A pure inductance of 1.273 mH is connected in series with a pure resistance of 30 Ω. If the
frequency of the sinusoidal supply is 5 kHz and the p.d. across the 30 Ω resistor is 6 V,
determine the value of the supply voltage and the voltage across the 1.273 mH inductance.
Draw the phasor diagram.

Ans:
The circuit is shown in Fig. Q(1a).
Fig. Q(1a)
Supply voltage, V = IZ
V 6
Current I = R = =0.20 A
R 30
Inductance reactance XL = 2πf L = 2π (5 x 103)(1.273 x 10-3)
= 40 Ω
Impedance, Z = √ ( R + X )= √( 30 + 40 )= 50Ω
2 2
L
2 2

Supply voltage V = IZ = (0.20)(50) = 10V

Voltage across the 1.273 mH inductance, VL = IXL = (0.2)(40) = 8 V
The phasor diagram is shown in Fig. Q(1b), take I as a reference. Fig. Q(1b)

Q2. A metal-filament lamp, rated at 750 W, 100 V, is to be connected in series with a capacitor
across a 230 V, 60 Hz supply. Calculate:
(a) The capacitance required;
(b) The phase angle between the current and the supply voltage.
(c) Draw the phasor diagram.

Ans:
(a) The voltage VR across R is in phase with the current I, while the
voltage VC across C lags I by 90°. The resultant voltage V is the phasor sum of VR and VC,
V2 = VR2 + VC2
∴ (230)2 = (100)2 + VC2
∴ VC = 207 V
750W
Rated current I of lamp = = 7.5 A
100 V
V 1
By Xc = C =
I 2 πfC
2πf CVC = I
2 x 3.14 x 60 x C x 207 = 7.5
C = 96 x 10-6 F = 96µF
(c) Phasor diagram
(b) If Φ is the phase angle between the current and the supply voltage
From the phasor diagram,
V
cos Φ= R
V
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 100
= = 0.435
230
Φ= 64.23

Q3. A circuit having a resistance of 12 Ω, an inductance of 0.15 H and a capacitance of 100 μF in

series, is connected across a 100 V, 50 Hz supply. Calculate:
(a) The total impedance;
(b) The current;
(c) The voltages across R, L and C;
(d) The phase angle between the current and the supply voltage.

Ans:
(a) Total impedance:
√
Z = { R 2+ ( X L −X C )2 }

√√ { ( )}
2
Z = (12 )2 + 2 ×3.14 ×50 × 0.15− 106
2 ×3.14 × 50× 100
= {144+ ( 47.1−31.85 )2 } = 19.4 Ω
V 100
(b) Current = = = 5.15 A
Z 19.4
(c) Voltage across R = VR = IR = 5.15 x 12 = 61.8 V
Voltage across L= VL = IXL = 5.15 x 47.1 = 242.5 V
Voltage across C = VC = IXC = 5.15 x 31.85 = 164V

(d) Phase angle between current and supply is

−1 V R −1 61.8
∅ =cos =cos =¿ 51.83 Phasor diagram
V 100

Q4. A coil having a resistance of 6 Ω and an inductance of 0.03 H is connected across a 50 V, 60

Hz supply. Calculate:
(a) The current;
(b) The phase angle between the current and the applied voltage;
(c) The apparent power;
(d) The active power.

Ans:
(a) The phasor diagram for such a circuit is given in Fig. Q(4a).
Reactance of circuit = 2πfL = 2 ×3.14 ×60 ×0.03
= 11.31 Ω
Impedance = √ {6 +(11.31) } = 12.8 Ω
2 2

50
And current =  = 3.91 A
12.8
X 11.31
(b) Tan∅ = L = =¿ 1.885
R 6
∅ = 62.05
(c) Apparent power S = 50 × 3.91 = 195.5 VA
(d) Active power P = apparent power × cos Φ
= 195.5 × cos 62.05 = 92 W
Fig. Q(4a)
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Alternatively:
Active power P = I2 R = (3.91)2 × 6 = 92 W

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Q5. A 20 Ω resistor is connected in parallel with an inductance of 2.387 mH across a 60 V, 1 kHz
supply. Calculate (a) the current in each branch, (b) the supply current, (c) the phase angle
between the supply current and the supply voltage, (d) the circuit impedance, and (e) the power
consumed.

Ans:

Phasor diagram

Q6. A 30µF capacitor is connected in parallel with an 80Ω resistor across a 240 V, 50 Hz supply.
Calculate (a) the current in each branch, (b) the supply current, (c) the phase angle between the
supply current and the supply voltage, (d) the circuit impedance, (e) the power dissipated, and
(f) the apparent power.

Ans:

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Q7. Three branches, possessing a resistance of 50Ω, an inductance of 0.15 H and a capacitance of
100 µF respectively, are connected in parallel across a 100 V, 50 Hz supply. Calculate:
(a) the current in each branch;
(b) the supply current;
(c) the phase angle between the supply current and the supply voltage.

Ans:
(a)

(b)

Phasor diagram

(c)

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Q8. A coil of inductance 159.2 mH and resistance 40Ω is connected in parallel with a 30µF
capacitor across a 240 V, 50 Hz supply. Calculate (a) the current in the coil and its phase angle,
(b) the current in the capacitor and its phase angle, (c) the supply current and its phase angle,
(d) the circuit impedance, (e) the power consumed, (f) the apparent power, and (g) the reactive
power. Draw the phasor diagram.

Ans:

Phasor diagram

(c) The current ILR and IC may be resolved into their horizontal and vertical components. The
horizontal component of ILR is
ILR cos(5120’) = 3.748 cos 5120’ = 2.342 A
The horizontal component of IC is IC cos 90 = 0
Thus the total horizontal component, IH = 2.342 A
The vertical component of ILR = − ILR sin(5120’)
= −3.748 sin 5120’
= −2.926 A
The vertical component of IC = IC sin 90°
= 2.262 sin 90° = 2.262 A
Thus the total vertical component, IV = −2.926 + 2.262
= −0.664 A
IH and IV are shown in Figure in right side, from which,
I =√ [(2.342)2 + (−0.664)2] = 2.434 A
0.664
Angle ϕ = tan-1 = 15.83 = 1550’ lagging
2.342
Hence the supply current I = 2.434 A lagging V by 1550’.
V 240
(d) Circuit impedance, Z = = = 98.60 Ω
I 2.434
(e) Power consumed, P = VI cos ϕ =( 240)(2.434) cos 1550’
= 562 W
(Alternatively, P = IR R = ILR2R (in this case)
2

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 = (3.748)2(40) = 562 W
(f) Apparent power, S = VI = (240)(2.434) = 584.2 VA
(g) Reactive power, Q = VI sin ϕ = (240)(2.434)(sin 1550’)
= 159.4 VAR

Q9. Consider the below circuit. (a) Calculate the values of C and R for the circuit to have a resonant
frequency of 580 kHz and a bandwidth of 10 kHz.
(b) Use the above values to determine the power dissipated by the circuit at resonance.
(c) Find the vout across the inductor L.
Ans:
1
fs=
(a) By 2 π √ LC
1
=> C = = 7.53x10-12 F =
4 π ( 580 k ) ( 10× 10−3 )
2 2

7.53pF
X C =X L=2 π (580 k )(10×10−3 )=36 . 442 k
ωs ωs R
fs = = = ( rad / s )
BW = ( Hz) Qs ωs L L
By Qs R
580 k 36 .442 k
Q= =58 R=
Given fs = 580k, BW = 10k => 10 k => 58 = 628.3
(b) Erms = 0.707Ep = 0.707x 35.4 = 25 V
2 2
E 25
By P = = = 0.995 W
R 628.3
(c) By vout = VL = QSE = 58 x 25 = 1450V

Q10. For the below circuit, determine:

(a) The resonant frequency expressed as ω (rad/s) and f (Hz).
(b) Total impedance at resonance.
(c) Current at resonance.
(d) VL and VC.
(e) Reactive powers, QC and QL.
(f) Quality factor of the circuit, QS.
Ans:
1
(a) ω S=
√ LC
1
¿ = 10 000 rad/s
√(10 m)(1 μ)
ω 10000
f S= = = 1592 Hz
2π 2π
(b) Since at resonance, XL = XC
Total impedance Z= √ { R +( X −X ) }
2
L C
2

Z= √ {8 +( 0 ) } = 8Ω
2 2

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 E 10
(c) I = = = 1.25A
Z 8
(d) Reactance XL = ωL = 10 000 x 10m = 100Ω
VL = (100)(1.25) = 125 V
VC = (100)(1.25) = 125 V

(e) QL = (1.25)2(100) = 156.25 VAR

QC = (1.25)2(100) = 156.25 VAR
Q L 156.25
(f) QS = = = 12.5
P 1.252 ×8

Q11. Refer to the series resonant circuit. Suppose the circuit has a resonant frequency of 600 kHz
and a quality factor of QS = 60.
(a) Determine the value of inductor L in henries.
(b) Calculate the value of resistor R in ohms.
(c) Find I, VL, and power, P, at resonance.
Ans:
1
fs=
(a)By 2 π √ LC
1
=> L = = 3.20x10-4F = 320 H
4 π ( 600 k ) ( 220 ×10−12 )
2 2

XL
(b) By QS =
R
2 πfL 1206.4
=> R= = = 20.1 
60 60
E 20 m
(c) I= = = 0.995 mA
R 20.1
VL = I XL = 0.995m x 1206.4 = 1.2V
Or VL = QSE = 60 x 20m = 1.2V
P = (0.995 m)2 x 20.1 = 20W

End

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