MEC3123 Fundamentals of Electricity and Electronics
Tutorial 4b 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.
Solution:
The circuit is shown in Fig. Q(1a).
Supply voltage, V = IZ
Fig. Q(1a)
Fig. Q(1b)
Fig. Q(1b), take I as a reference.
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.
Solution:
(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
Rated current I of lamp =
By Xc =
= 7.5 A
=
2πfCVC = I
2 x 3.14 x 60 x C x 207 = 7.5
C = 96 x 10-6 F = 96µF
Phasor diagram
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(b)
If ϕ is the phase angle between the current and the supply voltage
From the phasor diagram,
cos ϕ =
=
= 0.435
ϕ = 6413
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 difference between the current and the supply voltage.
Solution:
(a) Total impedance:
(b) Current =
=
= 5.15 A
Phasor diagram
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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.
Solution:
(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 =
and Current =
(b)
{62 + (11.31)2} = 12.8 Ω
= 3.91 A
tan
= 623
(c) Apparent power S = 50 × 3.91 = 196 VA
(d) Active power = apparent power × cos ϕ
= 195.5 ×0.469 = 92 W
Alternatively:
Active power = I2R = (3.91)2 × 6 = 92 W
Fig. Q(4a)
Q5. A circuit consists of a 115 Ω resistor in parallel with a 41.5 µF capacitor and is connected to a
230 V, 50 Hz supply. Calculate:
(a) the branch currents and the supply current;
(b) the phase angle between the supply current and
the supply voltage.;
(c) the circuit impedance.
Solution:
(a)
(b)
(c)
Phasor diagram
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Q6. 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.
Solution:
(a)
(b)
Phasor diagram
(c)
Q7. 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.
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Solution:
Phasor diagram
Q8. 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.
Solution:
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Q9. 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.
Solution:
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
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IH and IV are shown in Figure in right side, from which,
I = [(2.342)2 + (−0.664)2] = 2.434 A
= 15.83 = 1550’ lagging
Angle ϕ = tan-1
Hence the supply current I = 2.434 A lagging V by 1550’.
(d) Circuit impedance, Z =
= 98.60 Ω
=
(e) Power consumed, P = VI cos ϕ =( 240)(2.434) cos 1550’
= 562 W
2
(Alternatively, P = IR R = ILR2R (in this case)
= (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
Q10. 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. [7.53 pF 628.3
(b) Use the above values to determine the power dissipated by the circuit at resonance.
[0.995W]
3
(c) Solve for  out at resonance. [ vout  2051sin(1160 10 t  900 ) ]
Solution:
(a) By fs 
1
2 LC
=> C 
1
 7.53 pF
4 (580k ) 2 (10 10 3 )
2
X C  X L  2 (580k )(10 10 3 )  36.442 k
By
BW 

s
fs
(Hz)  s 
Qs s L
Qs
Given fs = 580k, BW = 10k
(b) Erms = Ep / 2
=> E 
=> Q 
35.4
 25 V
2

R
R
(rad / s)
L
580k
36.442k
 58 => R 
= 628.3
58
10k
;
By P=V2/R
 36.442 103 90 0 
(25)  145090 0 ;
(c) Vout  
628.3


=>
P
252
 0.995 W
628.3
vout  2051sin(1160 103 t  900 )
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Q11. Refer to the below circuit: (a) Determine the apparent power for each box.
[Top Left: 1476VA, Right 1KVA;
Mid: 100VA;
Bottom Left: 80VA, Right 1206VA]
(b) Sum the apparent powers that you just computed. Why does the sum not equal 1803 VA?
[Total 3862VA]
Solution:
(a)
S1  P12  Q12  1476VA; S 2  1000VA
S3  100VA; S4  80VA; S5  1206VA
(b)
S1  S 2  S3  S 4  S5  3862VA
Actual ST  1803VA
 ST  S1  S 2  S3  S 4  S5
Because it is a vector sum, not arithmetic sum
Q12. A motor with an efficiency,  =87% supplies 10 hp to a load as showed in the below circuit.
Its power factor is 0.65 (lag). (1 horse power (hp) = 746W).
(a) What is the power input to the motor?
[8575W]
(b) What is the reactive power to the motor?
[10026 VAR]
(c) Draw the motor power triangle. What is the apparent power to the motor?
[13193 VA]
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Solution:
(a)
P0  (10hp)(746W/hp)  7460W
Pin 
(b)
Po


7460W
 8575W
0.87

  cos 1 (0.65)  49.5o
Sin
Qin  Pin x tan  8575  tan 49.5o  10026VAR
(c)
S2
Q2
Qin – Q2
Qin
Sin  Pin2  Qin2  85752  10026 2  13193VA
(d) To improve the power factor (p.f.) to 0.9 by adding capacitors, find the capacitive KVAR
required:
θ2 = cos-1 0.9 = 25.80 ; S2 = Pin / 0.9 = 8575/ 0.9 = 9527.8 VAR ;
Q2 = S2 sin25.8 = 4146.8 VA
Capacitive KVAR required = Qin – Q2 = 10026 – 4146.8 = 5879.2VA = 5.8792 KVAR
End
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