Module NF2FSLT SP025

Solution Non Face-to-face SLT
(a) Mutual inductance of the coils:

Using:

Coil X Coil Y

Non Face-to-face SLT

Solution

(b) magnitude of the emf induced in coil X:
Value of M for both coil is same

Using:

254

Non Face-to-face SLT

LO 5.5 Exercise 106

Primary coil of a cylindrical former with the length of 50 cm
and diameter 3 cm has 1000 turns. If the secondary coil
has 50 turns, calculate :

(a) its mutual inductance
(b) the induced emf in the secondary coil if the current

flowing in the primary coil is decreasing at the rate of
4.8 A s–1.

Solution

Given : N1 = 1000 turns ; l = 50×10–2 m,
d1 = 3×10–2 m, N2 = 50 turns and

255

Non Face-to-face SLT

Solution

(a) Mutual inductance

Given : N1 = 1000 ; l = 50×10–2 m ; d1 = 3×10–2 m
N2 = 50 and

Using :

256

Solution Non Face-to-face SLT

(b) The induced e.m.f in secondary coil

257

LO 5.5 Exercise 107 Non Face-to-face SLT

A current of 3.0 A flows in coil C and produces a
magnetic flux of 0.75 Wb in it. When a coil D is moved
near to coil C coaxially, a flux of 0.25 Wb is produced in
coil D. If coil C has 1000 turns and coil D has has 5000
turns.
a. Calculate self-inductance of coil C and the energy

stored in C before D is moved near to it.
b. Calculate the mutual inductance of the coils.
c. If the current in C decreasing uniformly from 3.0 A to

zero in 0.25 s, calculate the induced emf in coil D.

Solution
Given :

258

Non Face-to-face SLT

Solution
a. Calculate self-inductance of coil C and the energy stored in

C before D is moved near to it.

259

Non Face-to-face SLT

Solution

(b) The mutual inductance of the coils is given by

260

Non Face-to-face SLT

Solution

(c) Induced e.m.f in coil D when current in C decreasing
uniformly from 3.0 A to zero in 0.25 s

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TOPIC 6
ALTERNATING CURRENT

NF2F SLT

262

Exercise 108

(1) Copy and complete the table below.

Magnitude of Alternating Direct
current current current

Direction of Vary with time Remain
current constant
Change
periodically Always flow
from positive to
negative

263

Exercise 109

(i)

  2

T

 2

0.02

 100 rad s1

(ii) I  3sin(100t)

264

Exercise 110

An AC voltage source has an output of :

V  200 sin 2 f t

This source is connected to a 100 Ω resistor. Find the rms current
and maximum current in the resistor.

Solution

Applying Ohm’s Law :Vo  Io R
Io  Vo  200  2A
R 100

Using : I rms Io  2  1.4142 A
22

265

Exercise 111

Given : R  250 

By comparing V  500sin t to the V  V0 sin t

Thus the peak voltage is V0  500 V

a. By applying the formulae of rms current, thus

I rms  I0 and I0  V0
2 R

I rms  V0
R2

 500 Irms  1.4142 A
250 2

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b. The peak current of AC is given by

I rms  I0
2

1.4142  I0
2

I0  2.0 A

c. Mean power = Average power

Pav  I rms 2 R

 1.41422250

Pav  499.9904 W 267

Exercise 112

From the graph, R  10  10 3  I 0  0.02 A ; T  40  10 3 s

a. By applying the formulae of rms current, thus

I rms  I0 I rms  0.02
2 2
Irms  1.4142 102 A

b. The frequency of the AC is 1
40 10 3
f 1 f 
T

f  25 Hz

c. The mean power dissipated from the resistor is given by

Pav  I rms 2 R

    1.4142 102 2 10103

Pav  2.0 W 268

Exercise 113

A heating element is connected to an a.c voltage source of
an r.m.s. voltage of 110 V. If an r.m.s. current of 8.0 A occurs,
calculate the resistance and the average power input in the
heating element.

Solution

Given : Vrms = 110 V, Irms = 8 A , R = ? ; Pav = ?

R  Vrms  110  13.75 
I 8rms

Pav  I R2  82 (13.75)  880 W
rms

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Exercise 114

A 15.0 μF capacitor is connected to a 120 V, 60 Hz source.
What are (a) the capacitive reactance and (b) the current
(rms & peak) in the circuit ?

Solution Find : (a) XIoC,
(b)
Given : C = 15.0 μF = 15x10–6 F Irms
Vrms = 120 V
f = 60 Hz

(a) The capacitive reactance :

XC  1 C  1  176.8388

2 f 2 (60) (15 10 6 )

270

(b) The rms current :

V C  I X c  V rms  I Xrms c

Vrms 120  0.6786 A
I rms
XC 176.8388

The peak current :

I rms Io  Io  Irms 2 2
2  0.6786

 0.9597 A

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Exercise 115

A 125 mH inductor is connected to a 120V, 60 Hz source.
What are (a) the inductive reactance and (b) the rms
current in the circuit ? (c) If the frequency is doubled, what
happens to the inductive reactance and the current ?

Solution

Given : L = 125 mH = 125×10–3 H Find : (a) XIrmL s
Vrms = 120 V (b)

f = 60 Hz

(a) The inductive reactance :

XL  2 f L 272
 2 (60) (125  10 3 )  47.1239

(b) The rms current : 273

VL  I XL
Ir ms Vrms  120  2.5465A

XL 47.1239

(c) If the frequency is doubled :

From : X L  2fL

 XL is proportional to f .
Thus when f is doubled, XL is doubled.

From : Irms Vrms
XL

 Current is inversely proportional to XL.
When XL doubled, current is halved.

Exercise 116

A 2 F capacitor and a 1000  resistor are placed in series
with an alternating voltage source of 12 V and frequency of
50 Hz. Calculate
a. the current flowing,
b. the voltage across the capacitor,
c. the phase angle of the circuit.

Solution

Given : C  2 106 F; R  1000 ;V  12 V; f  50 Hz

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(a) Capacitive reactance

 1  1

2fC 50 2 106
XC  XC 2

X C  1.5915103 

and the impedance of the circuit

Z  R2  XC2  Z  10002  1.5915103 2

Z  1.8796103 

The current flowing in the circuit

V  IZ

 12  I 1.8796103

I  6.3843103 A 275

(b) Voltage across the capacitor

  VC  IXC
 6.3843103 1.5915103
VC  10.1606 V

(c) Phase angle between the current and supply voltage

tan  X C   tan1 X C 
R
R

 tan 1  1.5915 103 
1000

  57.8573

276

Exercise 117

An AC current of angular frequency of 377 rad s1 flows
through a 425  resistor, a 3.50 F capacitor and a 1.25 H
inductor which are connected in series. If the peak voltage of
the circuit is 150 V, determine the

(a)impedance of the circuit .
(b)rms current in the circuit.
(c)phase angle between the current & total voltage. Sketch
the phasor diagram of the circuit. Which leads ?
(d)resonant frequency.

277

(a) Capacitive reactance

XC  1  377 1 106 )  757.86 28
(3.50
C

Inductive reactance

XL   L  (377) (1.25)  471.25 

Impedance of the circuit

Z  R 2  ( XL  XC) 2

 (425)2  (471.25  757.8628)2  512.6128

278

(b) rms current

V  I Z  Io  Vo  150  0.2926 A
Z 512.6128

I rms Io  0.2926  0.2069A
2 2

(c) Phase angle   tan1 (471.25  757.8628)

tan   XL  XC 425

R   33.9951

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Phasor diagram
XL

ω

XL  XC  R

Z

XC

circuit is more capacitive than inductive,  is negative

 the current leads voltage by 33.9951º

280

(d) resonant frequency.

fr  1

2 LC

1

2 (1.25)3.50106

fr  76.0906Hz

281

Exercise 118

fr  1

2 LC

99.7 106  1

2 1.4 106 C

C  1.4 1 106 ]2

106[(2 )99.7

 1.8202 1012 F

282

Exercise 119

(a) XL  2fL  2 (600)(0.75)  2.8274103

At resonance, Xc  XL  2.8274 103 

(b) Value of C is given :

Xc  1

2fC

2.8274103  1

2 (600)C

C  9.3817 108 F
C  93.82nF

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(c) Z  R  200
(d) I  V  250  1.25A

Z 200

ω

V
I

284

Exercise 120

Given : R  1.15103 ; L  505103 H;
V  14.2 V; f  1250 Hz

(a) The inductive reactance is

 X L  2fL  2 1250 505103  3.9663103

Impedance of the circuit is

Z  R2  XL2
 1.151032  3.96631032

Z  4129.6532   4.13103

285

V  IR  I V  14.2  3.4383103 A
Z 4.13103

(b) Value rms current is reduced to half

I '  1 (3.4383103)  1.7192103 A
2

Z V  14.2  8.2597 103
I 1.7192 103

Z  R2  X L  X C 2

   8.2597 103  1.15103 2  3.9663103  XC 2 286

XC  12145.5508 

XC  1

2fC

12141.5508  2 1
(1250)C

C  10.4866109 F  10.4866 nF

287

Exercise 121

(a) Phasor diagram of the circuit is

VL V ω

VL VC   I

VC VR

From the phasor diagram, the applied voltage V is

V   VR2  VL VC 2  1532  314 1152

V  251.0179 V

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and the phase angle  is

tan  VL VC   tan 1  VL VC 
VR
VR
b. Given f  50 Hz  tan1 314 115 
 153 

  52.4453

Since R, C and L are connected in series, hence the
current passes through each devices is the same.
Therefore

289

VR  IR

153  I 26

I  5.8846 A

(c) The inductive reactance is

VL  IX L 314  5.8846X L

X L  53.3596 

Inductance of the inductor is

X L  2fL

53.3596  2 50L

L  0.1698 H

290

the capacitive reactance is

VC  IX C 115  5.8846X C

X C  19.5425 

the capacitance of the capacitor is

XC  1 19.5425  2 1

2fC 50C

C  1.6288104 F

(d) The resonant frequency is given by

 1 1
LC
2 0.16981.6288104
 fr 2

fr  30.2634 Hz 291

Exercise 122

(a) When f is varied, the impedance Z of the circuit decreases to a
minimum value (resonance) and then increases. Z is a minimum
when XL = XC, so that Z=R at resonance. Since the current
flowing in the circuit increases to a maximum and then
decreases, the brightness of the lamp increases to a
maximum at resonance and then decreases.

(b) (i) The resonant frequency is given by

 1 1
LC
2 0.4 0.4106
 fr 2

fr  397.8874 Hz

292

(ii) Maximum current

I  V  0.01  0.001A
R 10

(iii) Voltage across the capacitor

VC  IXC  I  1 

2fC

 0.001 2 1 106 ) 
(400)(0.4

 0.9947 V

293

Exercise 123 294

A 100 F capacitor, a 4.0 H inductor and a 35  resistor are
connected in series with an alternating source given by the
equation below:

V  520sin100t

Calculate:
a. the frequency of the source,
b. the capacitive reactance and inductive reactance,
c. the impedance of the circuit,
d. the peak current in the circuit,
e. the phase angle,
f. the power factor of the circuit.

ANS : (a) 15.9155 Hz (b) 100Ω ; 400Ω (c) 302.0348 Ω (d) 1.72A (e) 83.3456º
(f) 0.1159

By comparingV  520sin100t with V  V0 sint
Thus V0  520 V;   100 rad s1

(a) The frequency of AC source is given by

  2f 100  2f
f  15.9155 Hz

(b) The capacitive reactance is

 1  1  100 

2fC 2 15.9155100106
 XC

and the inductive reactance is

X L  2fL  2 15.91554.0  400 

295

(c) The impedance of the circuit is

Z  R2  X L  X C 2
 352  400 1002

Z  302.0348 

(d) The peak current in the circuit is

V0  I0Z

520  I0 302.0348

I0  1.7217 A

296

(e) The phase angle between the current and the supply

voltage is

tan  X L  X C   tan1 X L  X C 

R R

 tan1 400 100   83.3456
 35 

(f) The power factor of the circuit is given by

power factor  cos

 cos83.3456

power factor  0.1159

297

Exercise 124

A 22.5 mH inductor, a 105  resistor and a 32.3 F
capacitor are connected in series to the alternating
source 240 V, 50 Hz.
(a) Sketch the phasor diagram for the circuit.
(b) Calculate the power factor of the circuit.
(c) Determine the average power consumed by the

circuit.

298

Given : R  105 ; C  32.3106 F; L  22.5103 H
V  240 V; f  50 Hz

(a) The capacitive reactance is

XC  1  1

2fC 2 50 32.3106
 XC

X C  98.548 

The inductive reactance is

 X L  2fL

 2 50 22.5103

X L  7.0686 

299

Phasor diagram for the circuit is

XL R ω

XL  XC  

XC Z

(b) Impedance of the circuit is

Z  R2  X L  X C 2
 1052  7.0686  98.5482

Z  139.2605 

300

power factor of the circuit is

cos  R cos  105

Z 139.2605

cos  0.7540

(c) Average power consumed by the circuit

Pav  IV cos and I V
 V 2 cos Z

Z
 2402 0.7540
139.2605

Pav  311.8645 W

301

Exercise 125

An RLC circuit has a resistance of 105 , an
inductance of 85.0 mH and a capacitance of 13.2 F.

(a) What is the power factor of the circuit if it is
connected to a 125 Hz AC generator?

(b) Will the power factor increase, decrease or
stay the same if the resistance is increased?
Explain.

302