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8/21/2019 AC CIRCUIT .docx
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CHAPTER 15 SINGLE-PHASE SERIES A.C. CIRCUITS
Exercise 83, Page 236
1. Calculate the reactance of a coil of inductance 0.2 H when it is connected to (a) a 50 Hz,
(b) a 600 Hz and (c) a 40 kHz suppl.
(a) I!"c#i$e reac#ace,( ) ( )!" 2 f ! 2 50 0.2= π = π # 62.83
(b) I!"c#i$e reac#ace,( ) ( )!" 2 f ! 2 600 0.2= π = π # %5&
(c) I!"c#i$e reac#ace,( ) ( )$!" 2 f ! 2 40 %0 0.2= π = π × # 5'.2% (
2. & coil has a reactance of %20 Ω in a circuit with a suppl fre'uenc of 4 kHz. Calculate the
inductance of the coil.
!" 2 f != π hence, i!"c#ace, L # ( )!
$
" %20
2 f 2 4 %0=
π π × # &.%% )H
3. & suppl of 240 , 50 Hz is connected across a pure inductance and the resultin current is %.2 &.
Calculate the inductance of the coil.
*nducti+e reactance,!
240"
* %.2= =
# 200
!" 2 f != π hence, i!"c#ace, L # ( )!
" 200
2 f 2 50=π π # '.63% H
&. &n e.-.f. of 200 at a fre'uenc of 2 kHz is applied to a coil of pure inductance 50 -H.
eter-ine (a) the reactance of the coil, and (b) the current flowin in the coil.
(a) I!"c#i$e reac#ace,( ) ( )$ $!" 2 f ! 2 2 %0 50 %0
−= π = π × × # 628
* + ir! P"/0ise! / Ta0r a! racis %//
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(b) C"rre#, * # !
200
" 62=
# '.318 A
5. & %20 -H inductor has a 50 -&, % kHz alternatin current flowin throuh it. 1ind the p.d.
across the inductor
*nducti+e reactance,( ) ( )$ $!" 2 f ! 2 % %0 %20 %0
−= π = π × × # /5$.2 Ω
P.!. acrss i!"c#r,$
! ! * " 50 %0 /5$.2−= × = × ×
# 3%.%
6. Calculate the capaciti+e reactance of a capacitor of 20 31 when connected to an a.c. circuit of
fre'uenc (a) 20 Hz, (b) 500 Hz, (c) 4 kHz
(a) Ca4aci#i$e reac#ace,C 6
% %"
2 f C 2 20 20 %0−= =
π π× × × # 3%.
(b) Ca4aci#i$e reac#ace,C 6
% %"
2 f C 2 500 20 %0−= =
π π× × × # 15.2
(c) Ca4aci#i$e reac#ace,C 6
% %"
2 f C 2 4000 20 %0−= =
π π× × × # 1.8
%. & capacitor has a reactance of 0 Ω when connected to a 50 Hz suppl. Calculate the +alue of
the capacitor.
Capaciti+e reactance,C
%"2 f C
= π fro- which, ca4aci#ace, C #
( ) ( )C% %
2 f " 2 50 0=π π
# 3.% µ
8. Calculate the current taken b a %0 µ1 capacitor when connected to a 200 , %00 Hz suppl.
Capaciti+e reactance,C 6
% %" 2 f C 2 %00 %0 %0−= =π π× × ×
# %5.%55 Ω
* + ir! P"/0ise! / Ta0r a! racis %/
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C"rre#, I # C
200
" %5.%55=
# 1.25% A
. & capacitor has a capaciti+e reactance of 400 Ω when connected to a %00 , 25 Hz suppl.
eter-ine its capacitance and the current taken fro- the suppl.
Capaciti+e reactance,C
%"
2 f C=
π fro- which, ca4aci#ace, C #
( ) ( )C
% %
2 f " 2 25 400=
π π
# 15.2µ
C"rre#, * # !
%00
" 400=
# '.25 A
1'. wo si-ilar capacitors are connected in parallel to a 200 , % kHz suppl. 1ind the +alue of
each capacitor if the current is 0.62 &.
C
200
" * 0.62= = # $%.4/ Ω
i.e.
%$%.4/
2 f C=
π, hence, total capacitance,
( ) ( ) $
%C
2 %0 $%.4/=
π # 0.50 µ1
ince for parallel connection of capacitors, % 2 %C C C 2C= + =
, then%
0.50C
2=
# 0.25 µ1
i.e. eac ca4aci#r as a ca4aci#ace '.25 µ
* + ir! P"/0ise! / Ta0r a! racis %/
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Exercise 8&, 4age 23
1. eter-ine the i-pedance of a coil which has a resistance of %2 Ω and a reactance of %6 Ω
I)4e!ace, 7 #2 2 2 2
! " %2 %6+ = + # 2'
2. & coil of inductance 0 -H and resistance 60 Ω is connected to a 200 , %00 Hz suppl.
Calculate the circuit i-pedance and the current taken fro- the suppl. 1ind also the phase anle
between the current and the suppl +oltae.
*nducti+e reactance,( ) ( )$!" 2 f ! 2 %00 0 %0
−= π = π × # 50.265 Ω
I)4e!ace, 7 #
2 2 2 2
!
" 60 50.265+ = + # %8.2% (see i-pedance trianle in the diara-
below)
C"rre#, I #
200
7 /.2/=
# 2.555 A
1ro- the i-pedance trianle,
!"tan
φ =
hence the circ"i# 4ase ag0e, #
% 50.265tan60
− ÷ # 3.5
°
0aggig
3. &n alternatin +oltae i+en b + # %00 sin 240t +olts is applied across a coil of resistance $2 Ω
and inductance %00 -H. eter-ine (a) the circuit i-pedance, (b) the current flowin, (c) the p.d.
* + ir! P"/0ise! / Ta0r a! racis %0
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across the resistance, and (d) the p.d. across the inductance.
(a) *nducti+e reactance,( ) ( )$!" 2 f ! ! 240 %00 %0
−= π = ω = × # 24 Ω
Circ"i# i)4e!ace, 7 #
2 2 2 2
! " $2 24+ = + # &'
(b) C"rre# 0ig, I #
0./0/ % 00
7 40
×=
# 1.%% A
(8ote r.-.s. current # 0./0/ × -a9i-u- +alue)
(c) P.!. acrss #e resis#ace, * %.// $2= = ×
# 56.6&
(d) P.!. acrss #e i!"c#ace, ! ! * " %.// 24= = ×
# &2.&8
&. & coil takes a current of 5 & fro- a 20 d.c. suppl. :hen connected to a 200 , 50 Hz a.c.
suppl the current is 25 &. Calculate the (a) resistance, (b) i-pedance and (c) inductance of the
coil.
(a) 1ro- a d.c. circuit, resis#ace, R #
20
* 5=
# &
(b) 1ro- an a.c. circuit, i)4e!ace, 7 #
200
* 25=
# 8
(c) 1ro- the i-pedance trianle,2 2 2
!7 "= +
fro- which,2 2 2 2
!" 7 4= − = − # 6.22 Ω
&lso, !" 2 f != π
fro- which, i!"c#ace, L #( )
!" 6.22
2 f 2 50=
π π # 22.'5 )H
5. & resistor and an inductor of neliible resistance are connected in series to an a.c. suppl. he
p.d. across the resistor is % and the p.d. across the inductor is 24 . Calculate the suppl +oltae
* + ir! P"/0ise! / Ta0r a! racis %%
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and the phase anle between +oltae and current.
S"440 $0#age, #2 2 2 2
! % 24+ = + # 3'
an #ϕ
!
24
%= fro- which,
4ase ag0e /e#ee $0#age a! c"rre#, ϕ #
% 24tan%
− ÷ # 53.139 0aggig (i.e. current las
+oltae in an inducti+e circuit)
6. & coil of inductance 6$6.6 -H and neliible resistance is connected in series with a %00 Ω
resistor to a 250 , 50 Hz suppl. Calculate (a) the inducti+e reactance of the coil, (b) the
i-pedance of the circuit, (c) the current in the circuit, (d) the p.d. across each co-ponent, and
(e) the circuit phase anle.
he circuit is shown in the diara- below.
(a) I!"c#i$e reac#ace ci0,( ) ( )$!" 2 f ! 2 50 6$6.6 %0
−= π = π × # 2''
(b) I)4e!ace, 7 #2 2 2 2
! " %00 200+ = + # 223.6 (fro- i-pedance trianle)
(c) C"rre#, I #
250
7 22$.6=
# 1.118 A
(d) 0#age acrss resis#ace, * %.%% %00= = ×
# 111.8
0#age acrss i!"c#ace, ! ! * " %.%% 200= = ×
# 223.6
* + ir! P"/0ise! / Ta0r a! racis %2
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(e) 1ro- i-pedance trianle,
!"tan
φ =
fro- which, circ"i# 4ase ag0e, # # 63.&3° 0aggig
Exercise 85, 4age 2&1
1. & +oltae of $5 is applied across a C; series circuit. *f the +oltae across the resistor is 2% , find the
+oltae across the capacitor.
uppl +oltae, #2 2
C + i.e.2 2 2
C = +
i.e.2 2 2
C$5 2% = +
fro- which, $0#age acrss #e ca4aci#r,2 2
C $5 2%= − # 28
2. & resistance of 50 Ω is connected in series with a capacitance of 20 µ1. *f a suppl of 200 ,
%00 Hz is connected across the arrane-ent find (a) the circuit i-pedance, (b) the current
flowin, and (c) the phase anle between +oltae and current.
he circuit diara- is shown below.
(a) Capaciti+e reactance,( ) ( )
C 6
% %"
2 f C 2 %00 20 %0−= =
π π × # /.5// Ω
I)4e!ace, 7 #
2 2 2 2
C " 50 /.5//
+ = + # 3.8
* + ir! P"/0ise! / Ta0r a! racis %$
% %!" 200
tan tan %00
− − = ÷
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(b) C"rre#, I #
200
7 $.=
# 2.128 A
(c)
C"tan
φ = fro- which, 4ase ag0e, #
% %C" /.5//
tan tan 50
− − = ÷ # 5%.86
°
0ea!ig
3. & 24./ 31 capacitor and a $0 Ω resistor are connected in series across a %50 suppl. *f the current
flowin is $ & find (a) the fre'uenc of the suppl, (b) the p.d. across the resistor and (c) the p.d. across the
capacitor.
(a) *-pedance, 7 #
%50
* $=
# 50
&lso, i-pedance, 7 #2 2
C "+
i.e. 50 #2 2
C$0 "+
fro- which,2 2 2
C50 $0 "= +
and2 2
C" 50 $0= − # 40
Capaciti+e reactance,C
%"
2 f C=
π fro- which,
re:"ec, #( ) ( )6C
% %
2 " C 2 40 24./ %0−=
π π × # 16' H;
(b) P.! acrss #e resis#r, * $ $0= = ×
# '
(c) P.! acrss #e ca4aci#r, C C * " $ 40= = ×
# 12'
&. &n alternatin +oltae + # 250 sin 00t +olts is applied across a series circuit containin a $0 Ω
and 50 µ1 capacitor. Calculate (a) the circuit i-pedance, (b) the current flowin, (c) the p.d.
across the resistor, (d) the p.d. across the capacitor, and (e) the phase anle between +oltae and
current.
* + ir! P"/0ise! / Ta0r a! racis %4
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he circuit is shown below.
(a) Capaciti+e reactance,( ) ( )
C 6
% % %"
2 f C C 00 50 %0−= = =
π ω × # 25 Ω
I)4e!ace, 7 #2 2 2 2
C " $0 25+ = + # 3.'5
(b) C"rre#, I #
0./0/ 250
7 $.05
×=
# &.526 A
(c) P.! acrss #e resis#r, * 4.526 $0= = ×
# 135.8
(d) P.! acrss #e ca4aci#r, C C * " 4.526 25= = ×
# 113.2
(e)
C"tan
φ = fro- which, 4ase ag0e, #
% %C" 25
tan tan $0
− − = ÷ # 3.81° 0ea!ig
5. & 400 Ω resistor is connected in series with a 2$5 p1 capacitor across a %2 a.c. suppl.
eter-ine the suppl fre'uenc if the current flowin in the circuit is 24 -&.
he circuit is shown below.
*-pedance, 7 #
$
%2
* 24 %0−=
× # 500 Ω
* + ir! P"/0ise! / Ta0r a! racis %5
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1ro- the i-pedance trianle (as in the pre+ious proble-),2 2 2
!7 "= +
fro- which, capaciti+e reactance,2 2 2 2
C" 7 500 400 $00= − = − = Ω
Hence, $00 #( )%2
% %
2 f C 2 f 2$5 %0−=
π π ×
fro- which, s"440 re:"ec, #( ) ( )%2
%
2 $00 2$5 %0−π × # 225 (H;
Exercise 86, Page 2&&
1. & 40 31 capacitor in series with a coil of resistance Ω and inductance 0 -H is connected to a 200 ,
%00 Hz suppl. Calculate (a) the circuit i-pedance, (b) the current flowin, (c) the phase anle between
+oltae and current, (d) the +oltae across the coil, and (e) the +oltae across the capacitor.
he circuit diara- is shown below.
(a) *nducti+e reactance,( ) ( )$!" 2 f ! 2 %00 0 %0 50.265−= π = π × = Ω
Capaciti+e reactance,( ) ( )
C 6
% %" $./
2 f C 2 %00 40 %0−= = = Ω
π π ×
! C" " 50.265 $./ %0.4/6− = − = Ω
I)4e!ace, 7 #( )
22 2 2
! C " " %0.4/6+ − = + # 13.18
* + ir! P"/0ise! / Ta0r a! racis %6
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(b) C"rre# 0ig, I #
200
7 %$.%=
# 15.1% A
(c)
! C" "tan
−φ =
fro- which,
4ase ag0e, #
% %! C" " %0.4/6
tan tan
− −− = ÷ # 52.63° 0aggig
(d) %2 2 2 2
coil % !7 " 50.265 50.= + = + = Ω
0#age acrss ci0,( ) ( )coil coil * 7 %5.%/ 50.= = # %%2.1
(e) 0#age acrss ca4aci#r,( ) ( )C C * " %5.%/ $./= = # 6'3.6
2. 1ind the +alues of resistance and inductance ! in the circuit shown.
Circuit i-pedance, 7 #
240 0%60 $5 (%$%
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co-prise= (i) an inductance of 0.45 -H and resistance 2Ω
(ii) an inductance of 5/0 µH and 5 Ω resistance, and
(iii) a capacitor of capacitance %0 µ1 and resistance $ Ω
&ssu-in no -utual inducti+e effects between the two inductances calculate (a) the circuit
i-pedance, (b) the circuit current, (c) the circuit phase anle and (d) the +oltae across each
i-pedance. raw the phasor diara-.
he circuit is shown below.
otal resistance,
# 2 > 5 > $ # %0 Ω
otal inductance, ! 0.45 -H 5/0 H %.02 -H= + µ =
he si-plified circuit is shown below.
*nducti+e reactance,
( ) ( )$!" 2 f ! 2 2000 %.02 %0 %2.%−= π = π × = Ω
Capaciti+e reactance,( ) ( )
C 6
% %" /.5
2 f C 2 2000 %0 %0−= = = Ω
π π ×
! C" " %2.% /.5 4.6− = − = Ω
* + ir! P"/0ise! / Ta0r a! racis %
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(a) I)4e!ace, 7 #( )
22 2 2
! C " " %0 4.6+ − = + # 11.12
(b) C"rre#, I #
%00
7 %%.%2=
# 8. A
(c)
! C" "
tan
−
φ = fro- which,
4ase ag0e, #
% %! C" " 4.6tan tan %0
− −− = ÷ # 25.2
°
0aggig
(d)( ) ( )
%
$
!" 2 2000 0.45 %0 5.655−= π × = Ω
%2 2 2 2
% % !7 " 2 5.655 5.= + = + = Ω
0#age acrss irs# i)4e!ace, ( ) ( )% % * 7 . 5.= = # 53.2
( ) ( )
2
6
!" 2 2000 5/0 %0 /.%6$−= π × = Ω
22 2 2 2
2 2 !7 " 5 /.%6$ ./$5= + = + = Ω
0#age acrss sec! i)4e!ace,( ) ( )2 2 * 7 . ./$5= = # %8.53
$C" /.5= Ω
fro- earlier
22 2 2 2
2 2 !7 " $ /.5 .505= + = + = Ω
0#age acrss #ir! i)4e!ace,( ) ( )$ $ * 7 . .505= = # %6.&6
&. 1or the circuit shown below deter-ine the +oltaes %
and 2
if the suppl fre'uenc is % kHz.
raw the phasor diara- and hence deter-ine the suppl +oltae and the circuit phase anle.
( ) ( )%
$
!" 2 f ! 2 %000 %.% %0 %2−= π = π × = Ω
%
2 2 2 2
% % !7 " 5 %2 %$= + = + = Ω
and
% %!" %2tan tan 6/.$ 5
− − φ = = = ° ÷ lain
* + ir! P"/0ise! / Ta0r a! racis %
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0#age, 13
#( ) ( )%* 7 2 %$= # 26.' a# 6%.38° 0aggig
( ) ( )2C 6
% %" $2
2 f C 2 %000 4./4 % 0−= = = Ω
π π ×
2
2 2 2 22 2 C
7 " %0 $2 $$.526= + = + = Ω and
% %C" $2
tan tan /2.65 %0
− − φ = = = ° ÷ leadin
0#age, 23
#( ) ( )2* 7 2 $$.526= # 6%.'5 a# %2.65
°
0ea!ig
he +oltaes are shown in the phasor diara- (i) below.
(i) (ii)
he suppl +oltae is the phasor su- of +oltaes %
and 2
. is shown b the lenth ac in
diara- (ii).
*n trianle abc, ∠ b # %0° ; /2.65° ? 6/.$° # $./°
@sin the cosine rule,( ) ( )2 2 2ac 6/.05 26.0 2 6/.05 26.0 cos$./= + − °
fro- which, ac # 50
@sin the sine rule,
26.0 50
sin sin $ ./=
φ ° fro- which,
26.0sin$./sin 0.$$40
50
°φ = =
fro- which,%sin 0.$$40 %.5%
−φ = = ° and fro- diara- (ii), /2.65 %.5% 5$.%4α = ° − ° = ° leadin.
Hence, s"440 $0#age, < 5' a# 53.1&°
0ea!ig
* + ir! P"/0ise! / Ta0r a! racis %0
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Exercise 8%, Page 2&%
1. 1ind the resonant fre'uenc of a series a.c. circuit consistin of a coil of resistance %0 Ω and inductance
50 -H and capacitance 0.05 31. 1ind also the current flowin at resonance if the suppl +oltae is %00 .
Resa# re:"ec,( ) ( )r $ 6
% %
f 2 !C 2 50 %0 0.05 %0− −= =π π × × # 3.183 (H;
&t resonance, c"rre#, I #
%00
%0=
# 1' A
2. he current at resonance in a series !;C; circuit is 0.2 -&. *f the applied +oltae is 250 - at
a fre'uenc of %00 kHz and the circuit capacitance is 0.04 µ1, find the circuit resistance and
inductance.
* + ir! P"/0ise! / Ta0r a! racis %%
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&t resonance, current, * #
i.e. resis#ace, R #
$
$
250 %0
* 0.2 %0
−
−
×=
× # 1.25 (
&t resonance, resonant fre'uenc,r
%f
2 !C=
π i.e.
%2 f
!Cπ =
and( )
2 %2 f
!Cπ =
Hence, i!"c#ace, L #( ) ( )
2 26 $
% %
C 2 f 0.04 %0 2 %00 %0−=
π × π× × # 63.3
µ
H
3. & coil of resistance 25 Ω and inductance %00 -H is connected in series with a capacitance of
0.%2 µ1 across a 200 , +ariable fre'uenc suppl. Calculate (a) the resonant fre'uenc, (b) the
current at resonance and (c) the factor b which the +oltae across the reactance is reater than
the suppl +oltae.
(a) Resa# re:"ec,( ) ( )
r $ 6
% %f
2 !C 2 %00 %0 0.%2 %0− −= =
π π × × # 1.&53 (H;
(b) &t resonance, c"rre#, I #
200
25
= # 8 A
(c) =-ac#r #
$
6
% ! % %00 %0
C 25 0.%2 %0
−
−
×=
× # 36.51
&. & coil of 0.5 H inductance and Ω resistance is connected in series with a capacitor across a 200 , 50 Hz
suppl. *f the current is in phase with the suppl +oltae, deter-ine the capacitance of the capacitor and the
p.d. across its ter-inals.
*f the current is in phase with the suppl +oltae, then the circuit is resonant.
&t resonance, ! C" "=
i.e.
%2 f !
2 f Cπ =
π fro- which,
ca4aci#ace, C #( ) ( ) ( )
2 2
% %
2 f ! 2 50 0.5=
π π× # 2'.26 >
* + ir! P"/0ise! / Ta0r a! racis %2
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P.!. acrss #e ca4aci#r #er)ia0s,C C 6
% 200 % *"
2 f C 2 50 20.26 %0− = = = ÷ ÷ ÷ ÷π π× × ×
< $2 < 3.28 (
5. Calculate the inductance which -ust be connected in series with a %000 p1 capacitor to i+e a resonant
fre'uenc of 400 kHz.
esonant fre'uenc,r
%f
2 !C=
π
fro- which, r
%2 !C
f π =
and r
%!C
2 f =
π
and !C #
2
r
%
2 f
÷π and i!"c#ace, L #
2 2
%2 $
r
% % % %
C 2 f %000 %0 2 400 %0− = ÷ ÷π × π× ×
# 158µ
H or '.158 )H
6. & series circuit co-prises a coil of resistance 20 Ω and inductance 2 -H and a 500 p1 capacitor.
eter-ine the A;factor of the circuit at resonance. *f the suppl +oltae is %.5 , what is the
+oltae across the capacitorB
=-ac#r #
$
%2
% ! % 2 %0
C 20 500 %0
−
−
×=
× # 1''
= <
C
hence, $0#age acrss #e ca4aci#r,( ) ( )
C
A %.5 %00= = < 15'
* + ir! P"/0ise! / Ta0r a! racis %$
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Exercise 88, Page 251
1. & +oltae + # 200 sin ωt +olts is applied across a pure resistance of %.5 k Ω. 1ind the power dissipated in the
resistor.
ower dissipated in the resistor, # 2*
Current, * #
( )200 D 2 %500
=# 0.042 & (note that in the for-ula for power * has to be the r.-.s. +alue)
Hence, 4er !issi4a#e! # ( ) 22
* 0.042 (%500)= < 13.33 ?
* + ir! P"/0ise! / Ta0r a! racis %4
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2. & 50 µ1 capacitor is connected to a %00 , 200 Hz suppl. eter-ine the true power and the
apparent power.
Capaciti+e reactance,( ) ( )
C 6
% %" 2 fC 2 200 50 %0−= =π π ×
# %5.%5 Ω
Current, * # C
%00
" %5.%5=
# 6.2$ &
Tr"e 4er, P # * cos φ # (%00)(6.2$) cos 0° # '
A44are# 4er, S # * # (%00)(6.2$) # 628.3 A
3. & -otor takes a current of %0 & when supplied fro- a 250 a.c. suppl. &ssu-in a power
factor of 0./5 lain find the power consu-ed. 1ind also the cost of runnin the -otor for %
week continuousl if % k:h of electricit costs %2.20 p.
# * cos φ # (250)(%0)(0./5) since power factor # cos φ
# 18%5 ?
Ener # power × ti-e # (%./5 k:)(/ × 24) # $%5 k:h
Hence, cs# r"ig )#r r 1 ee( # $%5 × %2.20 # $4$ p # @38.&3
&. & -otor takes a current of %2 & when supplied fro- a 240 a.c. suppl. &ssu-in a power
factor of 0./0 lain find the power consu-ed.
ower consu-ed, # * cos φ # (240)(%2)(0./0) since power factor # cos φ
# 2'16 ? or 2.'16 (?
5. & transfor-er has a rated output of %00 k& at a power factor of 0.6. eter-ine the rated power output and
the correspondin reacti+e power.
* + ir! P"/0ise! / Ta0r a! racis %5
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* # %00 k& # %00 F %0$ and p.f. # 0.6 # cos φ
Per "#4"#, P # * cos φ # (%00 F %0$)(0.6) # 6' (?
eacti+e power, A # * sin φ
*f cos φ # 0.6, then φ # cos −1 0.6 # 5$.%$°
Hence sin φ # sin 5$.%$o # 0.
Hence reac#i$e 4er, = # (%00 F %0$)(0.) # 8' ($ar
6. & substation is supplin 200 k& and %50 k+ar. Calculate the correspondin power and power
factor.
&pparent power, # * #$
200 %0× & and reacti+e power, A # * sin φ #$
%50 %0× +ar
Hence,$
200 %0× sin φ #$%50 %0× fro- which, sin φ #
$
$
%50 %00./5
200 %0
×=
×
and φ #%
sin 0./5− # 4.5°
hus, 4er, P # * cos φ #$200 %0× cos 4.5° # 132 (?
and 4er ac#r # cos φ # cos 4.5° # '.66
%. & load takes 50 k: at a power factor of 0. lain. Calculate the apparent power and the reacti+e power.
rue power # 50 k: # * cos φ and power factor # 0. # cos φ
A44are# 4er, S # * #
cos φ #
50
0. # 62.5 (A
&nle φ # cos %−
0. # $6./o hence sin φ # sin $6./o # 0.6
Hence, reac#i$e 4er, = # * sin φ # 62.5 F %0$ F 0.6 # 3%.5 ($ar
8. & coil of resistance 400 Ω and inductance 0.20 H is connected to a /5 , 400 Hz suppl.
Calculate the power dissipated in the coil.
* + ir! P"/0ise! / Ta0r a! racis %6
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*nducti+e reactance,( ) ( )!" 2 f ! 2 400 0.20= π = π # 502.65 Ω
*-pedance, 7 #2 2 2 2
! " 400 502.65+ = +
# 642.$ Ω
Current, * #
/5
7 642.$= # 0.%%6/5 &
1ro- the i-pedance trianle,
!"tan
φ = and
% %!" 502.65
tan tan 400
− − φ = = ÷ # 5%.4°
Hence, 4er !issi4a#e! i ci0, P # * cos φ # (/5)(0.%%6/5) cos 5%.4° # 5.&52 ?
<ernati+el, P # ( ) ( )
22* 0.%%6/5 400= < 5.&52 ?
. &n 0 Ω resistor and a 6 31 capacitor are connected in series across a %50 , 200 Hz suppl. Calculate
(a) the circuit i-pedance, (b) the current flowin and (c) the power dissipated in the circuit.
(a) Capaciti+e reactance,C 6
% %"
2 f C 2 (200)(6 %0 )−= =
π π × # %$2.6$ Ω
I)4e!ace, 7 #2 2 2 2
! " 0 %$2.6$+ = + # 15&.
(b) C"rre#, I #
%50
7 %54.=
# '.68 A
(c) 1ro- the i-pedance trianle,
C"tan
φ = and
% %C" %$2.6$
tan tan 0
− − φ = = ÷ # 5.0°
Hence, 4er !issi4a#e! i ci0, P # * cos φ # (%50)(0.6) cos 5.0° # %5 ?
<ernati+el, P # ( ) ( )
22* 0.6 0= < %5 ?
1'. he power taken b a series circuit containin resistance and inductance is 240 : when
connected to a 200 , 50 Hz suppl. *f the current flowin is 2 & find the +alues of the
resistance and inductance.
* + ir! P"/0ise! / Ta0r a! racis %/
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ower, #2* i.e. 240 #
( ) 2
2 fro- which, resis#ace, R #
( ) 2
240
2 # 6'
*-pedance, 7 #
200
* 2=
# %00 Ω
1ro- the i-pedance trianle,2 2 2
!7 "= +
fro- which,2 2 2 2
!" 7 %00 60= − = − # 0 Ω
i.e. 2π f ! # 0 fro- which, i!"c#ace, L # ( )0
2 50π # '.255 H or 255 )H
11. he power taken b a C; series circuit, when connected to a %05 , 2.5 kHz suppl, is 0. k: and the
current is %5 &. Calculate (a) the resistance, (b) the i-pedance, (c) the reactance, (d) the capacitance,
(e) the power factor, and (f) the phase anle between +oltae and current.
(a) ower, #2* i.e.
( ) 2$0. %0 %5 × =
fro- which, resis#ace, R #( )
2
00
%5 # &
(b) I)4e!ace, 7 #
%05
* %5=
# %
(c) 1ro- the i-pedance trianle,2 2 2
!7 "= +
fro- which, ca4aci#i$e reac#ace,2 2 2 2
C" 7 / 4= − = −
# 5.%&5
(d) Capaciti+e reactance,C
%"
2 f C=
π i.e. 5./45 #
%
2 (2500)Cπ
fro- which, ca4aci#ace, C #
%
2 (2500)(5./45)π # 11.'8 >
(e) ower factor, 4.. #
4
7 /=
# '.5%1
(f) an #ϕ
C" 5./45
4=
and #e 4ase ag0e /e#ee $0#age a! c"rre#, ϕ #
% 5./45tan
4
− ÷
# 55.159 0ea!ig
* + ir! P"/0ise! / Ta0r a! racis %
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12. & circuit consistin of a resistor in series with an inductance takes 2%0 : at a power factor of
0.6 fro- a 50 , %00 Hz suppl. 1ind (a) the current flowin, (b) the circuit phase anle, (c) the
resistance, (d) the i-pedance and (e) the inductance.
(a) ower, # * cos φ i.e. 2%0 # (50) * (0.6) since p.f. # cos φ
Hence, c"rre#, I #( ) ( )
2%0
50 0.6 # % A
(b) *f cos φ # 0.6 then circ"i# 4ase ag0e, #%
cos 0.6−
# 53.13°
0aggig
(c) ower, # 2* i.e. 2%0 # ( )
2
/ fro- which, resis#ace, R # ( )
2
2%0
/ # &.286
(d) I)4e!ace, 7 #
50
* /=
# %.1&3
(e) 1ro- the i-pedance trianle,2 2 2
!7 "= +
fro- which,2 2 2 2
!" 7 /.%4$ 4.26= − = − # 5./%425 Ω
i.e. 2π f ! # 5./%425 fro- which, i!"c#ace, L # ( )5./%425
2 %00π # .'5 )H
13. & 200 , 60 Hz suppl is applied to a capaciti+e circuit. he current flowin is 2 & and the
power dissipated is %50 :. Calculate the +alues of the resistance and capacitance.
ower, #2* i.e. %50 # fro- which, resis#ace, R # # 3%.5
*-pedance, 7 #
200
* 2=
# %00 Ω
1ro- the i-pedance trianle,2 2 2
C7 "= +
fro- which,2 2 2 2
C" 7 %00 $/.5= − = − # 2./02 Ω
* + ir! P"/0ise! / Ta0r a! racis %
( )2
2
( ) 2
%50
2
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i.e. 2./02 #
%
2 f Cπ fro- which, ca4aci#ace, C #
( ) ( )
%
2 60 2./02π # 28.61 µ
* i i i 200