AC CIRCUIT .docx

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 %//


2/24

(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 %/


3/24

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.


%"

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



4/24

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


5/24

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 %%


6/24

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



7/24

(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

− − = ÷


8/24

(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


%"

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.



9/24

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 -&.


*-pedance, 7 #

$

%2

* 24 %0−=

× # 500 Ω



10/24

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



11/24

(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 $5 (%$%


12/24

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-.


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


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 %


13/24

(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



14/24

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



15/24

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 %%


16/24

&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 >



17/24

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 %$


18/24

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 ?



19/24

2. & 50 µ1 capacitor is connected to a %00 , 200 Hz suppl. eter-ine the true power and the

apparent power.


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.



20/24

* # %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.



21/24

*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 ?

&lternati+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 ?

&lternati+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.



22/24

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



23/24

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


( )2

2

( ) 2

%50

2


24/24

i.e. 2./02 #

%

2 f Cπ fro- which, ca4aci#ace, C #

( ) ( )

%

2 60 2./02π # 28.61 µ

* i i i 200

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