Chapter # 39
Alternating Current
SOLVED EXAMPLES
Example 39.1. The peak value of an alternating current is 5 A and its frequency is 60 Hz. Find its rms value. How long will the current take to reach the peak value starting from zero? Solution : Therms current is irms =
i0
2 The time period is
5A 2
3.5 A
1 1 s v 60 The current takes one fourth of the time period to reach the peak value starting from zero. Thus, the time required is
T=
i=
T 1 s 4 240
Example 39.1 Find the reactance of a capacitor (C = 200 F) when it is connected to (a) 10 Hz AC soiurce, (b) a 50 Hz AC source and (c) a 500 Hz AC source. Sol.
The reactance is XC = (a)
XC =
1 1 = . C 2vC
1 2(10 Hz ) ( 200 10 6 F)
= 80 Similarly, the reactance is 16 for 50 Hz and 1.6 for 500 Hz. Example 39.2 An inductor (L = 200 mH) is connected to an AC source of peak current. What is the intantaneous voltage of the source when the current is at its peak value? Sol. The reactance of the inductor is XL = L = (2 × 50 s–1) × (200 × 10–3 H) = 62.8 . The peak current is i0 =
0 210 V = 3.3 A. XL 62.8
As tge current lags behind the voltage by /2, the voltage is zero when the current has its peak value. Example 39.3 An LCR series circuit with L = 100 mH, C = 100 F , R = 120 is connected to an AC source of emf = (30 V) sin (100 s–1) t. Find the impedance , the peak current and the resonant frequency of the circuit. Sol. The reactance of the circuit is X=
1 – L C
1 =
(100 s )(100 10 6 F) -1
– (100 s–1) × (100 × 10–3 H )
= 100 – 10 = 90 . The resistance is R = 120. The impedance is Z= =
R 2 X2 (120 )2 (90)2 = 150 .
The peak current is manishkumarphysics.in
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Chapter # 39
Alternating Current i0 =
=
0 Z 1 2
1 LC
1 (100 10
3
H)(100 10 6 F)
50 Hz.
QUESTIONS
FOR
SHORT
ANSWER
1.
What is the reactance of a capacitor connected to a constant DC source ?
2.
The voltage and current in a series AC circuit are given by V = V0 cost and i = i0sint. What is the power dissipated in the circuit ?
3.
Two alternating currents are given by i2 = i0sin t . 3 Will the rms values of the currents be equal or different ?
i1 = i0 sint
and
4.
Can the peak voltage across the inducator be greater than the peak voltage of the source in an LCR circuit ?
5.
In a circuit conataining a capacitor and as AC source , the circuit is zero at the stant the source voltage is maximum. Is it consistent with Ohm's law ?
6.
An AC source is connected to a capacitor . Will the rms current increase , decrease or remain constant if a dielectric slab is inserted into the capacitor ?
7.
When the frequency of the AC source in an LCR circuit equals the resonant frequency , the reactance of the circuit is zero. Does it mean that there is no current through the inductor or the capacitor ?
8.
When an AC source is connected to a capacitor there is a steady-state current in the circuit. Does it mean that the charges jump from one plate to the other to complete the circuit ?
9.
A current i1 = i0 sint passes through a resistor of resistance R. How much thermal energy is produced in one time period ? A current i2 = – i0 sint passes through the resistor. How much thermal energy is produced in one time period ? If i1 and i2 both pass through the resistor simultaneously, how much thermal energy is produced ? Is the principle of superposition obeyed in this case ?
10.
Is energy produced when a transformer steps up the voltage ?
11.
A transformer is designed to concert an AC voltage of 220 V to an AC voltage of 12 V. If the input terminals are connected to a DC voltage of 220 V , the transformer usually burns . Explain.
12.
Can you have an AC series circuit in which there is a phase difference of 1800 between the emf and the current ? 1200 ?
13.
A resistance is connected to an AC source. If a capacitor is included in the series circuit , will the average power absorbed by the resistance increase or decrease ? If an inductor of small inductance is also included in the series circuit , will the average power absorbed increase or decrease further ?
14.
Can a hot-wire ammeter be used to measure a direct current having a constant value ? Do we have to change the graduations ?
manishkumarphysics.in
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Chapter # 39
1.
Alternating Current
Objective - I
A capacitor acts as an infinite resistance for (A*) DC (B) AC (C) DC as well as AC fuEu ds fy;s la/kkfj=k vuar izfrjks/k dh Hkkafr O;ogkj djrk gS (A*) fn"V /kkjk (B) izR;korhZ /kkjk (C) fn"V /kkjk ,oa izR;korhZ
(D) neither AC nor DC
/kkjk (D) u rks izR;korhZ /kkjk u gh fn"V /kkjk
2.
An AC source producing emf = 0[cos(100 s -1)t + cos(500 s -1)t] is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be i = i1 cos[(100 s -1)t + 1]+i2 cos[(500 s -1)t+ 1] ,d izR;korhZ /kkjk lzkrs dk fo|qr okgd cy = 0[cos(100 s -1)t + cos(500 s -1)t] gSA bldks ,d la/kkfj=k rFkk ,d izfrjks/k ds lkFk Js.khØe esa tksM+k x;k gSA ifjiFk esa LFkk;h /kkjk dk eku gS i = i1 cos[(100 s -1)t + 1]+i2 cos[(500 s -1)t+ 2] (A) i1 > i2 (B) i1 = i2 (C*) i1 < i2 (D) i1 rFkk i2 ds e/; laca/k Kkr djus ds fy;s nh x;h lwpuk vi;kZIr gSA
3.
The peak voltage in a 200 V AC source is 200 V izR;korhZ /kkjk lzkr s dh f'k[kj oksYVrk dk eku (A) 220 V (B) about yxHkx 160 V
gS -
(C*) about yxHkx 310 V
(D) 440 V
4.
An AC source is rated 220 V, 50 Hz. The average voltage is calculated in a time interval of 0.01 s. It (A) must be zero (B*) may be zero (C) is never zero (D) is (220/2)V ,d izR;korhZ /kkjk lzkrs 220 V, 50 gV~Zt eku dk gSA 0.01 ls- le;kUrj esa vkSlr oksYVrk dh x.kuk djus ij ;g (A) fuf'pr :i ls 'kwU; gksxk (B*) 'kwU; gks ldrk gSA (C) dHkh Hkh 'kwU; ugha gksxk (D) (220/2) V gksxkA
5.
The magnetic field energy in an inductor changes from maximum value to minimum value in 5.0 ms when connected to an AC source. The frequency of the source is izR;korhZ /kkjk lzkrs ls tksM+us ij ,d izjs dRo esa pqEcdh; {ks=k dh ÅtkZ 5.0 feyh lsd.M esa vf/kdre eku ls U;wure eku rd ifjofrZr gks tkrh gSA lzkrs dh vko`fÙk gS (A) 20 Hz (B*) 50 Hz (C) 200 Hz (D) 500 Hz
6.
Which of the following plots may represnet the reactance of a series LC combination ? fuEu esa ls dkSulk ys[kkfp=k LC Js.kh la;kstu dh izfr?kkr dks O;Dr djrk gS -
(D*) 7.
A series AC circuit has resistance of 4 and a reactance of 3 . the impedence of the circuit is ,d Js.kh izR;korhZ /kkjk ifjiFk dk izfrjks/k 4 rFkk izfr?kkr 3 gSA ifjiFk dh izfrck/kk gS (A*) 5 (B) 7 (C) 12/7 (D) 7/12
8.
Transformers are used (A) in DC circuits only (C) in both DC and AC circuits
(B*) in AC circuits only (D) neither in DC nor in AC circuits
VªkalQkWeZj dk mi;ksx fd;k tkrk gS (A) dsoy DC ifjiFk esa (C) DC rFkk AC nksuksa ifjiFkksa esa
(B*) dsoy AC ifjiFk es a (D) u rks DC ifjiFk esa u
9.
An alternating current is given by i = i1 cos t + i2 sin t . The rms current is given by izR;korhZ /kkjk fuEu lw=k }kjk O;Dr dh tkrh gS - i = i1 cos t + i2 sin t
i1 i2 (A) 2 10.
(B)
i1 i2 2
(C*)
i12 i12 2
gh AC ifjiFk esa
oxZ ek/; ewy /kkjk fuEu }kjk O;Dr dh tk;sxh (D)
i12 i12 2
An alternating current having peak value 14 A is used to heat a metal wire. To produce the same heating effect, a constant current i can be used where is /kkrq ds ,d rkj dks xeZ djus ds fy;s 14 A f'k[kj eku dh izR;korhZ /kkjk dk mi;ksx fd;k tkrk gSA leku rki izHkko mRiUu manishkumarphysics.in
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Chapter # 39
Alternating Current
djus ds fy;s fu;r /kkjk i dk mi;ksx fd;k tk ldrk gS] tgk¡ i dk eku gS (A) 14 A (B) about yxHkx 20 A (C) 7 A 11.
(D*) about yxHkx 10 A
A constant current of 2.8 A exists in a resistor. The rms current is (A*) 2.8 A (B) about 2 A (C) 1.4 A (D) undefined for a direct current ,d izfrjks/k esa 2.8 A fu;r /kkjk izokfgr gks jgh gSA oxZ ek/; ewy /kkjk gS (A*) 2.8 A (B) about yxHkx 2 A (C) 1.4 A (D) fn"V /kkjk ds fy;s ifjHkkf"kr
ugha
Objective - II 1.
An inductor, a resistor and a capacitor are joined in series with AC source. As the frequency of the source is slightly increased from a very low value, the reactance (A*) of the inductor increase (B) of the resistor increase (C) of the capacitor increases (D) of the circuit increases
,d izR;korhZ lzkrs ds lkFk Js.khØe esa ,d izjs dRo] ,d izfrjks/k rFkk ,d la/kkfj=k la;ksftr gSA lzkrs dh vko`fÙk dk eku vko`fÙk ds vR;Yi vkjfEHkd eku ls FkksM+k lk c<+kus ij (A*) izjs dRo dh izfr?kkr c<+rh gSA (B) izfrjks/k dh izfr?kkr c<+rh gSA (C) la/kkfj=k dh izfr?kkr c<+rh gSA (D) ifjiFk esa izfr?kkr c<+rh gSA 2.
3.
The reactance of a circuits is zero. It is possible that the circuit contains (A*) an inductor and a capacitor (B) an inductor but no capacitor (C) a capacitor but no inductor (D*) neither an inductor nor a capacitor ifjiFk esa izfr?kkr 'kwU; gSA ifjiFk esa fuEu ?kVd gks ldrs gSa (A*) ,d izjs .k dq.Myh rFkk ,d la/kkfj=kA (B) ,d izjs .k dq.Myh gS] fdUrq la/kkfj=k (C) ,d la/kkfj=k fdUrq izjs .k dq.Myh u gh la/kkfj=k (D*) u rks izjs .k dq.Myh u gh la/kkfj=k
ugha gSA
In an AC series circuit, the instantaneous current is zero when the instantaneous voltage is maximum. Connected to the source may be a (A*) pure inductor (B*) pure capacitor (C) pure resistor (D*) combination of an inductor and capacitor
,d izR;korhZ Js.kh ifjiFk esa tc oksYVrk dk rkR{kf.kd eku vf/kdre gS] /kkjk dk rkR{kf.kd eku 'kwU; gSA lzkrs ls la;ksftr vo;o gks ldrs gSa (A*) 'kq) izsj.k dq.Myh (B*) 'kq) la/kkfj=k (C) 'kq) izfrjks/k (D*) ,d izjs .k dq.Myh rFkk ,a la/kkfj=k 4.
An inductor-coil having some resistance is connected to an AC source. Which of the following quantities have zero average value over cycle ? (A*) current (B*) induced emf in the inductor (C) Joule heat (D) magnetic energy stored in the inductor
,d izfrjks/k ;qDr izjs .k dq.Myh izR;korhZ lzkrs ds lkFk la;ksftr gSA ,d iw.kZ pØ ds fy;s fuEu esa ls dkSulh jkf'k;ksa dk vkSlr eku 'kwU; gS (A*) /kkjk (B*) izjs dRo esa izfs jr fo|qr okgd cy (C) twy Å"ek (D) izsjdRo esa lafpr pqEcdh; ÅtkZ 5.
The AC voltage across a resistance can be measured using (A) a potentiometer (B*) a hot-wire voltmeter (C) a moving-coil galvarometer (D) a moving-magnet galvarometer izfrjks/k ds fljksa ij izR;korhZ oksYVrk dk eku fuEu dk mi;ksx djds ekik tk ldrk gS (A) xeZ rkj okyk oksYV ehVj (B*) xeZ rkj okyk oksYV ehVj (C) py&dq.My /kkjkekih (D) py&pqEcd /kkjkekih
6.
To convert machanical energy into electrical energy, one can use (A*) DC dynamo (B*) AC dynamo (C) motor
(D) transformer
;kaf=kd ÅtkZ dks fo|qr ÅtkZ esa ifjofrZr djus ds fy, mi;ksx fd;k tkrk gS (A*) DC Mk;ukeks (B*) AC Mk;ukeks (C) eksVj (D) VªkalQkWeZj 7.
An AC source rated 100 V (rms) supplies a current of 10 A (rms) to a circuit. The average power delivered by the source (A) must be 1000 W (B*) may be 1000 W (C) may be greater than 1000 W (D*) may be less than 1000 W ,d izR;korhZ lzkrs dh oksYVrk 100 V (oxZ ek/; ewy) ,d ifjiFk dks 10 ,Eih;j /kkjk nsrk gSA lzkrs }kjk nh xbZ vkSlr 'kfDrmanishkumarphysics.in
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Chapter # 39
Alternating Current
(A) fuf'pr :i ls 1000 okWV gSA (C) 1000 okWV ls vf/kd gks ldrh
(B*) 1000 okWV (D*) 1000 okWV
gSA
gks ldrh gSA ls de gks ldrh gSA
WORKED OUT EXAMPLES EXERCISE Q.1
Find the time required for a 50 Hz alternating current to change its value from zero to the rms value. 50 gV~tZ dh izR;korhZ /kkjk dk eku 'kwU; ls oxZ ek/; ewy eku rd ifjofrZr gksus esa yxk le; Kkr dhft;sA Ans : 2.5 ms
Q.2
The household supply of electricity is at 220 V rms value ) and 50 Hz Find the peak voltage and the least possible time in which the voltage can change from the rms value to zero. ?kjsyw fo|qr izokg 220 V ¼oxZ ek/; ewy½ rFkk 50 gV~tZ dk gksrk gSA f'k[kj eku rFkk og U;wure le; Kkr dhft;s] ftlesa
Ans:
3.11 V, 2.5 ms
Q.3
A bulb rated 60 W at 220 V is connected across a household supply of alternating voltage of 220 V calculate the maximum instantaneous current through the filament. 220 oksYV ij 60 okWV {kerk dk ,d cYc] 220 oksYV dh ?kjsyw forj.k ykbu ls tksM+k tkrk gSA cYc ds rarq ls izokfgr
Ans:
0.39 A
Q.4
An electric bulb is designed to operate at 12 volts. DC If this bulb is connected to an AC source and gives normal brightness what would be the peak voltage of the source ? ,d fo|qr cYc 12 oksYV dh fn"V oksYVrk ds fy;s cuk;k x;k gSA ;fn ;g cYc ,d izR;korhZ lzkrs ls tksMs+ tkus ij lkekU;
Ans:
1.7 volts
Q.6
The dielectric strength of air is 3.0 × 106 V/m A parallel plate air capacitor had area 20 cm and plate separation 0.10 mm. Find the maximum rms voltage of an AC source which can be safely connected to this capacitor. ok;q dh ijkoS|rq {kerk 3.0 × 106 oksYV@eh- gSA ,d lekukUrj iV~V la/kkfj=k dh IysVksa dk {ks=kQy 20 cm 2 rFkk IysVksa dk varjky 0.10 mm gSA bl la/kkfj=k ls tksM+s x;s AC lzkrs dh oksYVrk dk vf/kdre oxZ ek/; ewy eku Kkr dhft;s] rkfd
Ans:
4.0 kj
Q.7
The current in a discharging LR circuit is given by i = io e–t/T where t is the time constant of the circuit calculate the rms current for the period t = 0 t = LR ifjiFk esa foltZu dky esa /kkjk i = io e–t/T }kjk O;Dr dh tkrh gS] tgk¡ ifjiFk dk le;&fLFkjkad gSA le;karj t = 0 ls t = ds fy;s /kkjk ds oxZ ek/; ewy eku dh x.kuk dhft;sA
oksYVrk oxZ ek/; ewy eku ls 'kwU; gks tk;sA
vf/kdre rkR{kf.kd /kkjk dk eku Kkr dhft;sA
:i ls ped jgk gS rks lzkrs dh oksYVrk dk f'k[kj eku fdruk gS\
la/kkfj=k lqjf{kr jgsA
Ans: Q.8
Ans: Q.9
io (e 2 ) / 2 e
A capacitor of capacitance 10 µF is connected to an oscillator giving an output voltage= (10V) sin Find the peak currents in the circuit for = 10 s – 1 , 100 s – 1 , 500 s – 1 , 1000 s –1 . ,d nksfy=k dh fuxZr oksYVrk = (10V) sin gS] blls 10 µF /kkfjrk dk ,d la/kkfj=k tksMk+ x;k gSA = 10 ls– 1, 100 ls ] 500 s – 1 rFkk 1000 s– 1 ds fy;s f'k[kj /kkjk,¡ Kkr dhft;sA 1.0 × 10 – 3 A, 0.01 A, 0.05 A, 0.1 A A coil of inductance 5.0 mH and negligible resistance is connected to the oscillator of the previous problem Find the peak currents in the circuit for = 100 s – 1 , 500s – 1 , 1000s –1. fiNys iz'u esa of.kZr nksfy=k ls 5.0 mH izjs dRo rFkk ux.; izfrjks/k okyh dq.Myh tksM+h tkrh gSA = 100 s– 1,500s– 1, manishkumarphysics.in
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Chapter # 39
Alternating Current
rFkk 1000s ds fy;s f'k[kj /kkjk,¡ Kkr dhft;sA –1
Ans:
20 A , 4.0 A, 0.20 A
Q.10
A coil has a resistance of 10 and an inductance of 0.4 henry. It is connected to an AC source of 6.5 V ,
30 Hz Find the average power consumed in the circuit.
,d dq.Myh dk izfrjks/k 10 rFkk izsjdRo 0.4 gsujh gSA bldks 6.5 V ,
30
gV~Zt ds AC lzkrs ls tksM+k x;k gSA ifjiFk
}kjk mi;ksx dh x;h vkSlr 'kfDr Kkr dhft;sA Ans:
5/8 W
Q.11
A resistor of resistance 100 is connected to an AC source = (12V) sin (250 s – 1) t. Find the energy dissipated as heat during t = 0 to t = 1.0 ms. = (12V) sin (250 s – 1)t ds izR;korhZ lzkr s ls ,d 100 izfrjks/k tksM+k x;k gSA le;karj t = 0 ls t = 1.0 feyh lsd.M
esa m"ek ds :i esa u"V gqbZ ÅtkZ Kkr dhft;sA –4
Ans:
2.61 × 10
J
Q.12
In a series RC circuit with an AC source R = 300 C =m 25µF eo = 50 V and v = 50/ Hz Find the peak current and the average power dissipated in the circuit. ,d AC lzkrs ds lkFk tksM+s x;s Js.kh RC ifjiFk esa R = 300 , C =m 25µF o = 50 V rFkk v = 50/ gV~t Z gSA ifjiFk
esa f'k[kj /kkjk rFkk O;f;r vkSlr 'kfDr Kkr dhft;sA Ans:
0.10 A, 1.5 W
Q.13
An electric bulb is designed to consume 55 W when operated at 110 volts It is connected to a 220 V , 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the bulb gets correct voltage ? ,d 55 okWV dk fo|qr cYc] 110 oksYV ij mi;ksx gsrq fufeZr fd;k x;k gSA blds Js.khØe esa ,d pkWd dq.Myh dk mi;ksx djds bldks 220 , 50 gV~t Z dh ykbu ls tksMk+ x;k gSA cYc ij lgh foHkoikr izkIr djus ds fy;s dq.Myh dk izjs dRo fdruk
gksuk pkfg;s\ Ans:
1.2 H
Q14.
In a series LCR circuit with an AC source R = 300 C = 20 µ F, L = 1.0 henry e0 = 50 V and v = 50/ Hz. Find (a) the rums current in the circuit and (b) the rms potential differences across the capacitor the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater that the rms voltage of the source. ,d AC lzksr ds lkFk tksM+s x;s Js.kh RC ifjiFk esa R = 300 C = 20 µ F, L = 1.0 H rFkk v = 50/ gV~Zt gSA Kkr dhft;s - (a) ifjiFk esa /kkjk dk oxZ ek/; ewy eku rFkk (b) la/kkfj=k] izfrjks/k rFkk izjs .k dq.Myh ds fljksa ij oxZ ek/; ewy
foHkokUrjA uksV dhft;s fd rhuksa vo;oksa ij oxZ ek/; ewy foHkokUrjksa dk ;ksx] lzkrs dh oxZ ek/; ewy foHkokUrjksa dk ;ksx] lzkrs dh oxZ ek/; ewy oksYVrk ls vf/kd gSA
Ans:
(a) 0.01 A (b) 50 V, 30 V 10 V
Q.15
Consider the situation of the previous problem find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil .
fiNys iz'u esa of.kZr ifjfLFkfr dks /;ku esa j[krs gq,] la/kkfj=k esa lafpr vkSlr fo|qr {ks=k ÅtkZ rFkk dq.Myh esa lafpr vkSlr fo|qr {ks=k ÅtkZ rFkk dq.Myh esl a fa pr vkSlr pqEcdh; {ks=k ÅtkZ Kkr dhft;sA 25 mJ 5mJ Q.16
An inductance of 2.0 H, a capacitance of 18µ F and a resistance of 10k are connected to an AC source of 20 V with adjustable frequency (a) What frequency should be chosen to maximise the current in the circuit ? (b) What is the value of this maximum current ? ,d izjs dRo 2.0 H, ,d la/kkfj=k] 18µF rFkk ,d izfrjks/k 10k ] ,d ifjorZu'khy vko`fÙk rFkk 20 V okys AC lzkrs ds lkFk tksMs+ x;s gSAa (a) vko`fÙk ds fdl eku dk p;u fd;k tk;s fd ifjiFk esa /kkjk vf/kdre gks\ (b) bl vf/kdre /kkjk dk
Ans:
(a) 27 Hz (b) 2 mA
Q.17
An inductor-coil a capacitor and an AC source of rms voltage 24 V are connected in series When the frequency of the source is varied a maximum rms current of 6.0 A is observed If this inductor coil is connected to a battery of emf 12 V and internal resistance 4.0 what will be the current? ,d izjs d dq.Myh] ,d la/kkfj=k rFkk 24 V oxZ ek/; ewy eku dk AC lzkrs Js.khØe esa tqMs+ gq, gSAa tc lzkrs dh vko`fÙk ifjofrZr
eku fdruk gksxk\
manishkumarphysics.in
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Chapter # 39
Alternating Current
dh tkrh gS rks vf/kdre oxZ ek/; ewy /kkjk 6.0 A izfs {kr dh tkrh gSA ;fn bl izjs .k dq.Myh dks 12 V fo|qr okgd cy rFkk 4.0 vkarfjd izfrjks/k okyh cSVjh ls tksM+ tk;s rks /kkjk dk eku fdruk gksxk\ Ans : Q. 18
1.5 A Figure shows a typical circuit for low-pass filter, An AC input Vi = 10mV is aplied at the left end and
the output Vo is received at the right end Find the output voltages for v = kHz 100 Hz, 1.0 MHz and 10.0 MHz Note that as the frequency is increased the output decreases and hence the name low -pass filter. fp=k esa ,d fof'k"V vYi vko`fÙk pkyd (low-pass filter) n'kkZ;k x;k gSA cka;s fljs ij izR;korhZ fufo"V oksYVrk Vi = 10mV gS rFkk nka;s fljs ij fuxZr oksYVrk Vo izkIr dh tkrh gSA v = kHz 100 Hz, 1.0 MHz rFkk 10.0 MHz ds fy;s fuxZr
oksYVrk,¡ Kkr dhft;sA ;g /;ku nsus ;ksX; gS fd vko`fÙk c<+kus ij fuxZr oksYVrk de gksrh gSA vr% bldks vYi vko`fÙk pkyd (low -pass filter) dgrs gSaA Q.19
Ans:
A transformer has 50 turns in the primary and 100 in the secondary If the primary is connected to a 220 V DC supply, what will be the voltage across the secondary ? ,d Vªkl a QkWejZ dh izkFkfed dq.Myh esa 50 Qsjs rFkk f}rh;d dq.Myh esa 100 Qsjs gSA ;fn izkFkfed dq.Myh dks 220 V fn"V
oksYVrk lzkrs ds lkFk tksM+k tk;s rks f}rh; ds fljksa ij fdruh oksYVrk izkIr gksxh\ zero 'kwU;
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