Chapter # 39
1.
Sol.
Sol.
[1]
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 /kkjk A XC
2.
Alternating Current
in
1 1 c O C
DC 0
(D) neither AC nor DC (D) u rks izR;korhZ /kkjk u gh fn"V /kkjk
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 lzksr 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 C Q = C=0C [cos (100 s-1)t + cos (500 s-1)t] i
d 0 c 100 sin 100 s1 t 0 C 500 sin 500 s 1 t dt
= 100 0 C cos 100 s 1 t 1 500 0 C cos 500 s 1 t 2
i1 100 0 C
i2 500 0 C
&
i 2 > i1 3.
Sol.
The peak voltage in a 200 V AC source is 200 V izR;korhZ /kkjk lzksr dh f'k[kj oksYVrk dk eku gS (A) 220 V (B) about yxHkx 160 V C Vrms = 220 v
(C*) about yxHkx 310 V
(D) 440 V
VP 2 Vrms 220 1.414 = 311 volt
4.
Sol.
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 lzksr 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 B (I) V = V0 sint = 2f = 2 × 3.14 × 50 = 314 t = 3.14 = 0.01
Vavg
vdt
0 0.01
dt
0.01
1 cos t v0 0
0
V0 1 cos 0.1 0.01
V0 1 cos 314 0.01 314 0.01
manishkumarphysics.in
Chapter # 39
Alternating Current
V0 1 cos 314 3.14
V0 1 cos 3.14
[2]
2V0 140.127 volt (II) may be zero if v = v0 cos t
vavg
5.
Sol. 6.
Sol.
v0 +
vd 0 dt
_
t
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 lzksr ls tksM+us ij ,d izsjdRo esa pqEcdh; {ks=k dh ÅtkZ 5.0 feyh lsd.M esa vf/kdre eku ls U;wure eku rd ifjofrZr gks tkrh gSA lzksr dh vko`fÙk gS (A) 20 Hz (B*) 50 Hz (C) 200 Hz (D) 500 Hz B Frequency of the source is remain constant = 50H2 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*) D
x
frequres 'd' X = XL - Xc = wL 7.
Sol.
1 1 2fL wc 2fc
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 A
Z R 2 x2 R = 4 X = 3
Z 8.
42 32 5
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
manishkumarphysics.in
Chapter # 39
Sol.
9.
Alternating Current
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 gh AC ifjiFk esa
B Transformers are used in AC circuits only.
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 Sol.
[3]
(B)
i1 i2 2
(C*)
i12 i12 2
oxZ ek/; ewy /kkjk fuEu }kjk O;Dr dh tk;sxh (D)
i12 i12 2
C i i1 cos t i i2 sin t T
I dt 2
Irms
0 T
dt 0
if I = cosf
than
I20 2 i = i1 cost + i2 sint 2 Irms
2 irms
i12 i22 2 2 l12 i22 2
irms 10.
Sol.
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 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 (D*) about yxHkx 10 A D IP= 14 Amp
Irms 11.
Sol.
IP 2
14 2
9.9 10 Amp.
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 A A constant current exists in a resistor is rms current it is equal to 2.8 Amp.
manishkumarphysics.in
Chapter # 39
1.
Sol.
Alternating Current
Objective - II
[4]
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 lzksr dh vko`fÙk dk eku vko`fÙk ds vR;Yi vkjfEHkd eku ls FkksMk+ lk c<+kus ij (A*) izsjdRo 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 A XL = L
1 t c 2 If frequency increases, is increases that causes ‘XL’ Reaction of inductor increases & ‘Xc’ Reactance of capacitor decreses. XC
2.
Sol.
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 izsj.k dq.Myh rFkk ,d la/kkfj=kA (B) ,d izsj.k dq.Myh gS] fdUrq la/kkfj=k ugha gSA (C) ,d la/kkfj=k fdUrq izsj.k dq.Myh u gh la/kkfj=k (D*) u rks izsj.k dq.Myh u gh la/kkfj=k A X = 0 (Given) X = XL + XC
=
L
1 0 wC
It is possible that the circuit contains an inductor and a capacitor. 3.
Sol.
4.
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 lzksr 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 izsj.k dq.Myh rFkk ,a la/kkfj=k ABD Instantaneous current is zero when the intantaneous voltage is maximum. mean Resistance = 0
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 izsj.k dq.Myh izR;korhZ lzksr 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*) izsjdRo esa izsfjr fo|qr okgd cy (C) twy Å"ek (D) izsjdRo esa lafpr pqEcdh; ÅtkZ
manishkumarphysics.in
Chapter # 39 Sol. AB I = I0 sint
Alternating Current
[5]
L
R
I AC
+ t
_ average value of current over a cycle = 0 V = v0 cost = v0 sin (t + /2) I
+
+ _
t
average value of induced emf in inductor over a cycle = 0 5.
Sol.
6.
Sol.
7.
Sol.
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 B The AC voltae across a resistance can be measured using a hot-wore voltmeter. 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 AB DC dynamo or AC dynamo use to convert mechanical Energy into electrical energy. 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 lzksr dh oksYVrk 100 V (oxZ ek/; ewy) ,d ifjiFk dks 10 ,Eih;j /kkjk nsrk gSA lzksr }kjk nh xbZ vkSlr 'kfDr(A) fuf'pr :i ls 1000 okWV gSA (B*) 1000 okWV gks ldrh gSA (C) 1000 okWV ls vf/kd gks ldrh gSA (D*) 1000 okWV ls de gks ldrh gSA BD Average power Pav = Vrms Irms cos = 100 × 10 cos Pav = 1000 cos
cos lies " o to1" 0 Pav 1000
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