18-Current Electricity
SECTION - I : STRAIGHT OBJECTIVE TYPE
[k.M-I: lh/ks oLrqfu"B izdkj
18.1 In the circuit shown, each resistances is 2 . The potential V1 as indicated in the circuit, is equal to
fn,x,ifjiFkesaizR;sdizfrjks/kdkeku2gSAifjiFkesaçnf'kZrfoHkoV1dkekugSA
(A) 11 V (B) – 11V (C) 9 V (D*) – 9 V
7 V
Sol. i = 7 = 1 A.
Current flows in anticolockwise direction in the loop. Therefore 0 – 1 × 2 – 1 × 2 – 5 = V1
V1 = – 9 V.
7 V
gy i = 7
= 1 A.
ywiesa/kjkokekorZfn'k esaizokfgrgksrhgS] blfy, 0 – 1 × 2 – 1 × 2 – 5 = V1
V1 = – 9 V.
18.2 In the circuit shown, the value of R in ohm that will result in no current through the 30 V battery, is :
fn[k ,x,ifjiFkesa]Rdkeku¼vkseesa½D;kgksxkrkfd30VcSVjhlsdksbZ/k jkizokfgrughagks&
R
10
(A) 10 (B) 25 (C*) 30 (D) 40
A C i E
Sol. i =
50
20 R
B 10
50V
i R
20
D i F
Potential drop across R = Potential drop across AB
50
20 R
.R = 30 R = 30
A C i E
gy i =
50
20 R
B 10
50V
i R
20
D i F
R ds fljksa ij foHko iru = AB ij foHko iru
50
20 R
.R = 30 R = 30
18.3 The maximum current in a galvanometer can be 10 mA. It’s resistance is 10. To convert it into an ammeter of 1 Amp. a resistor should be connected in
,d /kkjkekih esa vf/kdre /kkjk 10 m A izokfgr gks ldrh gSA bldk izfrjks/k 10 gSA bldks 1 ,fEi;j ds vehVj esa cnyus ds fy;s blds lkFk çfrjks/k tksM+uk pkfg;s &
(A) series, 0.1 (B*) parallel, 0.1 (C) series, 100 (D) parallel, 100.
(A)0.1izfrjks/kJs.khØeesa (B*) 0.1 izfrjks/klekUrjØeesa
(C)100izfrjks/kJs.khØeesa (D)100izfrjks/klekUrjØeesa
Sol. G = 10 mA
G = 10
S ( – G) = GG where S isshunt is parallel ;gk¡S lekUrj Øe esa 'kaV çfrjks/k gSA
S = 0.1
18.4 When a galvanometer is shunted with a 4 resistance, the deflection is reduced to one - fifth. If the galvanometer is further shunted with a 2 wire, the further reduction (find the ratio of decrease in current to the previous current) in the deflection will be (the main current remains the same). tc/k jkekihdks4çfrjks/kdslkFk'kUVdjrsgSrksfo{ksiik¡posHk xrd?kVtkrkgSA;fn/k jkekihdks2vfrfjDr çfrjks/kdslkFkvkSj'kUVdjfn;ktk;srksfo{ksiesavfrfjDrdeh(/k jkesadehrFk iwoZ/k jkdkvuqikrKkrdjks) gksxh& (eq[; /kkjk leku jgrh gS)
(A*) (8/13) of the deflection when shunted with 4 only
fo{ksi dk (8/13) tc dsoy 4 dslkFk 'kUV fd;k tk;sA
(B) (5/13) of the deflection when shunted with 4 only
fo{ksi dk (5/13) tc dsoy 4 dslkFk 'kUV fd;k tk;sA
(C) (3/4) of the deflection when shunted with 4 only
fo{ksi dk (3/4) tcdsoy 4 dslkFk 'kUV fd;k tk;sA
(D) (3/13) of the deflection when shunted with 4 only
fo{ksi dk (3/13) tc dsoy 4 dslkFk 'kUV fd;k tk;sA
Sol. Case I fLFkfr I
S = 4
R × = × 4
g 5 5
Rg = 16
Case II fLFkfrII
4 2
16 1 = 6
1 = /13
2
( – 1)
so decrease in current to previous current
vr%izkjfEHkd/kjkls/kjkesadeh
/ 5 /13
= / 5
8
= 13
Ans.
18.5 In the figure shown the current flowing through 2 R is : (fp=kesaizfrjks/k2Resaizokfgrgksusokyh/kjkgksxh)
(A) from left to right (ck¡;h ls nk;h rjQ) (B*) from right to left (nk¡;h ls ck¡;h rjQ) (C)nocurrent ('kwU;/kkjk) (D) Noneof these(buesals dksbZugha)
Sol.
In figure all resistance are connected in parallel.
iznf'kZrfp=kesalHkhizfrjks/klekUrjØeesatqM+sgSaA
2R R / 2
So Req =
2R R / 2
and current in all resistance flow from positive terminal of battery (means A end) to
negative terminal of battery (means B end).
vr%Req
2R R / 2
= 2R R / 2 rFk lHkhizfrjks/k saesa/k jk/kukRedfljsals(vFk Zr~Afljsls)_.k Redfljsals(vFk Zr~B
fljk)dhrjQizokfgrgksrhgSA
18.6 In a practical wheat stone bridge circuit as shown, when one more resistance of 100 is connected is parallel with unknown resistance ' x ', then ratio 𝑙1/𝑙2 become ' 2 '. 𝑙1 is balance length. AB is a uniform wire. Then value of ' x ' must be:
,d izk;ksfxd OghV LVksu lsrq fp=k esa iznf'kZr gSA tc ,d 100 dk izfrjks/k vKkr izfrjks/k ' x ' ds lkFk lekUrj Øe esa tksM+ fn;k tkrk gS rks 𝑙1/𝑙2 ds vuqikr dk eku ' 2 ' gks tkrk gSA tgk¡ 𝑙1 larqyu yEckbZ gSA AB ,d le:i rkj gS rc ' x' dk eku gksuk pkfg, &
(A) 50 (B*) 100 (C) 200 (D) 400
100
Sol.
𝑙1 𝑙2
wheat stone bridge is in balanced condition
OghVLVksulsrqlUrqfyrgS
100 x
So vr%
100
𝑙1
= 100 x
𝑙 2
𝑙1
𝑙 2
= 2 x = 100
18.7 A battery of internal resistance 2 is connected to a variable resistor whose value can vary from 4 to 10
. The resistance is initially set at 4 If the resistance is now increased then
2 vkUrfjd izfrjks/k dh ,d cSVjh ,d ifjorhZ izfrjks/kd ls tqM+h gqbZ gS ftldk izfrjks/k 4 ls 10 rd ifjofrZr gks ldrk gSA izkjEHk esa izfrjks/k dk eku 4 gSA vc ;fn izfrjks/k c<+k;k tkrk gS rks (A*) power consumed by it will decrease
blds}kjkmiHk sxdhxbZ'kfDr?kVsxhA
(B) power consumed by it will increase
blds}kjkmiHk sxdhxbZ'kfDrc<+sxh
(C) power consumed by it may increase or may decrease
blds}kjkmiHk sxdhxbZ'kfDrc<+HkhldrhgSsvkSj?kVHkhldrhgSaA
(D) power consumed will first increase then decrease.
miHk sxdhxbZ'kfDrigysc<+sxhvkSjfQj?kVsxhA
Sol.
R
Power maximum when r = R.
So, power consumed by it will decrease. for R > r. 'kfDrvf/kdregksxhtcr=Rgks vr% blds fy,O;rhr 'kfDr R>rdsfy,?kVrhgSA
18.8 A cell of emf E having an internal resistance ' r ' is connected to an external resistance R. The potential difference ' V ' across the resistance R varies with R as shown by the curve:
vkUrfjd izfrjks/k ' r ' rFkk fo-ok- cy E okyh ,d cSVjh cká izfrjks/k R ls tksM+h tkrh gSA ' R ' ds fljks ij foHkokUrj ' V ' rFkk R ds e/; oØ dk
izn'kZu gksxk&
(A) A (B*) B (C) C (D) D
Er R r r
Hint : V = E – ir = –
R r
ER
= E
R r
V = (R r)
V = 0 at R = 0 R = 0 ij V = 0
V = E at R =
R = ij V = E
so (B) is correct option.
vr% lgh mÙkj (B) gSA
18.9 ' n' identical light bulbs, each designed to draw P power from a certain voltage supply are joined in series and that combination is connected across that supply. The power consumed by one bulb will be-
' n' ,d leku izdk'k cYc] izR;sd fuf'pr oksYVst l¶ykbZ ls P 'kfä ds fy, cuk;s x;s gSA ;s Js.kh Øe esa tksM+s x;s gS rFkk la;qXeu dks bl l¶ykbZ ls tksM+k x;k gS rks ,d cYc }kjk [kpZ dh xbZ 'kfDr gksxh&
(A) n P (B) P (C) P/n (D*) P/n2
Sol. Voltage across each bulb will be
V1 = iR =
V . R = (V/n)
nR
V V V 2 V 2
so power developed by each bulb = iV1 = nR . n
P
= n2R
& P =
R
so power consumed by one bulb =
n
gy izR;sdcYcdsfljksaijfoHko
V1 = iR =
V . R = (V/n)
nR
V V V 2 V 2
vr% izR;sd cYc }kjk mRiUu 'kfDr = iV1 =
P
vr% ,d cYc }kjk O;f;r 'kfDr = n2
nR . n
= n2R
rFkkP = R
18.10 To get maximum current through a resistance of 2.5 , one can use ' m ' rows of cells, each row having ' n ' cells. The internal resistance of each cell is 0.5 . What are the values of n & m, if the total number of cells is 45. 2.5dsizfrjks/klsvf/kdre/k jkizkIrdjusdsfy,,dO;fDrlsykasdh'm'iafDr;kacukrkgSizR;sdiafDresa'n' lsygSAizR;sdlSydkvkUrfjdizfrjks/k0.5gSAvxjlSyksadhdqyla[;k45gksrksm rFk ndkekuD;kgksxk& (A) 3, 15 (B) 5, 9 (C) 9, 5 (D*) 15, 3
Hint : For maximum current, net resistance of cells must be equal to 2.5
i.e.
n (0.5)
m
= 2.5 .........(1)
& m × n = 45 (2)
solving, we get n = 15, m = 3
gy vf/kdre /kkjk ds fy, lsyks dk dqy izfrjks/k 2.5 ds cjkcj gksuk pkfg,
vFkZr~
n (0.5)
m
= 2.5 .........(1)
& m × n = 45 (2)
gy djus ij ge izkIr djrs gSa n= 15, m = 3
18.11 Two circular rings of identical radii and resistance of 36 each are placed in such a way that they cross each other’s centre C1 and C2 as shown in figure. Conducting joints are made at intersection points A and B of the rings. An ideal cell of emf 20 volts is connected across AB. The power delivered by cell is nkslekuf=kT;krFk 36¼izR;sd½izfrjks/kdhnkso`Ùk dkjoy;blizdkjj[khtkrhgSrkfdos,dnwljsdsdsUnzksa C1oC2lsfp=k uqlkjxqtjrhgSAnksuksadsizfrPNsnfcUnqvksaAoBijpkydcancuk;sx;sgSA20oksYVfo0ok0cy cyds,dvkn'kZlsydksABdse/;tksM+kx;kgSAlsy}kjkiznku'kfDrgksxh&
(A) 80 watt (B*) 100 watt (C) 120 watt (D) 200 watt
Sol. From thefigure. fp=k ls
AC1 = AC2 = C1C2 = radius f=kT;k
AC1B = 120°
Hence the resistance of four sections are
vr%pkjksaHkxksadkizfrjks/k
Hence equivalent resistance R across AB is
vr%ABdsfljksaijrqY;izfrjks/kR
1 = 1 + 1 +
1 + 1 or ;k R = 4 Power 'kfDr = V 2 = (20)2 = 100 watt.
R 24 12 12 24 R 4
18.12 Circuit for the measurement of resistance by potentiometer is shown. The galvanometer is first connected at point A and zero deflection is observed at length PJ = 10 cm. In second case it is connect at point C and zero deflection is observed at a length 30 cm from P. Then the unknown resistance X is foHkoekih}kjkizfrjks/kekiusdkifjiFkn'k Z;kx;kgSAxsYosuksehVjdksigysAfcUnqlstksM+rsgSrFk 'kwU;fo{ksi dhfLFkfr]yEckbZPJ=10cmijizkIrgksrhgSAf}rh;fLFkfresablsCfcUnqlstksM+rsgSrFk fo{ksidhfLFkfr P fcUnq ls 30 cm dh yEckbZ ij izkIr gksrh gSA rks vKkr izfrjks/k x gSA
P
C
(A*) 2R (B) R
2
R
(C) 3
(D) 3R
Sol. In potentiometer wire potential difference is directly proportional to length
foHkoekihrkjesafoHkokUrjlh/ksyEckbZdslekuqikrhgksrkgSA
Let potential drop per unit length a potentiometer wire be K.
For zero deflection the current will flow independently in two closed circuits ekukfoHkoekihrkjdh,dkadyEckbZijfoHkoiruKgSA 'kwU;fo{ksidsfy,nksuksacUnifjiFksaesa/kjkLorU=k:ilsvyxvyxizHkforgksrhgSA
R 1
x = 2R
X 2
18.13 Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity and a constant potential difference V is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder the other is : aobf=kT;kdhnksyEchlev{kh;pkydufydkvkasdkspkydrkokysinkFkZ}kjki`Fkd `rfd;kx;kgSArFk buds e/;,dcSVjh}kjkVfu;rfoHkokUrjcuk;sx;kgSArks,dufydklsnwljhufydkesaizfr,dkadyEckbZesacgusokyh
/kjkdkekugksxk&
(A)
4 V
𝑙n(b / a)
(B)
4 V (b a)
(C*)
2 V
𝑙n(b / a)
(D)
2 V (b a)
Sol. E =
2 r , where is the linear charge density on the inner cylinder.
;gk¡=vkUrfjdcsyuijjs[kh;vkos'k?kuRogSA
b
and rFkk V =
E.d𝑙
= 2
𝑙n b
a
.... (1)
a 0
→ →
Now vc ; = J.dA =
. 2 r 𝑙
2 0r
E.dA
Current peruntilengthwillbe: ,dkad yEckbZ esa /kkjk %
= 0
.... (2)
From (1) (1)ls
=
NewSol. vU; fof/k
V
2 0 V
0 𝑙n(b / a)
2 V
𝑙n(b / a)
b R
b
1 dx
1 b
2V
R =
2x.1 =
𝑙n
2 a
=
𝑙 n(b / a)
x a
18.14 50 V battery is supplying current of 10 amp when connected to a resistor. If the efficiency of battery at this current is 25%. Then internal resistance of battery is :
50VdhcSVjh10,Eih;jdh/k jkizokfgrdjrhgS]tc;gizfrjks/klstqM+hgSA;fnbl/k jkijcSVjhdh{kerk25%
gSaArccSVjhdkvkUrfjdizfrjks/kgS%
(A) 2.5 (B*) 3.75 (C) 1.25 (D) 5
Sol. 50 = 10 [R + r]
R + r = 5
R
= R r
0.25 =
R
R r
then vr% R =
R + r = 4 R
r = 3 R
5 = 1.25 , and rFkk r = 3.75 .
18.15 A battery is supplying power to a tape-recorder by cable of resistance of 0.02 . If the battery is generating 50 W power at 5V, then power received by tape-recorder is : (neglect internal resistance of battery)
,dcSVjhVsifjdkMZjdks0.02izfrjks/kdhdscy}kjk'kfDriznkudjrhgSA;fncSVjh5Vij50W 'kfDrmRiUudjrh gksrksVsifjdkMZj}kjkxzg.kdhxbZ'kfDrgksxh%(cSVjhdkvkUrfjdizfrjks/kux.;gSA)
(A) 50 W (B) 45 W (C) 30 W (D*) 48 W
Sol. P = V, 50 = 5 ×
= 10 A
Power lost in cable = 2 R = 10 × 10 × 0.02 Power supplied to T.R. = 50 W – 2 W = 2 W
= 48 W
gy P = V, 50 = 5 ×
= 10 ,fEi;j dscy esa 'kfDr {k; = 2 R = 10 × 10 × 0.02
= 2 W
VsifjdkMZjdksiznkudhxbZ'kfDr= 50W – 2W = 48 W
18.16 In the shown wire frame, each side of a square (the smallest square) has a resistance R. The equivalent resistance of the circuit between the points A and B is :
fn[k ,x,rkjdsÝseesaoxZ¼lclsNksVkokyk½dhizR;sdHkqtkdkizfrjks/kRgSAfcUnqAoBdse/;ifjiFkdk rqY;izfrjks/kgSA
A B
(A) R (B*) 2R (C) 4R (D) 8R
Sol. The circuit can be folded about B and redrawn as
ifjiFkdksABdsifjr%eksM+k(folded)tkldrkgSrFk fp=k uqlkjiqu%vkjsf[krfd;ktkldrkgS&
Hence equivalent resistance between A and B is 2R.
blfy,A rFk BdsrqY;izfrjks/k2RgksxkA
109
18.17 A spherical shell, made of material of electrical conductivity (-m)–1, has thickness t = 2 mm and radius
r = 10 cm. In an arrangement, its inside surface is kept at a lower potential than its outside surface.
The resistance offered by the shell is equal to -
109
,dxksyh;dks'k (vkse-eh-)–1fo|qrpkydrkokysinkFkZlscukgSAbldheksVkbZt=2fe h-rFk f=kT;kr=10lseh-gSA,d
la;kstuesa]bldhvkUrfjdlrgdksckgjhlrglsdefoHkoijj[krsgSAdks'k}kjkfd;kx;kizfrjks/kcjkcjgSA
(A) 5 x 10–12 (B) 2.5 x 10–11 (C) 5 x 10–12 (D*) 5 x 10–11
Ans. (D)
1
Sol. R = ×
t
4r 2
Using values R = 5 x 10–11
R= 5x 10–11 ds eku dk iz;ksx djus ij
18.18 Two cylindrical rods of uniform cross-section area A and 2A, having free electrons per unit volume 2n and n respectively are joined in series. A current flows through them in steady state. Then the ratio of drift velocity
vL
of free electron in left rod to drift velocity of electron in the right rod is
is :
R
A o 2A ,d leku vuqizLFk dkV {ks=kQy dh nks csyukdkj NM+s ftuesa eqDr bysDVªkWu izfr bdkbZ vk;ru esa Øe'k% 2n o n gS, Js.khØe esa tksM+h xbZ gSA lkE;koLFkk esa muesa ls /kkjk izokfgr gSA rks ck;h NM+ esa eqDr bySDVªku ds vuqxeu osx dk nk;h NM+ esa eqDr bysDVªkWu ds vuqxeu osx ls
vL
vuqikr
gSA
R
(A)
1 (B*) 1 (C) 2 (D) 4
2
Sol. Since current = neAv through both rods is same
pwafd/kjk=neAvd nksuksaNM+ksaesalekugS
2 (n) e A vL = n e (2A) vR
vL
or;k vR = 1
18.19 A charge passing through a resistor is varying with time as shown in the figure. The amount of heat generated in time ' t ' is best represented (as a function of time) by:
,dizfrjks/klsxqtjusokyh/k jkdksle;dslkFkfuEuizdkjlsiznf'kZrdjrsgSAtle;esamRiUuÅ"ekdksle; ds Qyu ds :i esa iznf'kZr djus okyk loZJs"B xzkQ gksxk&
(A) (B) (C*) (D)
dq
Sol. i = dt
dq
i = dt
= slope of q - t graph
= q- t vkjs[k dh izo.krk
= – 5 (which is constant tks fu;r gSA)
Amount of heat generated in time t t le; esa mRiUu Å"ek
H = i2RT
H t.
18.20 Two cylindrical rods of same cross-section area and same length are connected in series to an ideal cell as shown. The resistivity of left rod is and that of right rod is 2. Then the variation of potential at any point P distant x from left end of combined rod system is given by. nks csyukdkj NM+ksa dk vuqizLFk dkV {ks=k leku gS rFkk mudh yEckbZ leku gS rFkk Js.khØe esa
,d vkn'kZ lsy ls fp=kkuqlkj tqM+s gq, gSaA ck;ha rjQ dh NM+ dh izfrjks/kdrk gS rFkk nk;ha rjQ dh NM+ dh izfrjks/kdrk 2 gSA rc fdlh fcUnq P ij foHko dk ifjorZu la;qDr NM+ fudk; ds ck;sa fljs ls nwjh x ds lkFk fuEu xzkQ }kjk fn;k tk;sxkA
(A) (B*) (C) (D)
Sol. From relation E = J, the magnitude of electric field is greater in right rod as compared to left rod. There fore magnitude of potential gradient in the right rod is greater. (remember potential is continous).
Therefore the variation is shown by figure. lEcU/kE=Jls,nka;hNM+esafo|qr{ks=kdkifjek.kcka;hNM+dhrqyukesavf/kdgksxkAblfy;snka;hNM+esafoHkoizo.krk vf/kdgksxhA¼;knjf[k;sfdfoHkolr ~gS½blfy,ifjorZufp=k uqlkjfn[k ;kx;kgSA
18.21 A copper sphere of 10cm diameter is lowered into a water filled hemispherical copper vessel of 20 cm diameter so that the sphere and the vessel becomes concentric. Electrical conductivity of water is = 10–3 ( – m)–1. The electrical resistance between the sphere and the vessel is :
10cm O;kl ds rkez xksys dks 20 cm O;kl ds rkez v}Z xksyh; ik=k esa Hkjs ty esa bl izdkj mrkjk tkrk gS fd xksyk o ik=k ladsUnzh; gks tk;sA ty dh fo|qr pkydrk = 10–3 ( – m)–1 gSA xksys o ik=k ds chp oS|qr izfrjks/ k gSA
(A*) 1591.6 (B) 1450 (C) 1682.4 (D) 1489.6
Sol. The arrangement is shown in figure. Consider the hemispherical shell of radius r and thickness dr as shown.
Resistance of this shell is ; fudk;fp=kesafufnZ"VgSAekukrf=kT;kodreksVkbZdk,dv}Zxksyh;dks'kvo;ofy;kx;kgS¼fp=k uqlkj½bl dks'kdkizfrjks/kgSA
dR =
dr
2r 2
1 r10cm dr
R = 2
r5cm r 2
= 1591. 6 .
18.22 In the shown circuit the resistance R can be varied :
fp=kesaizfrjks/kRifjofrZgSA
The variation of current through R against R is correctly plotted as :
Rlsizokfgr/k jkdkRdslkFkifjorZudklghvkjs[kgksxkA
(A) (B) (C*) (D)
Sol. Redrawing the given circuit diagram as shown below :
fn;sx;sifjiFkdksfuEukuqlkjcukldrsgSA
Usingpointpotential theory,fcUnq foHko fl)kUr ls
V E V E V E 0
r r R
2 1
(V – E)
= 0
AstSlkfd
2 1 0
so blfy, V – E = 0
r R
V E
So, current through R, i = R
= 0 whatever be the value of R.
vr% Rls izokfgr /kkjk i= V E =0 RdkekupkgsdqNHkhgks
18.23 A cylindrical solid of length L and radius a is having varying resistivity given by = x where is a positive
0 0
constant and x is measured from left end of solid. The cell shown in the figure is having emf V and negligible internal resistance. The electric field as a function of x is best described by :
yEckbZ L o f=kT;k a dk Bksl csyu dh izfrjks/kdrk = 0x gS tgka 0 / kukRed fu;rkad gS o x dks Bksl ds cka;s fljs ls ekik x;k gSA fp=k esa fufnZ"V cSVjh dk fo|qr okgd cy V o vkUrfjd izfrjks/k ux.; gSA x ds Qyu :i esa fo|qr {kS=k dk mi;qDr lw=k gSA
2V
(A*)
L2
× x (B)
2V
0L2
V
× x (C) L2
× x (D)Noneof thesebuesa ls dksbZugh
Sol. Consider an elemental part of solid at a distance x from left end of width dx.
csyudscka;sfljslsxnwjhijdxpkSMkbZds Resistance of this elemental part is, vo;odkizfrjks/kgSA
dR =
dx
a2
0xdx
= a2
L 0xdx 0L2
R = dR
0
a2
2a2
V
Currentthroughcylinder is, csyu ls izokfgr /kkjk = R =
0L2
2V
Potential drop across element is, vo;o ij foHko iru dV = IdR = L2
x dx
dV
E(x) = dx
2V
= L2 x.
SECTION - II : MULTIPLE CORRECT ANSWER TYPE
[k.M- II: cgqy lgh fodYi izdkj
18.24 Two cells of unequal emfs E1 and E2 and internal resistances r1 and r2 are joined as shown in figure. VP and VQ are the potential at P and Q respectively.
vlekufo-ok-cyE1 oE2 rFk vkUrfjdizfrjks/kr1 or2 dsnkslsyfp=kesan'k Z;savuqlkjtqM+sgSAVP oVQ Øe'k%PoQ ij foHko gSA
(A*) The potential difference across both the cells will be equal
nksuksalsyksaijfoHkokUrjlekugksxk
(B*) One of the cell, will supply energy to the other cell.
,dlsynwljsdksÅtkZiznkudjsxk
(C*) The potential difference across one of the cells will be greater than its emf.
,d lsy ij foHkokUrj blds fo-ok-cy ls vf/kd gksxk
E1r2 E2r1
(D*) VP – VQ =
r1 r2
Sol. (A) p.d. across each cell = VP – VQ
(B) If i is clockwise then E2 is source and for anti–clockwise current E1 is source.
(C) P.D. = E – ir (when battery supplies energy)
= E + ir (when battery consumes energy).
E1 E2
By KVL i =
r1 r2
(Anticlockwise)
VP
– VQ
= E1
– i r1
E1r2 E2r1
= r1 r2
Sol. (A) izR;sd lsy ij foHkokUrj = VP – VQ (B);fninf{k.k orZgSrksE2lzksrgSvkSjokekorZ/k jkdsfy,E1lzksrgSA (C)foHkokUrj = E – ir (tc cSVjh ÅtkZ nsrh gS)
= E + ir (tc cSVjh ÅtkZ ysrh gS).
E1 E2
KVLls i =
r1 r2
(nf{k.ko`r)
VP
– VQ
= E1
– i r1
E1r2 E2r1
= r1 r2
18.25 In the circuit shown in the figure
fp=kesan'kZ;sifjiFkesa&
(A*) power supplied by the battery is 200 watt
cSVjh}kjkçnku'kfDr200WgSA
(B) current flowing in the circuit is 5 A
ifjiFkesa/kjkçokg5AgSA
(C*) potential difference across 4 resistance is equal to the potential difference across 6 resistance 4çfrjks/kdsfljksaijfoHkokUrj]6çfrjks/kdsfljksaijmRiUufoHkokUrjdscjkcjgSA
(D) current in wire AB is zero
ABrkjesa/kjk'kwU;gSA
A
Sol.
20V
there is zero potential difference across 4 and 6 resistance.
;gk¡4rFk 6izfrjks/kijfoHkokUrj'kwU;gSA
i = 20 = 10 A
2
power by battery cSVjh }kjk 'kfDr
pb = i = 20 × 10 = 200 W Ans.
18.26 In the figure a conductor of nonuniform cross-section is shown. A steady current flows in it.
fp=kesa,dvlekuvuqizLFkdkV{ks=kQydkpkydn'k Z;kx;kgSAblesalsfLFkj/k jkizokfgrgksrhgS
(A*) The electric field at A is more than at B.
Aij fo|qr {ks=k B ls T;knk gSA
(B) The electric field at B is more than at A.
B ijfo|qr {ks=kAlsT;knkgSA
(C*) The thermal power generated at A is more than at B in an element of small same width.
lekuNksVhpkSM+kbZdsHk xdsfy;sAijBdhvis{k T;knkrkih;'kfDrmRiUugksxhA
(D) The thermal power generated at B is more than at A in an element of small same width.
lekuNksVhpkSM+kbZdsHk xdsfy;sBijAdhvis{k T;knkrkih;'kfDrmRiUugksxhA
Sol. (A,C)
The area of cross-section of conductor at point A is less than that at point B. So current density at A is higher. Hence, the electric field at A is more than at B and the thermal power generated at A is more than at B in an element of small same width. since resistance at A is greater
Sol. (A,C)
pkyddkAfcUnqijvuqizLFkdkVdk{ks=kQyBfcUnqdhvis{k degSAblfy,A ij/kkjk?kuRomPpgksxkAvr%A fcUnqijfo|qr{ks=kdhrhozrkBdhrqyukesaT;knkgSArFk lekuNksVhpkSM+kbZdsvo;odsfy,AfcUnqijrkih;'kfDr B fcUnqdhrqyukesavf/kdmRiUugksxhApqafdA ijizfrjks/kT;knkgSA
18.27 In the figure shown: (All batteries are ideal) n'kkZ;s x;s fp=k esa ¼lHkh cSVªh vkn'kZ gS½
(A) current through 25 V cell is 20 A (25oksYVdslsyesa/k jk20AgSA) (B*) current through 25 V cell is 12.5 A (25oksYVdslsyeas/k jk12.5AgSA)
(C) power supplied by 20 V cell is 20 W (20oksYVdslsy}kjknhx;h'kfDr20okVgSA)
(D*) power supplied by 20 V cell is 20 W (20 oksYV ds lsy }kjk nh x;h 'kfDr – 20 okV gSA)
Sol. (B,D)
25 V 25 V 25 V 25 V
12.5A
25V
Power supplied by 20 V cell = (–1) (20) = – 20 W
as the cell is not supplying the power, it is eating the power (getting charged) 20V ds lsy }kjk iznÙk 'kfDr =(–1) (20) =– 20W tSlkfdlsy'kfDriznkuughdjjgkgS, ;gvkos'kxzg.kdjrkgSA
18.28 Consider a resistor of uniform cross sectional area connected to a battery of zero internal resistance. If the length of the resistor is doubled by stretching it then ekuk,dlekuvuqizLFkdkV{ks=kQydk,dizfrjks/k'kwU;vkUrfjdizfrjks/kdh,dcSVjhlstqM+kgqvkgSA;fn izfrjks/kdks[khapdjbldhyEckbZdksnqxqukdjfn;ktk;srksa
(A) current will become four times.
(B*) the electric field in the wire will become half.
(C*) the thermal power produced by the resistor will become one fourth.
(D) the product of the current density and conductance will become half. (A)/k jkpkjxqukgkstk;sxhA
(B*)rkjesafo/kqr{ks=kvk/k gkstk;sxkA
(C*)izfrjks/k}kjkmRiUuÅ"eh;'kfDr,dpkSFk bZgkstk;sxh
(D) /kkjk?kuRoopkydrkdkxq.kuvk/k gkstk;sxkA
Sol. (Moderate) As the length is doubled, the cross section area of the wire becomes half. thus the resistance
of the wire R L
A
becomes four times the previous value. Hence after the wire is elongated the current
becomes one fourth. Electric field is potential difference per unit length and hence becomes half the
V 2
initial value. The power delivered to resistance is P and hence becomes one fourth.
R
gy % (Moderate) tcyEckbZnqxuhgkstkrhgS]rksrkjdkvuqizLFkdkVdk{ks=kQyvk/kkgkstkrkgSAblizdkj]rkj
dkizfrjks/k R L igysdkpkjxqukgkstkrkgSAblfy;srkjdhyEckbZc<+usdsckn/k jk,dpkSFk bZgkstkrh A
gSAfo|qr{ks=k]foHkokUrjizfr,dkadyEckbZgksrkgSvkSjblfy;sizkjfEHkdekudkvk/k gkstkrkgSAizfrjks/kdksnh
xbZ'kfäP
V 2 gksrhgSvkSjblfy;s,dpkSFk bZgkstkrhgSA
R
18.29 A variable current flows through a 1 resistor for 2 seconds. Time dependence of the current is shown in the graph.
,difjorhZ/k jk1izfrjks/kesa2lsd.MrdcgrhgSA/kjkdhle;ijfuHkZjrkxzkQesan'k Z;hgSA
(A)
10
O 2 t(s)
(A*) Total charge flown through the resistor is 10 C.
izfrjks/k ls cgus okyk dqy vkos'k 10 C. (B*) Average current through the resistor is 5A.
izfrjks/klscgusokyhvkSlr/kjk5AgSA
(C) Total heat produced in the resistor is 50 J.
izfrjks/kesamRiUudqym"ek50JgSA
(D*) Maximum power during the flow of current is 100 W.
/k jkizokgdsnkSjkuegRre'kfDr100WgSA
Ans. A,B,D
Sol. Total charge = dt = Area under the curve
dqyvkos'k = dt =oØ ds vUnj {ks=kQy
= 10 C
dt
Average current =
dt
= 5A
Total heat produced = 2Rdt
dt
vkSlr/kjk =
dt
= 5A
dqy mRiUu m"ek = 2Rdt
2
= (5t 10)2
0
. 1 . dt =
200
3 J
Maximum Power = 2R when I is maximum current. = 100 x 1 = 100 W
egRre 'kfDr= 2R tc /kkjk I egRre gS. = 100x 1= 100W
18.30 A conductor of truncated conical (frustum) is connected to a battery of emf as shown in the figure.
,ddVsgq,'kaDokdkjpkyddksfo|qrokgdcydhcsVªhlsfp=kkuqlkjtksMkx;kgSA
If at a section distant x from left end, electric field intensity, potential and rate of generation of heat per unit length are E, V and H respectively, then which of the following graph(s) is/are correct ?
;fncka;sfljslsxnwjhijfo|qr{ks=krhozrk]foHkooizfrbdkbZyEckbZmRiUum"ekdhnjØe'k%E,VoHgSA rksfuEuesalsdkSulsvkjs[klghgSA
(A*) (B*) (C*) (D)
Sol. Let a be the radius of left end side cross–section, then radius of cross–section at distance x from left end is a + bx where b is a constant. ekukcka;sfljsdsvuqizLFkdkV{ks=kdhf=kT;kagSArkscka;sfljslsxnwjhijvuqizLFkdkV{ks=kdhf=kT;ka+bx gksxh tgkab,dfu;rkadgSA
From J = E ls
i = E
A
E =
i 1
as i and are same for all cross–section
A
i
E= A
1 pwafdiolHkhdkV{ks=kdsfy,lekugksxs
E 1
A
1
= (a bx)2
i2
H 1
Rateof heatgeneration per unit length, izfr bdkbZ yEckbZ m"ek mRiUu dh nj H= A , So A
H i2 A
E = A i i = constant fu;rkad
V X i dx
dV = →
→ dV = 2
E • dx
0 (a bx)
i 1
1
i bx
V =
b a bx
a =
ab a bx
18.31 AB is part of a circuit as shown, that absorbs energy at a rate of 50 W. E is an emf device that has no internal resistance.
(A) Potential difference across AB is 48 V. (B*) Emf of the device is 48 V.
(C) Point B is connected to the positive terminal of E.
(D*) Rate of conversion from electrical to chemical energy is 48 W in device E.
n'kkZ,uqlkj AB ,d ifjiFk dk Hkkx gS tks 50 W dh nj ls ÅtkZ vo'kks"k.k djrk gSA E ,d 'kwU; vkUrfjd izfrjks/k dh fo|qr okgd cy ;qfDr gSA
(A)AB ij foHkokUrj 48V gSA
(B*) ;wfDr dk fo|qr okgd cy 48 V gSA
(C) fcUnqB,Eds/kukxzlstqMkgSA
(D*) ;qfDrE esaoS|qrlsjklk;fudÅtkZesaifjorZudhnj48W gSA
Sol. Resistance absorbs energy at the rate of 2W. Potential difference across AB VAB. = 50 W
VAB = 50 V
Drop across resistor is 2V, therefore EMF of E is 48 V.
As AB is absorbing energy at the rate of 50 W, 48 W is being absorbed by E. Thus E is on charging mode i.e. current is entering from +ve terminal of E.
gy. izfrjks/k2W dhnjlsÅtkZvo'k s"k.kdjrkgSA
AB ij foHkokUrj VAB. = 50 W
VAB = 50 V
izfrjks/k ij foHko iru 2V gS vr% E dk fo|qr okgd cy 48 V gS
pwafd AB 50 W dh nj ls ÅtkZ vo'kks"k.k dj jgk gS] E 48 W vo'kksf"kr dj jgk gSA
vFk ZrEdkvkos'kugksjgkgSAEds/kukxzls/k jkizos'kgksjghgSA
18.32 The galvanometer shown in the figure has resistance 10 .It is shunted by a series combination of a resistance S = 1 and an ideal cell of emf 2V. A current 2A passes as shown.
fp=kesafn[k ,vuqlkj/k jkekih(galvanometer)dkizfrjks/k10gSAbldslekUrjØeesaizfrjks/kS=1rFk 2V
fo-ok-cyokyhvkn'kZcSVjhdsJs.khØela;kstudksfp=kuqlkjtksM+ktkrkgSAblesa2Adh/kjkfp=kuqlkjxqtjrhgS&
(A) The reading of the galvanometer is 1A
/kkjkekihdkikB~;kad1AgSA
(B*) The reading of the galvanometer is zero
/k jkekihdkikB~;kad'kwU;gSA
(C) The potential difference across the resistance S is 1.5 V
izfrjks/kSdse/;foHkokUrj1.5VgSA
(D*) The potential difference across the resistance S is 2 V
izfrjks/kS dse/;foHkokUrj2VgSA
Sol. (Moderate) Let the currents be as shown in the figure KVL along ABCDA
– 10 i – 2 + (2 – i)1 = 0
i = 0
Potential difference across S = (2 – i)1 = 2 × 1 = 2 V.
gy. (Moderate) ekuk/k jkfp=kesaiznf'kZrgS
ABCDA ds vuqfn'k KVL yxkus ij
– 10 i – 2 + (2 – i)1 = 0
i = 0
S ds fljks ij foHkokUrj= (2 – i)1 = 2 × 1 = 2 V.
SECTION - III : ASSERTION AND REASON TYPE
[k.M- III: dFku vkSj dkj.k izdkj
18.33 STATEMENT-1 : When an external resistor of resistance R (connected across a cell of internal resistance r) is varied, power consumed by resistance R is maximum when R = r.
STATEMENT-2 : Power consumed by a resistor of constant resistance R is maximum when current through it is maximum.
oDrO;-1: ,dckº;izfrjks/kftldkizfrjks/kRgSbldks,dlsyftldkvkUrfjdizfrjks/krgSdsfljksijyxk;k tkrkgStcckº;izfrjks/kRdksifjofrZrfd;ktkrkgSrksizfrjks/kR }kjkyh¼O;f;rdh½xbZ'kfävf/kdrerc gksxhtcR=rgksxkA
oDrO;-2: ,dfu;rizfrjks/kR}kjkyhxbZ'kfävf/kdregksrhgStcblesalsizokfgr/kkjkvf/kdregksxhA
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B*) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
(A) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA
(B*)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(C) oDrO;-1lR; gS] oDrO;-2vlR; gS ;
(D) oDrO;-1vlR; gS] oDrO;-2lR; gS
Sol. Both statements 1 and 2 are true. In statement-1 R is varied while in statement-2 R is kept constant.
Hence both statements are independent.
nksuksaoDrO;1o2lR;gSAoDrO;-1esaizfrjks/kRdsekudksifjofrZrfd;ktkrkgStcfdoDrO;-2esaRdksfu;r j[k tkrkgSAvr%nksuksaoDrO;LorU=kgSA
18.34 STATEMENT-1 : The current density J at any point in ohmic resistor is in direction of electric field E at that point.
STATEMENT-2 : A point charge when released from rest in a region having only electrostatic field always moves along electric lines of force.
oDrO; -1 : vkseh; izfrjks/k ds fdlh fcUnq ij /kkjk ?kuRo J ml fcUnq ij fo|qr {ks=k E dh fn'kk esa gksrk gSA
oDrO;-2:,dfcUnqvkos'kdkstcfLFkjfo|qr{ks=kesafojkelsNksM+ktkrkgS]rks;gges'k fo|qrcyjs[k vksads
vuqfn'kxfrdjrkgSA
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k gSA
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
oDrO;-1lR;gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dklgh Li"Vhdj.kugha gSA
(C*) Statement-1 is True, Statement-2 is False
oDrO; -1lR; gS, oDrO;-2vlR; gSA
(D) Statement-1 is False, Statement-2 is True.
oDrO; -1vlR; gS, oDrO;-2lR; gSA
→ →
Sol. From relation J E , the current density J at any point in ohmic resistor is in direction of electric field E at that point. In space having non-uniform electric field, charges released from rest may not move along ELOF. Hence statement 1 is true while statement 2 is false.
→ →
lEcU/k J E ls]vkseh;izfrjks/kesafdlhfcUnqij/k jk?kuRomlfcUnqijfo|qr{ks=kdhfn'k esagksrkgSAfdlh vlekufo|+qr{ks=kokysLFk uesa]fojkelsNksM+sx;svkos'kELOFdsvuqfn'kxfrughdjsaxsAblfy;soDrO;-1lR;
gS tcfd oDrO;-2vlR; gSA
18.35 STATEMENT-1 : A wire of uniform cross section and uniform resitivity is connected across an ideal cell. Now the length of the wire is doubled keeping volume of wire constant. The drift velocity of electrons after stretching the wire becomes one fourth of what it was before stretching the wire.
STATEMENT-2 : If a wire ( of uniform resitivity and uniform cross-section) of length 𝑙o is stretched to
length n𝑙o
, then its resistance becomes n2 times of what it was before stretching the wire( the volume
of wire is kept constant in stretching process). Further at constant potential difference, current is inversely proportional to resistance. Finally drift velocity of free electron is directly proportional to current and inversely proportional to cross section area of current carrying wire..
oDrO;-1: ,dlekuvuqizLFkdkVrFk ,dlekuizfrjks/kdrkds,drkjdks,dvkn'kZlsydsfljksaijtksM+ktkrk gSAvcrkjdsvk;rudksfu;rj[krsgq;s]rkjdhyEckbZdksnqxqukfd;ktkrkgSArkjdks[khapusdscknbysDVªkWuks dkvuqxeuosxrkjdks[khapusdsigysdk,dpkSFk bZgkstkrkgSA
oDrO;-2: ;fn¼,dlekuizfrjks/kdrkrFkk,dlekuvuqizLFkdkVds½𝑙o yEckbZds,drkjdksn𝑙o yEckbZrd[khapk
tkrkgSa]rksbldkizfrjks/krkjdks[khapusdsigysdkn2 xqukgkstkrkgS([khapusdsizfØ;kesarkjdkvk;rufLFkj j[k tkrkgS)Avkxs],dfu;rfoHkokUrjij/k jkizfrjks/kdsO;qRØekuqikrhgksrhgSAvUresa]eqäbysDVªkWudkvuqxeu osx/k jkdslh/kslekuqikrhgksrkgSrFk /k jkizokghrkjdsvuqizLFkdkVds{ks=kQydsO;qRØekuqikrhgksrkgSA
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO; -1 dk lgh Li"Vhdj.k gSA
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA
(C) Statement-1 is True, Statement-2 is False
oDrO; -1 lR; gS, oDrO;-2 vlR; gSA
(D*) Statement-1 is False, Statement-2 is True.
oDrO; -1 vlR; gS, oDrO;-2 lR; gSA
Sol. As the length of wire is doubled, the cross-section area of wire becomes half. Therefore resistance of wire
becomes four times and current becomes
also vd = neA
1 th of the initial value.
4
Since current becomes one fourth and cross-section area of wire becomes half, therefore from above equa- tion the drift velocity of electron becomes half. Hence statement is false. tcrkjdhyEckbZnqxquhdhtkrhgS]rksrkjdkvuqizLFkdkVdk{kS=kQyvk/k gkstkrkgSAblfy,rkjdkizfrjks/kpkj
1
xqukgkstkrkgSrFk /kjkizkjfEHkdekudh 4 HkxgkstkrhgSA
rFk vd = neA
pwafd/k jk,dpkSFkbZgkstkrhgSrFk vuqizLFkdkVdk{ks=kQyvk/k gkstkrkgS]blfy,mijksDrlehdj.klsbysDVªkWu dk vuqxeu osx vk/kk gks tkrk gSA blfy, oDrO;- vlR; gSA
18.36 STATEMENT-1 : In the circuit shown both cells are ideal and of fixed emf, the resistor of resistance R1 has fixed resistance and the resistance of resistor R2 can be varied( but R2 is always non-zero). Then the electric power delivered to resistor of resistance R1 is independent of value of resistance R2.
oDrO;-1: fn[k ;sx;sifjiFkesanksuksalsyvkn'kZgSrFk mudkfo-ok-cy(emf)fu;r(fixed)gSAR1dkizfrjks/kfu;r gSAizfrjks/kR2dsekudksifjofrZrfd;ktkldrkgS(ysfduR2ges'k v'kwU;gS)Aizfrjks/kR1dksiznkudhxbZfo|qr 'kfä]izfrjks/kR2dsekulsLorU=kgSA
STATEMENT-2 : If potential difference across a fixed resistance is unchanged, the power delivered to the resistor remains constant.
oDrO;-2: ;fnfu;r(fixed)izfrjks/kdsfljksaijfoHkokUrjvifjofrZrjgrkgSrksizfrjks/kdksiznkudhxbZ'kfä fu;rjgrhgSA
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
(A*) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(C) oDrO;-1lR; gS] oDrO;-2vlR; gS ;
(D) oDrO;-1vlR; gS] oDrO;-2lR; gS
Sol. The potential difference across the resistance is always |E1 – E2| in magnitude. Hence statement 1 and 2 are true and statement 2 is correct explanation of statement 1 . izfrjks/kdsfljksaijfoHkokUrjifjek.kesages'k |E1–E2| dscjkcjgksrkgSAblfy;soDrO;1rFk 2lR;gSrFk oDrO;
2 , oDrO; 1 dk lgh Li"Vhdj.k gSA
18.37 Statement–1 : The power delivered to a light bulb is more just after it is switched ON and the glow of the filament is increasing, as compared to when the bulb is glowing steadily, i.e., after some time of switching ON. oDrO;–1:daqthlaidZdjrsghizdk'kcYcdksiznÙk'kfDrdkQhle;i'pkriznku'kfDrdhrqyukesaT;knkgksrh gSAoizdk'kcYcdhiznhfIrlr ~iznhfIrdhrqyukesac RA (B) R < RA
(C) R = RA (D) dependent upon E and r. ErFk rijfuHkZj
R . RV
Sol. (Easy) RA =
R RV < R
18.40 The relation between RB and the actual value R is :
RB vkSjRds okLrfodeku dschp lEcU/kgS &
(A*) R < RB (B) R > RB
(C) R = RB (D) dependent upon E and r. ErFk rijfuHkZj
Sol. (Easy) RB = R + RG > R
18.41 If the resistance of voltmeter is Rv = 1 k and that of ammeter is R = 1 , the magnitude of the
percentage error in the measurement of R (the value of R is nearly 10 ) is :
;fnoksYVehVjdkizfrjks/kRv =1k gSvkSjvehVjdkRG =1gS]rks R(Rdkeku10 dsyxHkxgS) dsekiu esaizfr'kr=kqfVdkifjek.kgS&
(A) zero in both cases (B) non zero but equal in both cases
(C) more in circuit A (D*) more in circuit B (A)nksuksafLFkfr;ksaesa'kwU;gS (B)v'kwU;ijUrqnksuksafLFkfr;ksaesaleku
(C) ifjiFkAesavf/kd (D*) ifjiFkBesavf/kd
Sol. (Tough) % error in case A.
fLFkfrAesaizfr'kr=kqfV
RA R
RV R
R × 100 = R R
1 × 100 =
R RV
× 100 - 1%
% error in case B fLFkfrBesaizfr'kr=kqfV
RB R × 100 = RG× 100 10%
R R
Hence percentage error in circuit B is more than that in A.
blfy,]ifjiFkBesaizfr'kr=kqfVAlsvf/kdgksxhA
Comprehension # 2
Paragraph for Question Nos. 13 to 14
iz'u 13 ls 14 ds fy, vuqPNsn
In the arrangement shown in the figure when the switch S2 is open, the galvanometer shows no deflection for
𝑙 = L/2. When the switch S2
is closed, the galvanometer shows no deflection for 𝑙 = 5 L . The internal
12
resistance (r) of 6 V cell, and the emf E of the other battery are respectively. Wire AB is potentiometer wire and resistance of other conducting wires is negligible. (Internal resistance of cell E is negligible) fp=kesan'k Z;sx;svfHkfoU;klesatcdqathS2[kqyhgS]/kjkekih𝑙=L/2dsfy,dksbZfo{ksiughan'kZrkgSAtcdqathS2
cUngS]/k jkekih𝑙= 5 L dsfy,dksbZfo{ksiughan'k ZrkgSA6VlsydkvkUrfjdçfrjks/k(r)gSrFk vU;lsydkfo|qr
okgdcyEgSArkjABfoHkoekihrkjgSrFk vU;la;ksturkjksadkçfrjks/kux.;gSA(lsyEdkvkUrfjdçfrjks/kux.; gS)
18.42 Calculate emf of cell E : lsy Eds fo|qr okgd cy dh x.kuk dhft, &
(A) 6 V (B) 5 V (C*) 12 V (D) 10 V
18.43 Calculate the internal resistance 'r' : vkUrfjdçfrjks/k'r'dhx.kukdhft,&
(A) 1 (B*) 2 (C) 3 (D)zero'kwU;
Sol.(13 to 14)
casefLFkfr-I S2isopen[kqykgS
Potential gradient foHko ço.krk = E
so vr% 6 = E L
E = 12 V
casefLFkfr-I S2 is closed cUn gS
6
i = 10 r
6 – ir = E 5L
L 12
6
6 – 10 r r = 5
6r
r 10 = 1
6r = r + 10
r = 2
18.44 If the current in 8 resistance is 2A then the current through resistance 'R' (in ampere) would be :
;fn 8 izfrjks/k esa /kkjk 2A gS rks izfrjks/k 'R' ls izokfgr /kkjk
¼,Eih;j esa½ gksxhA
(A*) 6 (B) 7 (C) 8 (D) 9
Sol.
4 izfrjks/k esa/kkjkCurrentin4resistance=4A
dqy /kkjk Total current = 4 + 2 = 6A
Comprehension # 3
In the circuit shown, the resistances are given in ohms and the battery is assumed ideal with emf equal to 3.0 volts.
fn[k ;sx;sifjiFkesa]izfrjks/kvkseesafn;sx;sgSarFk cSVjhdksvkn'kZekfu;sftldkfo-ok-cy3.0oksYVgSA
18.45 The resistor that dissipates maximum power.
ogizfrjks/k]ftlesavf/kdre'kfä{k;gksxh]gS&
(A*) R1 (B) R2 (C) R4 (D) R5
18.46 The potential difference across resistor R3 is
izfrjks/kR3dsfljksaijfoHkokUrjgS&
(A*) 0.4 V (B) 0.6 V
(C) 1.2 V
(D) 1.5 V
18.47 The current passing through 3V battery is
3VcSVjhlstkusokyh/k jkgS&
(A) 10 mA (B) 30 mA
(C*) 40 mA
(D) 60 mA
Sol. 18 [Moderate],19 [Easy] ,20[Easy] The equivalent circuit can be redrawn as shown in figure1. From figure 1 it is obivious that power dissipated by R1 is maximum.
rqY;ifjiFkdksiqu%fp=k uqlkjfn[k ;ktkldrkgSAfp=klsLi"VgSfdR1 izfrjks/k}kjk{k;'kfDrvf/kdregSA
Potential difference across R2 is =
25 3 volt =1 volt
25 50
Therefore potential difference across R3
or R4 =
20 1volt =0.4 volt
20 30
The equivalent resistance of circuit across the cell is 50+25 = 75 ohms
Therefore current drawn through cell is
3 1000 mA = 40 mA.
75
R2 dsfljksaijfoHkokUrj=
25 3 volt =1oksYV
25 50
blfy;s R3
;kR4ij foHkokUrj =
20 1volt =0.4oksYV
20 30
blfy;slsydsfljksadschpifjiFkdkrqY;izfrjks/k=50+25=75vkse
blfy;s lsy ls yh xbZ /kkjk =
3 1000 mA = 40mA
75
SECTION - V : MATRIX - MATCH TYPE
[k.M-V:eSfVªDl&lqesyizdkj
18.48 Column I gives physical quantities of a situation in which a current i passes through two rods I and II of equal length that are joined in series. The ratio of free electron density (n), resistivity () and cross-section area (A)
of both are in ratio n1 : n2 = 2 : 1, : = 2 : 1 and A1 : A2 = 1 : 2 respectively. Column II gives corresponding
results. Match the ratios in Column with the values in Column . LrEHkImlfLFkfrdhHk Sfrdjkf'k;ksadksfn[k rkgSftlesa,d/k jkiJs.khØeesatqM+hcjkcjyEckbZdhnksNM+ksaI vkSjIIlsgksdjcgrhgSAnksuksadseqäbysDVªkWu?kuRo(n),izfrjks/kdrk()vkSjvuqizLFkdkVdk{ks=kQy(A)dkvuqikr Øe'k%n1 :n2 =2:1,1 :2 =2:1vkSj A1 :A2 =1:2gSA LrEHk II laxr ifj.kkeksa dks fn[kkrk gSA LrEHk esa fn;s x;sHk Sfrdjkf'k;ksadsvuqikrdksLrEHk esafn;sx;slaxrifj.k eksalslqesfyrdhft,A
i
A B C
Column Ι Column ΙΙ
LrEHkI LrEHkII
Drift velocity of free electron in rod
(A)
(B)
(C)
(D)
(A)
Drift velocity of free electron in rod
Electric field in rod
Electric field in rod
Potential difference across rod
Potential difference across rod
Average time taken by free electron to move from A to B Average time taken by free electron to move from B to C
NM eseaqäbysDkVWªduk vuqxeu¼vioguo½sx NM eseaqäbysDkVWªduk vuqxeu¼vioguo½s
(p) 0.5
(q) 1
(r) 2
(s) 4
(p) 0.5
(B) (q) 1
(C)
(D)
NM dsfljksij foHkoUkrj NM dsfljksij foHkoUkrj
eqDr bysªkDWV}kujk A ls B rd tkusesyaxk vkSlrle; eqDr bysªkDWV}kujk B ls C rd tkusesyaxk vkSlrle;
(r) 2
(s) 4
Ans. (A) q (B) s (C) s (D) q
Sol. (Moderate) (A) Since current in both rods is same.
pw¡fd/kjknksuksaNM+ksesalekugS
n ev A = n ev A
1 1 1 2 2 2
v1 n2 A 2 =
1 2 = 1
v2 n1 A1 2 1
(B) E = J =
E1 1 A 2
= 2 2 = 4
A E2 2 A1 1 1
(C)
p.d. across rod
p.d. across rod
NM ij foHkorkjU NM ij foHkorkjU =
E1 AB
E2 BC = 4
(D)
Average time taken by free electron to move from A to B
=
AB v2 = 1
Average time taken by free electron to move from B to C
v1 BC
eqDr bysªkDWV}kujk A ls B rd tkusesyaxk le;
= AB v2 = 1
eqDr bysªkDWV}kujk B ls C rd tkusesyaxk le;
v1 BC
18.49 In the circuit shown, battery, ammeter and voltmeter are ideal and the switch S is initially closed as shown. When switch S is opened, match the parameter of column I with the effects in column II .
iznf'kZr fp=k esa cSVjh] vehVj rFkk oksYVehVj vkn'kZ gS rFkk fLop S izkjEHk esa fp=kkuqlkj can gSA tc fLop S [kksy fn;k tkrk gS] LrEHk I esa fn;s x;s izkpyksa dks LrEHk II esa fn;s x;s izHkkoksa ls lqesfyr fdft;sA
(A) Equivalent resistance across the battery (p) Remains same
(B) Power dissipated by left resistance R (q) Increases
(C) Voltmeter reading (r) decreases
(D) Ammeter reading (s) Becomes zero.
(A) cSVjhdsfljksaijrqY;izfrjks/k (p)vifjofrZr
(B) ck;saizfrjks/kR}kjk{k;'kfDrdkeku (q)c<+sxk
(C) oksYVehVjdkikB~;kad (r)?kVsxk
(D) vehVjdkikB~;kad (s)'kwU;gkstk;sxk
Ans. (A) q, (B) r, (C) r , (D) r
When switch S is opened then right side resistance R which was short circuited earlier contributes to equivalent resistance. Hence, equivalent resistance across the battery increases, power dissipated by left resistance R decreases, voltmeter reading decreases and ammeter reading decreases.
18.49 (A) q, (B) r, (C) r , (D) r
tcfLopSdks[k syktkrkgSrksnk;hvksjdkizfrjks/kR tksigysy?kqifFkrFk vcrqY; izfrjks/kesala;ksftrgks tkrkgSAvr%cSVjhdsfljksaijrqY;izfrjks/kc<+rk gS]ck;saizfrjks/kR}kjk'kfDrdk{ks;degksrkgS]oksYVehVj rFk vehVjdkikB~;kad?kVrkgS
18.50 Consider the circuit shown. The resistance connected between the junction A and B is 60 including the resistance of the galvanometer. The switches have no resistance when shorted and infinite resistance when opened. All the switches are initially open and they are closed as given in column . Match the condition in column with the direction of current through galvanometer and the value of the current through the battery in column .
fn[kk;s x;s ifjiFk dks fyft;sA lfU/k;ksa A rFkk B ds chp tqM+k gqvk izfrjks/k xSYouksehVj ds izfrjks/k dks 'kkfey djrs gq;s 60 gSA fLopksa dks y?kqifFkr djus ij bldk dksbZ izfrjks/k ugh gS vkSj [kqyk NksM+us ij budk vuUr izfrjks/k gSA lHkh fLop izkjEHk esa [kqys gq, gSa vkSj os LrEHk esa fn;s vuqlkj cUn fd;s tkrs gSA LrEHk esa nh xbZ fLFkfr dks LrEHk esa xSYouksehVj esa izokfgr /kkjk dh fn'kk vkSj cSVjh ls izokfgr /kkjk ds eku ls lqesfyr dhft,A
Column Ι Column ΙΙ
(A) Only switch S1 is closed
(B) Only switch S2 is closed
(C) Only switch S3 is closed
(D) Only switch S4 is closed (p) Current from A to B
(q) Current from B to A
(r) Current through the battery is 12.0 A
(s) Current through the battery is 15.6 A
(t) Current through the Galvanometer is 1.2 A
LrEHkΙ LrEHkΙΙ
(A) dsoy fLopS1 cUn gS
(B) dsoy fLopS2 cUn gS
(C) dsoyfLop S3 cUn gS
(D) dsoyfLop S4 cUn gS
Ans. (A) p, s, t (B) q, r (C) q, t
(D) p (p) /k jkAlsB dhvksjA
(q) /kkjkBlsAdhvksjA
(r) cSVjhls/k jk12.0AgSA
(s) cSVjhls/k jk15.6AgSA (t)xsyosuksehVjls/k jk1.2AgSA
15
Sol. When switch S1
A
is closed
tc fLop S1 cUn gS
When switch S2 is closed
tc fLop S2 cUn gS
20
4.5A
When switch S3
is closed
tc fLop S cUn gS
When switch S4
is closed
tc fLop S4 cUn gS
SECTION - VI : INTEGER TYPE
[k.M-VI:iw.kkZadizdkj
SHORT SUBJECTIVE
y?kqÙkjkRediz'u
18.51 Find current in the branch CD of the circuit (in ampere).
fn[kk, x;s ifjiFk esa CD 'kk[kk esa /kkjk dk eku ¼,fEi;j esa½ Kkr djks &
Ans. 15 A
Sol.
R = 3/2
30
i = 3 / 2
= 20 Amp.
From figure current through B D branch = 5 Amp.
fp=k esa B D 'kk[kk ls izokfgr /kkjk = 5 ,Eih;j
18.52 The circuit shown in the figure contains three resistors R1
= 100, R
= 50 & R
= 20 and cells of
emf's E1= 2V & E2. The ammeter indicates a current of 50mA. Determine the current(mA) in the resistor R2
n'kkZ;s x;sifjiFk fp=k esa rhuizfrjks/k R1 =100, R2 = 50o R3 =20 rFkkE1= 2V o E2. fo|qrokgd cy
dslsytqM+sgq;sgaSAvehVj50mA/kjkn'kZrkgSAizfrjks/ksaR2esa/kjkmAesagksxhAvehVjolsyksadkvkUrfjdizfrjks/ kux.;gSA
Ans. 20
Ans. nk;h rjQ, I2=20mA towards left ck;h rjQ.
Sol. Applying KVL in loop ABCD ywi ABCD esa KVL ls
E = ( + 0.05) R + R
= – 20 mA
Current through R = 30 mA towards right
R1 ls izokfgr /kkjk = 30 mA nk;h rjQ
Current through R2 = 20 mA towards left
R2 ls izokfgr /kkjk = 20 mA ck;ha rjQ
Applying KVL in loop BGFE
ywi BGFE esa KVL ls
E = ( + 0.05) 100 + (0.05)20 = 4 volts
18.53 All batteries are having emf 10 volt and internal resistance negligible. All resistors are in ohms. Calcu- late the current in the right most 2 resistor. lHkhCkSVfj;ksadsvkUrfjdizfrjks/kux.;ofo|qrokgdcy10volt gSAlHkhizfrjks/kvkseesagSAnka;hvksjds2 izfrjks/kesa/kjkKkrdhft;sA
Ans. 70
Sol. Thesimplified circuit is. ljy ifjiFk esa ge Kkr djuk pkgrs gSa
We have to find .
Let potential of point ‘P’ be ‘0’. Potential at other points are shown in the figure apply kirchoff’s current law at point x.
ekukPfcUnqdkfoHko'kwU;gSAfp=kesaiznf'kZrnwljsfcUnqijfdjpkWQ/k jkfu;elsfoHkoxgS.
x 10
4
x 10
+ 2
x 20
+ 4
(x 10) 0
+ 2 = 0
x - 10 + 2x - 20 + x – 20 + 2x – 20 = 0 6x = 70
x =
=
35
3 volt.
20 35
3 = 25
4 12
A. [Ans: 25/12Amp]
18.54 In the circuit diagram shown if the current through the 1 resistor is
x A then x is
2
fn[kk;sa x;s ifjiFk fp=k esa 1 izfrjks/k ls xqtjus okyh /kkjk dk
eku
Ans. 5
5
Sol. 2 A
x A gks rks x gksxkA
v 20 v
2 + 2 + v 5 = 0 ; v 20 + v + 2 (v 5) = 0 4v 20 10 = 0
v = 30
= 15 ; v – 5 = 15
5 = 15 10
= 5 i = 5 / 2
= 5 amp. Ans.
4 2 2
2 2 1 2
18.55 The efficiency of a cell when connected to a resistance R is 60%. What will be its efficiency (in%) if the external resistance is increased to six times.
tc lsy dks R ls tksM+rs gS rks n{krk 60% gksrh gSA ;fn ckº; çfrjks/k dks N% xquk dj fn;k tk;s rks n{krk (in%) D;k gksxh
Ans. 90%
Sol. Efficiency = =
out put power input power
n{krk = =
fuxZr k' fDr fuosk'h k' fDr
i2R
= i
i =
R r
R
= R r
0.6 =
R
R r
3R + 3r = 5R ;k 2R = 3r
new efficiency ubZ n{krk =
6R
6R r
= 0.9 = 90% Ans.
18.56 Figure shows a cell in which unit positive charge experience a constant non electric force of 10N and a constant electric force of 8N in directions shown in the figure. Find the emf of the cell, difference across the cell (in volt) fp=kesafn[kk;k x;kgS fd,dlSyesa,d,dkad/kukRedvkos'kij10Nekudk,d fu;rvoS|qr cyo 8N eku dkfu;roS|qrcyfp=kesan'k Z,svuqlkjfn'kkesaeglwldjjgkgSAlSydkfo-ok-cy(involt) crkb;s
[3]
Sol. V = Potential difference across the cell = Electric field × width of the cell V= lsyijfoHkokUrj=fo|qr{ks=k×lsydhpksMkbZ
= 8 × 0.1 = 0.8 volt Ans.
= emf of the cell lsy dk fo|qr okgd cy = 10 × 0.1 = 1.0 volt Ans.
Also r is the internal resistance and i is the current drawn from the cell r lsydkvkUrfjdizfrjks/koilsylsyhxbZ/k jkgSA
V = – ir or;k 0.8 = 1 – 1r
r = 0.2 Ans.
LONG SUBJECTIVE
fucU/k Rediz'u
18.57 Consider the potentiometer circuit arranged as in figure. The potentiometer wire AB is 300 cm long. (a) At what distance from the point A should the jockey touch the wire to get zero deflection in the galvanometer ?
(b) If the jockey touches the wire at a distance of 275 cm from A, what will be the current in the galvanometer?
fp=kesan'k Z,vuqlkjfoHkoekihifjiFkdhdYiukdjksAfoHkoekihdkrkjAB,300cmyEckgSA(a)fcUnqAlsfdruh nwjhijtkWdh(jockey) dksrkjABlsLi'kZdjk;ktk,rkfdxsYosuksehVjdkikB;kad'kwU;vk,?(b)vxjtkWdh dksfcUnqAls275cmdhnwjhijLi'kZdjk;ktk,rksxsYosuksehVjesa/k jkdkekuD;kgksxk?
Sol. (a) When Jockey is not connected. tc
tkWdh tqM+h gqbZ ugha gSA
E
= 13 r
................(i)
Resistanceperunitlength ,dkad yEckbZ dk izfrjks/k
12r
= 300
/cm
Let 𝑙 be the length when we get zero deflection.
ekuk'kwU;fo{ksidhfLFkfresayEckbZ𝑙gS.
E
2 = ( 𝑙)
E E
2 13r
12r 𝑙
300
𝑙 = 157.5 cm
(b) Letpotentialat Ciszero ekuk Cij foHko 'kwU; gS Thenapply Kirchoff’sst lawfdjpkWQ ds izFke fu;e ls
x E 0
x 0
11r
2 (x E 0)
2 r 2r
11E
= 0 x = 16
E 11 E
x 16 2 3E
2
g 2 r
= 2 r =
32 r
Alternatemethod oSdfYid fof/k
𝑙 = 300 cm
r’ = (275) ×
r’ = 11r
12 r
300
Using KVLinloop (i) ywi (i) esa KVL ls
E – .11r – r – r = 0 (i)
andinloop(2) rFkk ywi(2)esa
– 11r + ( – )2r + E = 0 (ii)
1 1 2
Solving equation (i) and (ii) we have
lehdj.k (1) o (2) dks gy djus ij
E
1 16 r
5E
and rFkk = 32r
So current in galvanometer
Branch vr% /kkjkekih'kk[kk esa/kkjk =(–1) =
5E
32r
E
16 r
3E 32r
3E
g = 32r
Ans. (a) 157.5 cm (b)
3E
32r
18.58 In the circuit shown all five resistors have the same value 200 ohms and each cell has an emf 3 volts. if the
x
open circuit voltage is 10 volt then x is.
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