solutions of current electricity exercise-1...(ii) = 1 r dr dt 273 0 dt = 20 10 dr r .273 = n 20 10...

Current Electricity

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SOLUTIONS OF CURRENT ELECTRICITY EXERCISE-1

PART - I Section (A) :

A-1. i = io + sin t.

dq

dt = 10 +

2

sin t

q =

3

0

10 sin t dt2

= 10 × 3 +

2

×

1

× 2 = 31 C Ans.

Average current = q

t =

31

3 A Ans.

A-2 Mass of copper per unit volume = 9 × 103 kg

Atoms of Cu per unit volume =299 6.023 10

63.5

= n.

i = neAVd.

Vd = 19 7 29

1.5 63.5

1.6 10 10 9 6.023 10

= 1.1 0 × 10–3 ms–1.

A-3. (i) Q = i t = 5 × 4 × 60 = 1200C

(ii) Q = ne n = –19

Q 1200

e 1.6 10

= 75 × 1020

Section (B)

B-1. (a) 320 10

e

no of electrons passing per second

= 3

19

20 10

1.6 10

=

172 10

1.6

= 1.25 × 1017.

(b) j = 3

–3 2

20 10

( 0.2 10 )

=

1

2 × 106 A/m2.

B-2. Voltage the across the wire = E= 25 × 2 = 50 V

i = 50

5 = 10 A.

B-3. (i) R15 = 200

10 = 20 ., Rt =

200

9

Rt = R15 (1 + t) 200

9 = 20

t1

234

t

234

=

1

9 t =

234

9 = 26 t = 26 + 15 = 41º C.

(ii) =1

R

dR

dt

273

0

dt =

20

10

dR

R

.273 = n20

10

= n2

273 oC–1.

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Current Electricity




B-4. V

I = slope of given graph =

1

R or R =

1

slope

Resistance of a metallic wire increases with increase in temperature.

2T(slope) <

1T(slope)

2T

1

(slope) >

1T

1

(slope)

or 2TR >

1TR

or 2T > 1T

B-5. Resistance R = A

By partial differentiation

R

R

=

–

A

A

.........(1)

= Constant

Volume of wire remains constant

A × = Constant

By partial differentiation

A

A

+

= 0 .........(2)

By equation (1) and (2)

We get R

R

= 2

(% change in R) = 2 (% change in length) = 0.2 % Ans.

B-6 (i) R = –5 –2

–2 2

(3.5 10 ) (50 10 )

A (1.0 10 )

=

0.35

2 = 0.175

(ii) R = –5 –2

–2 –2

(3.5 10 ) (1.0 10 )

A (1.0 10 ) (50 10 )

= 7 × 10–5

Section (C) C-1. (a) In each case E.M.F. = 10 V (b) For case (A), battery is providing current to the circuit hence acting as source. For case (B) battery is taking current from external source, hence acting as a load. (c) For case (A) VA = E – ir = 10 – 1 × 1 = 9 V For case (B) VB = E + ir = 10 + 1 × 1 = 11 V (d) For case (A) PA = VA i = 9 × 1 = 9 W For case (B) PB = VBi = 11 × 1 = 11 W

(e) For case (A) AH

t = iA2 rA = (1)2 × 1 = 1 W

For case (B) BH

t = iB2 rB = (1)2 × 1 = 1 W

(f) For case (A) PA = EA i = 10 × 1 = 10 W For case (B) PB = EB i = 10 × 1 = 10 W (g) For case (A) VBox = VA = 9 V For case (B) VBox = VB = 11 V (h) For case (A) PBox = – 9 × 1 = – 9 W For case (B) PBox = 11 × 1 = 11 W



Current Electricity




C-2. V = W

Q

Q = I × t = 3 × 12 = 36 C

V = 500

36 =

125

9 V Ans .

C-3. (a) equal in all point a, b, and c (b) Vb > Vc = Va (c) (P.E)a = (P.E)c > (P.E)b

P.E = – eV C-4. (a) V = E – ir = 12 – 90 × 5 × 10–2 = 12 – 4.5 = 7.5 V

(b) max = E 12

r ' 500 = 24 mA

(c) For charging of battery V = E + ir , V > E V > 12 V

C-5. (a) i = P 1000 50

V 220 11 = 4.55 A

(b) R = 2 2V (220) 22 11

P 1000 5

= 48.4

(c) P = 1 kW

(d) H = Q 1000

J 4.2 = 240 cal/sec

(e) tH = mL m = Ht 240 60 80

L 540 3

gm.

C-6.

(a) VA = VB = VC = VD = 0 VE = VF = VG = VH = 10V VI = 10 + 5 = 15 V VJ = 15 V VK = 10 + 5 = 15 V (b) VBI = 15 V, VJG = 5 V, VKD = 15 V (c) Each battery is supplying the current hence each battery is acting as a source. (d) Let current through BF, CG, HP is respectively i1, i2, i3

i1 = 15

1 = 15 amp , i2 =

5

2 = 2.5 amp i3 =

15

3 = 5 amp

For 10 V Battery , current = i1 – i2 + i3 = 15 – 2.5 + 5 = 17.5 amp

(e) P1 = 2V

R =

2(15)

1= 225 W P2 =

2(5)

2 = 12.5 W P3 =

2(15) 225

3 3 = 75 W

Hence, 1 resistance consumes the maximum power. (f) PI = EI i1 = 10 × 17.5 = 175 W PII = E2 i1 = 5 × 15 = 75 W PIII = E3 i2 = 15 × 2.5 = 37.5 W PIV = E4 i3 = 5 × 5 = 25 W hence left most battery consume maximum power.



Current Electricity




C-7. The circuit shown above is a parallel circuit, and consists of a single node By assuming voltage V at

the node, we can find out the current in 10 branch.

According to Kinrchhoff’s Current Law,

I1 + I2 + I3 + I4 + 5 = 10

By using Ohm’s law, we have

I1 = 2

V;

5 = 3

V;

10 = 4

V;

2 =

V

1

V

5 +

V

10 +

V

2 + V + 5 = 10

V1 1 1

15 10 2

= 5

V [0.2 + 0.1 + 0.5 + 1] = 5

V = 5

1.8 = 2.78 V Ans

The voltage across 10 resistor is 2.78 V

and the current passing through it is

I2 = V 2.78

10 10 = 0.278 A Ans

C-8. According to Kirchhoffs Voltage Law, the sum of the potential drops equal to the sum of the potential

rises;

Therefore, 30 = 2 + 1 + V1 + 3 + 5

or V1 = 30 – 11 = 19 V Ans.

C-9. (a) H = I2 Rt 400 = 22 × R × 10 R = 10

(b) H = I2 Rt Now = 4A, t = 20s H = 42 × 10 × 20 = 3200 J

C-10. By applying Kirchhoff’s Current Law, we get the current in 10

I10 = I5 + I6 = 4 + 1 = 5 A Ans.

The voltage across 6 resistor is V6 = 24 V = V1 – 10 × 5 V1 = 74 V

Now, consider the loop ABFE

If we apply Kirchhoff’s Voltage Law, we get

Vs = 5 – 30 – 24 = –49 V



Current Electricity




Section : (D)

D-1. R =

2220

500 for the two bulbs,

110

2V will be the potential difference individually

P =1

4 ×

2

2

110 (500)

(220)

=

500

16 =

125

4 = 31.25 W Ans.

D-2. For quantitative purposes we assume that the resistances of the bulbs do not depend upon the voltages

across them. This is far from accurate, but will give the correct qualitative conclusion. If the (r.m.s.)

supply voltage is V, the resistance ri of a bulb is V2/wi, where wi is the nominal rating of the bulb. When

the two bulbs are connected in series across the supply, the (r.m.s) current drawn is V/(rA + rB) and the

power dissipated in bulb i (i = A or B ) is

Pi =

22

2 2i A B

V V

w (V / w ) (V / w )

According to the original agreement (wA = wB = 100 W), both PA and PB should be 25 W. Actually, PA =

8 W and PB = 32 W, and so A clearly failed his examinations. By comparison, student B might be

considered a double winner : he gets 32 W, but pays for only (8 + 32)/ 2 = 20 W. On the other hand, 32

W is still a very poor light to study by and B also could well have failed his examinations.

D-3.

3

3

3

3

3

3

66

6

A B

CD

33

3

3

6

A B

C

3

3

3

3

6

A B

Req = 2 Ans.

D-4. (a) Req = R1 + R2 + R3 + r = 10

(b) i = eqR

=

10

10 = 1 A

(c) V3 = 1 × 3 = 3 V, V4 = 1 × 4 = 4V, V2 = 1 × 2 = 2V

(d) Pconsumed = i = 10 × 1 = 10 W (e) Pgenerated =i2r = 1 W inside the battery

(f) Poutput = ( – ir)i = 10 – 1 = 9 W

(g) Vbattery = – ir = 9 V

(h) In seriesa, P R 4 consumes maximum power

(i) P3 = i2 R = 3 W.



Current Electricity




D-5. (a) Req = 2 + 1 = 3

(b) i = eqR

=

6

3 = 2 A

i1 + i2 + i3 = 2 i1 : i2 : i3 = 1

8 :

1

4 :

1

8 = 1 : 2 : 1 i1 = i3 =

1

2 A, i2 = 1 A

(c) Vacross battery = – ir = 6 – 2 × 1 = 4 V Vacross each cell = 4 V

(d) P of the cell consumed P = i = 12 W (e) P heat generated in cell P = i2r = 4 W

(f) Poutput = i – i2r = 8 W

(g) In parallel a, P 1

R 4 consumes max power

(h) P4 = 2v

R =

4 4

4

= 4 W

D-6.

i1 = 2

5 = 0.4 A i2 =

4

8 = 0.5 A

(a) Vx + 3i1 + 4 – 3i2 = Vy Vx – Vy = – 4 + 3 × 0.5 – 3 × 0.4

= – 4 + 1.5 – 1.2 = –3.7 V potential difference = 3.7 V. (b) Still same as No current flows in that cell

D-7. (i) RAB = 5 1 5

5 1 6

(ii) RCD =

3 66

2 6 33 6

3 6 2 6 26

3 6

= 1.5

(iii) REF =

3 64 2

3 6 (2 4) 2 3

3 6 2 4 2 24 2

3 6

= 1.5

(iv) RAF = RAB = 5

6 (v) RAC =

6 61 2

(3 1) 2 46 6

6 6 (3 1) 2 31 2

6 6



Current Electricity




D-8. (i) Let RAB = x. Then, we can break one chain and connect a resistance of magnitude x in place of it. Thus, the circuit remains as shown in figure.

26V

1A

B

x

Now, 2 and x are in parallel. So, their combined resistance is 2x

2 x

or RAB = 1 + 2x

2 x

But RAB is a assumed to x. Therefore,

x = 1 + 2x

2 x

Solving this equation, we get

x = 2 Hence proved.

(ii) Net resistance of circuit R = 1 + 2 2

2 2

= 2

Current through battery i = 6

2 = 3A

This current is equally distributed in 2 and 2 resistances. Therefore, the desired current is i

2 or 1.5

A.

D-9. (i)

As V R V2 = VBC = BC

AB BC

R

R R

V =

1.5 0.5

1.5 0.51.5 0.5

0.51.5 0.5

50

= 0.75

1 0.75

50 =

150

7 = 21.43 V

(ii) VBC = 40 V VBC = BC

BC AC

R

R R

V

40 = BCR

2000 50 RBC = 1600



Current Electricity




D-10.

VB = VE 1 (2 )

3

=

2 – = 3 – 2 2 – 4 + 2 = 0 = 4 16 8

2

= 2 – 2 . ( < 2)

CE

DE = CE

DE

R

R =

1

=

2 2

1 2 2

=

2 2

2 1

=

2( 2 1)

2 1

= 2 : 1

D-11. For maximum resistance, all resistance are in series

Rmax = 20 + 50 + 100 = 170 For minimum resistance, all resistance are in parallel

min

1 1 1 1 1 2 5 8

R 100 50 20 100 100

Rmin = 100

8 = 12.5

D-12. case (a) Req = 180

3 = 60 i =

60

60 = 1 amp

case (b) Req = 180

2 = 90 i =

60

90 =

2

3 amp

case (c) Req = 180 i = 60 1

180 3 amp

D-13. (a) S is open

i = 3

30= 0.1 A Ans.

(b) S is close

i = 3

10= 0.3 A Ans.

Section (E)

E-1.

(i) current i = 6E 6 2 12

R 6r 8.5 6 0.015 8.59

= 1.4 A

(ii) terminal voltage V = iR = 12

8.59 × 8.5 = 11.9 V



Current Electricity




E-2. (i) Effective emf = 4 × 1.5 – 1.5 = 4.5 V

(ii) Effective internal resistance

req = 4 0.4

2

+ 0.4 = 1.2 Req = 6 + 1.8 + 1.2 = 9

= eq

E 4.5 1

R 9 2 amp 36 =

eq

36

R 6 1 1

R 36 2 12

(ii) VAB = E + ir = 1.5 + 1

2 × 0.4 = 1.5 + 0.2 = 1.7 V

E-3. Eeq =

6 312 9 21600 400

1 1 2 3 5

600 400

= 4.2 V Ans

eq

1 1 1 2 3 5

r 600 400 1200 1200

req = 240

short circuit current in AB

i = eq

eq

E 21/5 21

r 240 5 240

= 17.5 × 10–3 amp i = 17.5 mA (from B to A)

E-4.

A A

AA

4 6

4V 2V

4i =0

i1

i1

i1

Potential difference across upper 4 resistance is zero so current is also zero.

Other two resistors are in series combination so current is same

= 4 2

4 6

= 0.2 A. Ans.

E-5. (a) i1 = 12

2 0.1 =

12

2.1

i2 = 6

0.5 0.1 =

6

0.6 = 10 A 1

2

i

i =

12

2.1 10 = 0.57 Ans.

(b) i1 = 12

2 1 = 4 A

i2 = 6

0.5 1 = 4A 1

2

i

i = 1 Ans.

(c) i1 = 12

2 10 =

12

12 = 1 A & i2 =

6

0.5 10 =

6

10.5

1

2

i

i =

1 10.5

6

= 1.75 Ans.



Current Electricity




Section (F)

F-1.

(I – Ig) S = Ig G S = g

g–

G

S = 0.002

0.3 – 0.002

30 = 0.002

300.298

S = 30

149 = 0.2013

V = Ig (R + G) 0.2 = 0.002 (R + 30) R = 70

F-2.

Reff = 400 100

500

+ 200 Reff = 280

i = 84

280 A. V =

84

280 ×

100 400

500

= 24 V

(b) i = 84

300 V =

84

300 × 100 = 28 V

F-3.

RGD = 400 400

400 400

= 200

Since GE EB

GD DB

R R

R R it is the case of balanced Wheatstone bridge.

Req = RGB = 300 300

300 300

= 150 , current I =

GB

V 10 1

R 150 15 amp

Potential difference across voltmeter. = 1 20

200 10030 15 3

volt



Current Electricity




F-4. (i)

(ii) Req = 2 + 4

3 +

100 r

100 r =

600 6 r 400 4 r 300 r

3(100 r)

=

310 r 1000

3(100 r)

i = V

R, 0.02 =

1.40

310 r 1000 × (300 + 3 r)

310 r + 1000 = 21000 + 210 r 10 r = 2000

r = 200 Ans.

(iii) V = i r = 0.02 × 200 100

200 100

=

4

3 = 1.33

Zero error ‘ = 1.1 – 1.33 = –0.23 V Ans.

F-5. Current in primary circuit. = 9r r 10r

Potential drop across length AB = VAB = .R

VAB = .9r10r

=

9

10

, x = ABV 9

L 10L

For balance point 2

= x =

9

10L

. =

5

9 L

F-6. r = R( – ')

= 9.5 70 – 60

60

= 9.5 70

–160

= 9.5

6

F-7. (a) 82.3

1.0267.3

= 1.25 V

(b) The high resistance to keep the initial current low when null point is being located. This saves the

standard cell from damage.

(c) This high resistance does not affect the balance point because then there is no flow of current

through the standard cell branch.

(d) The internal resistance of driver cell affects the current through the potentiometer wire. Since

potential gradient is changed, therefore, the balance point must be affected.

(e) No, it is necessary that the emf of the driver cell is more than the emf of the cells.

(f) This circuit will not work well for measurement of small emf (mV) because the balance point

will be very near to end A, and percentage error in EMF measured due to length measurement

would be very large E = V

100

dE

E =

dwill be large if is very small.



Current Electricity




F-8. x

y =

30

70 ........... (1)

x

12y

12 y

= 30 10

60

........... (2)

Solving (1) & (2) x = 20

7 & Y =

20

3

PART - II

Section (A) A-1. The drift velocity of electrons in a conducting wire is of the order of 1mm/s. But electric field is set up in

the wire very quickly, producing a current through each cross section, almost instantaneously.

A-2. In the presence of an applied electric field (E ) in a metallic conductor. The electrons also move

randomly but slowly drift in a direction opposite to E .

A-3. q ne

it t

n = i t

e

Substituting i = 3.2 × 10–3 A e = 1.6 × 10–19 C and t = 1 s we get n = 2 × 1016

A-4 j = A

and j =

E

jA > jB and EA > EB. Section (B) B-1. Copper is metal and germanium is semiconductor. Resistance of a metal decreases and that of a

semiconductor increases with decrease in temperature.

B-2.

a > b > c Let and a = 2c

Rmax = a

bc

Rmin = c

ab

max

min

R

R =

2

2

a

c = 4 Ans.

Section (C) C-1. In an electric circuit containing a battery, the positive charge inside the battery may go from the positive

terminal to the negative terminal



Current Electricity




C-2. Given A r i r = ki V = E – ir = E – i(ki) V = – i2 k + E

C-3. (a) V = E – ir , V < E (b) V = E + ir , V > E (C) V = E (D) V = E

C-4. P =

2E

R 5

R

dP

dR = 0 at R = 5, so power is maximum at R = 5]

R = 5 dP

dR = 0 R = 5

Therefore increase continuously till R = 5.

C-5.

Potential at C point may be greater than potential at point B. Therefore current flow in resistance may be from B to A.

C-6 (i-ii)

Apply KCL for circuit KCL V –10 V – 6 V – 5

10 20 30 = 0

6V – 60 + 3V – 18 + 2V – 10 = 0 11V = 88 V = 8 V Ans

current in resistance R1 = 10 – 8

10 = 0.2 amp

C-7. Current flow in 2R resistance is from right to left.

C-8.

Eq =

4.5 3543 10

1 1 13

3 10

= V, req = 3 10 30

13 13

i = 54 /13 54 1

30 108 26

13

amp.

V6 = i.R = 1

2 × 6 = 3V

There fore current in 10 is zero.



Current Electricity




C-9. = E – r

E

= 1 –

r R

r R r R

= 0.6 R = 0.6 r + 0.6 R

r = 4

6 R =

2R

3 =

6 R 6R 18R

2Rr 6R 2R 18R6R

3

= 0.9 = 90%

Section (D)

D-1. (a) R1 = R01 (1 + 1 ) = 600 (1 + 0.001 × 30) = 618

R2 = R02 (1 + 2 ) = 300 (1 + 0.004 × 30) = 336

Req = R1 + R2 = 618 + 336 = 954

(b) Req = R0eq (1 + eq ) 954 = 900 ( 1 + 30) = 54 1

900 30 500

degree–1

D-2.

Req = 7 + 4 + 9 = 20

V = IReq = 1 × 20 = 20 V

D-3. R2.5 W = 2(110)

2.5 , R100W =

2(110)

100 R2.5 > R100 .

In series current passes through both bulb are same

P2.5 = i2 R2.5, P100 = i2 R100

P2.5 > P100 due to R2.5 > R100 & P2.5 > 2.5W & P100 < 100 W (can be verified)

Therefore 2.5 W bulb will fuse

D-4.

Req =

20 2020

2 3

20 2020

2 3

= 50 20

110

Req =

100

11

P = 2 2V (10)

R 100 /11 = 11 W .

D-5. P = i2R

current is same, so P R

In the 1st case it is r

3, in 2nd case it is 3r, in 3rd case it is

3r

2 and in 4th case the net resistance is

2

3r.

R1 < R4 < R3 < R2

P1 < P4 < P3 < P2



Current Electricity




D-6. Initial Req = 5

Final Req = 3

Change in resistance = 5 – 3 = 2

D-7. R = 2(220)

100

Req = R

3 + R =

24 R 4 (220)

3 300

P = 2 2

2eq

V (220) 300 300

R 44(220)

= 75 W

D-8. This simplified circuit is shown in the figure.

30

30

30

2V=

30 602V =

202V

i

i

Therefore, current i = 2

20 =

1

10A

D-9. In series a , Req = 3 R P = 2V

3R = 10

2V

R = 30

in parallel Req = R/3

P = 2 2V 3V

R /3 R = 3 × 30 = 90 W

D-10. eq

100 200

R 40 Req = 20

Req = R 100

100 R

= 20 R = 25

D-11. For Pmax r = Req , Req = R/3

0.1 = R

3 R = 0.3



Current Electricity




D-12. Resistance of one side = 0.1 × 10 = 1

Req = 1 P = 2 2

eq

V (2)

R 1 = 4 watt

D-13. Since, resistance in upper branch of the circuit is twice the resistance in lower branch. Hence, current

there will be half. i/2

i

4 6

5 Now P4 = (i/2)2 (4) (P = i2R) P5 = (i)2 (5)

or 4

5

P 1

P 5

P4 = 5P

5 =

10

5 = 2 cal/s.

D-14.

For power across heater is maximum resistance of balb should be minimum.

Pheater =

2

H b

V

R R

RH

Rbulb is minimum for 200 W.

D-15. RAB = 9 12

9 12

+ 7

=85

7 Ans.

D-16. Condition for maximum power is r = R

4 = 6R 3R

9 R

R = 2 Ans.

D-17. 2v

Pr

PA = PD = PE > PB = PC



Current Electricity




Section (E)

E-1. Eeq =

1 2

1 2 2 11 2

1 2

1 2

E E

E r E rr r

1 1 r r

r r

req = 1 2

1 2

r r

r r

Therefore statement II is correct bul I is wrong.

E-2. Assume M cells are connected correct and N cells connected wrong. M + N = 12 .......(1)

(M + 2) E – NE = 3R M – N + 2 = 3R

E ......(2)

ME – (N + 2)E = 2R M – N – 2 = 2R

E ......(3)

from eq (1) and (2)

– M + N + 10 = 0 M – N = 10 ......(4) from eq (1) and (2) M = 11, N = 1

E-3. For ideal r 0

E =

10 1510 15 rr 1

1 1 1 r

r 1

E = 10 V

E-4.

i = 1 2 3

1 2 3

.....

R R R .......

= 1 2 3

1 2 3

(R R R .......)

R R R .......

i = . so that potential difference between any two points is zero Section (F)

F-1. Req = 200 + 300 600

300 600

+ 100 = 500

= 100 1

500 5 amp

600 =

11600

1 1 5

300 600

= 1

15 amp

Reading of volt meter = I600 R600 = 1

15 × 600 = 40 V



Current Electricity




F-2. Potential gradient = E

100 V/cm

e.m.f. of battery = 30 × E

100 Ans.

F-3. Req = 2 + 4 15

2 3 + RA = 9 + RA

= eq

V

R 1 =

A

10

9 R RA = 1

if 4 replace by 2 resistance then

Req = 2 + 2 15

2 3 + 1 = 9 I =

10

9 amp

F-4 Req = 10 + 480 20

480 20

= 10 +

96 146

5 5

current passes through the battery.

= 20 5 100 50

146 146 73

amp.

F-5. Case-I : g = 1 2

g

E E

R R 2r

1 =

g

3

R R 2r Rg + R + 2r = 3 .......... (1)

Case-II : Eeq = E = 1.5 V

g = eq

g

E

rR R

2

0.6 =

g

1.5

rR R

2

Rg + R + r

2 =

1.5

0.6= 2.5....(2)

from eq (1) and (2)

3r

2 = 0.5 r =

1

3

F-6. i = 2

10 R x =

V 2 10 1.

(R 10) 100

V1 = x 10 × 10–3 = 2 10 40

(R 10) 100

R + 10 = –3

8

10 10

R + 10 = 800 R = 790

F-7. 6

R x –

6 30

R 20 R = 4

F-8. By ohm's law V = iR (14.4 – x) = 1.75 R (22.4 – x) = 2.75 R

R = 8 14.4 – 1.75 × 8 = x

x = 0.4 volt Ans.

F-9. The ratio AC

CB will remain unchanged.

F-10. As there is no change in the reading of galvanometer with switch S open or closed. It implies that bridge

is balanced. Current through S is zero and IR = IG, IP = IQ.



Current Electricity




PART - III

1. Drift speed Vd = J

ne =

i

neA

i = V

R where R =

L

A

E =

V

L and P = I2 R

2.

short circuited resistor. In a resistor current always flows from higher potential to lower potential. In short circuited resistor or ideal cell, energy dissipated is always zero because in short circuited

resistor no current flow and in ideal cell no internal resistance. Potential difference may be zero across a resistor, non-ideal cell or short circuited resistor.

EXERCISE-2 PART - I

1. R 4r

1

%R = 4 × 0.1 = 0.4%

2. V = E – r 9.8 = 10 – I .1

= 0.2 amp R = V 9.8

0.2

= 49

3.

A

A

B

i

P

r

VA – VB = E – iR < E

4.

VR = E

r R R =

E

r1

R

R 0 VR = 0

& R VR = E

5. For maximum power across the resistance, R is equal to equivalent resistance of remaining resistance

R = 1 2

1 2

R R

R R

6. V = rR2

ER2R

2

rR

E



Current Electricity




7.

Due to input symmetry potential drop in AC, AD and AE part is same. Therefore potential at C, D and E point is same.

Req = 7

3

8. req= r12 = r

2

req = r34 = r

2

9.

RAB = 2 + 8 + 25

2 =

45

2 Ans

10.

Due to input output symmetry, here no current passes through resistance 2 to 6 and 4 to 8. Equivalent circuit is

eq

1

R =

1 1 1

3R 2R 2R

eq

1

R =

4

3R



Current Electricity




11. The circuit can be redrawn as follows :

P Q

2R 2R

2R2R

r r

P Q

4R

4R

2r

P Q

2RrR + r

12.

R = 2V

P

(i) R1 < R2

W1 < W2 (ii) R1 + R2 > R2

i' > i

W3 > W2 From (i) and (ii) W3 > W2 > W1 Alternate solution

P = 2V

R so R =

2V

P

R1 = 2V

100 and R2 = R3 =

2V

60

Now W1 = 2

21 2

(250)

(R R )· R1 W2 =

2

21 2

(250)

(R R )· R2

and W3 = 2

3

(250)

R

W1 : W2 : W3 = 15 : 25 : 64 or W1 < W2 < W3

13. (0 – 0

5

) 4 = 0

5

G ...(1)

(0 – g) 2 4

2 4

= gG ...(2)

from (1) and (2)

0 0

0 g g

16 6

5 8( – ) 5

12g = 0 – g g = 0

13

Ans



Current Electricity




14.

(4 – ) R = RV = 20 (4 – ) R = 20

So that, R is greater than 5 15. Case-I : For ideal voltmeter

V1 = E

4R. 3R =

3

4 E = 0.75 E

Case-II :

V2 = E

7R. 6R =

6

7 E = 0.857 E

Case-III :

V3 = E

3R. 2R =

2

3 E = 0.666 E

16. 0.75 × 60 = 15 × i2 i2= 3 amp

17. i1 = 0.75 + 3 (form KCL) i1 = 3.75 amp

18. For max , Rh is minimum which is zero .

max = 5.5

20 Amp.

for min , Rh is maximum which is 30 .

min = 5.5 5.5

20 30 50

Amp.

min

max

=

5.5 20 2

50 5.5 5 Amp. Ans

19. A

V

V VA

V

R

Equaivalent resistance decrease so current will increases. VA + VV = V Due to change, VA increases so voltmeter reading will decrease.

20.

( – x)

+ x = x =

1

Reading of voltmeter after connection of resistance is – x

= ( 1)

=

6

2 1 = 2V Ans



Current Electricity




21.

i = ig + is .........(1) igG = isS .........(2)

from (1) & (2) (putting ig = 0.1 mA, G = 100) we have i = 100.1 mA

22. BC, CD and BA are known resistance, The unknown resistance is connected between A and D. Hence, the correct option is (D)

23. Iv

v

RR. V

R R

I

v

v

RR V

R R

I

v

1 1

V R R

I

v

1 1

v R R

II

v v

1 VR

1 v

v R R

24. R = 10

2 2

av

(4) (2)0.1 10.1

10 10P0.2

= 1.6 0.4

1 watt2

25. 1 – 1 21

1 2

r 0r r R

r1 + r2 + R = 2r1 R = r1 – r2

26. P

Q =

R

X

10

100 =

600

X

X = 6000 For second case

P

Q =

R

X

10

100 =

630

X X = 6300

Rf = R0 (1 + T)

6300 = 6000 (1 + (100))

= 5 10–4 / C°

PART - II 1.

(a) J = J0 r

1R

shadded area dA = 2rdr di = 2J0

2rr dr

R

i = 2J0

R 2

0

rr dr

R

= 2 J0

2 2R R

2 3

= 2

02 J R

6

= 0J A

3 Put A = 4



Current Electricity




2. Let length of P = meter Resistance of P = A

.

Let length of Q = (1 – ) meter

Resistance of R = A2

2 = 4

A

Resistance of Q = (1 )

A

= 4

A 4 = 1 – =

1

5 1 – =

4

5

Q of Resistance

P of Resistance = A

1A

( )

=

1 =

15

45

= 1

4 Ans.

Q of Length

P of Length =

1 =

1

4 Ans.

3. For maximum power req = Req 2 + 6x

6 x = 4

12 8x

6 x

= 4

12 + 8x = 24 + 4x 4x = 12 x = 3 Ans.

4. Condition for maximum power is Net internal resistance = Net external resistance rnet = Rnet

300r

nn

= R

n2 = 300 r

R

n = 300 0.3

10

n = 3

5. (a) E = 6V, r =10 , r1 = 1

9V = 6 + i × 10 i = 3

10

i = 0.3 A Ans.

(b) When internal resistance r1 = 1

9 = 6 + i1 × 1 i1 = 3 A Ans.

6.

Reff = 1

4

ar ar

2

=

1

4 ar

2

2

= 2

8

ar.

7.

due to input output symmetry potential at point 2, 4, 5, are equal and potential at point 3, 6, 8 are equal

Req = R R R 5

3 6 3 6 R



Current Electricity




8. 1 = 2 sin 18º =2

3 RLK = 2 ×

2

3 =

4

3

RFC = 4

2 = 2

9. (i)

VAB = 2V i2 = 0 & i1 = i3

2 = 3 – i, r, i1 = 1A i3 = 1A

(ii)

Apply kvl in ABCD 2 – (I1 + I2 + I3)R = 0 I1 + I2 + I3 = 2 Ans.

10. (by folding)

Reft = R8

R8

7 R

7 R

=

7 R

15 i =

eft

U

R =

15 U

7 R

11. Applying Kirchhoff’s second law in loop ADBA :

2 – 2i1 – i1 –1 – 2(i1 – i2) = 0 .........(1) Similarly applying Kirchhoff’s second law in loop BDCB 2(i1 – i2) + 3 – 3i2 – i2 – 1 = 0 .........(2) Solving Equation (1) and (2) we get,

i1 = 5

13, i2 =

6

13 and i1 – i2 = –

1

13



Current Electricity




(i) Potential difference between B and D. VB + 2(i1 – i2) = VD

VB – VD = – 2(i1 – i2) = 2

13 volt

(ii) VG = EG – i2rG = 3 – 6

13× 3 =

21

13 volt

VH = EH + i2rH = 1 +6

13 × 1 =

19

13 volt

12. 2 = 12

10000 R R

R = 2000 = 2K, Galvanometer will show deflection, as the temperature of wire wound decreases, resistance decreases.

13. E = 1.52 V V = E – ir 1.45 = 1.52 – 1r

r = 0.07 Ans.

14. Case-I

120V

400300

– 1

1

RV

Current = 60 1

300 5 amp 1 =

60 3

400 20 amp.

60 = ( – 1) RV RV = 60

1 3–

5 20

= 1200

Case-

120V

400300

–

RV

1200

= 120

300 1200400

1200 300

=6

32amp.

0 300 = (– 0) 1200 0 = 1200

1500 =

4 3 3

5 16 20 amp

Reading of voltmeter = 3

20 × 300 =

900

20 = 45 V Ans

15. As the ammeters A1 and A2 are ideal, potential drop across AB and AC are zero. Hence point B and C

are at equal potential, so there will be no current through A3. I3 = 0 Resultant circuit may be drawn as

Applying KVL in the loop ABEFA

– 10 + i2 5 – 15 i1 + 20 = 0 3i1 – i2 = 2 ...(1) Applying KVL in the loop BCDEB



Current Electricity




8 – (i1 + i2) 3 – i2 5 + 10 = 0 3i1 + 8i2 = 18 ...(2)

i2 = 16

9 Amp, i1 =

34

27 Amp

Reading of ammeter A1, i1 + i2 = 82

27 amp Ans.

Reading of ammeter A2, i1 = 34

27 amp Ans.

iA1 + iA2 + iA3

82 34 116 2

0 5827 27 27 27

16.

6 V

800400

i A 10

(a)

6 V

800400

i

1000

(b)

V

Refer figure (a) : Current through ammeter,

i = net emf

net resistance =

6

400 800 10 = 4.96 × 10–3 A = 4.96 mA

Refer figure (b) : Combined resistance of 1000 voltmeter and 400 resistance is,

R = 1000 400

1000 400

= 285.71

i =

6

285.71 800 = 5.53 × 10–3 A

Reading of voltmeter = Vab = iR = (5.53 × 10–3) (285.71) = 1.58 volt

PART - III 1. Electrons are accelerated in opposite direction of electric field. Therefore speed of the electron is more

at B than at A.

2. In series current remain same = neAvd, J = /A, for constant current vd 1/A and J 1/A.

3. When no current is passed through a conductor the average velocity of a free electron over a large period of time is zero and the average of the velocities of all the free electrons at an instant is zero.

4. IR = V = E

EA

EA E

J

=

–25 10

10

= 5 × 10–3 –m

= –3

1 1

5 10

= 200 mho/m.

5. (i) Rbulb = –3

V 10

10 10

= 1. k

(ii) Rbulb = –3

220

50 10 = 4.4 k.

since increase in temperature increases resistance when it is connected to 220 V mains.



Current Electricity




6. (A) dR = 0

x dx

A, R =

00

x dx

A = 02 A

(B) I = 0V

R

(C) J = I

A

(D) J = E

E = 02

J 2Vx

7. Eeq =

KE KE KE N...........upto

r r r K1 1 1 N

........uptor r r K

Eeq = KE, eq

1 1 1

r Kr Kr ........ upto

N

K req =

2K r

N

For maximum power req = R 2K r

N = R K =

NR

r

Maximum power = Pmax = I2 R =

2

eqER

2R

=

2EK

R2R

= 2 2E K

4R =

2 NRE

r

4R

=

2NE

4r

8*. Equivalent resistance Req = 10 so current passing through battery and 3 resistance is

i = 10

10 = 1 A Ans.

and current passing through 4 is 0.25 A Ans.

1

2A 1

2A

14

A

10 V

1 =r

2 2 2

488

3 3 214

A1A

8 8 9*. Let potential of point D is x. by KCL at point D.

I1 + I2 + I3 = 0

x 70

10

+

x 0

20

+

x 10

30

= 0

6x – 420 + 3x + 2x – 20 = 0

11x = 440 x = 40 volt

I1 = 40 70

10

= – 3A, I2 =

40

20 = 2A



Current Electricity




I3 = 40 10

30

= 1A

P = i2R P = 32 × 10 + 22 × 20 + 12 × 30 P = 200 W

10.*

Current flow in circuit is = 10 amp

Power supplied by the battery is = 2R = (10)2 × 2 = 200 W

Potential drop across 4 & 6 are equal and it is equal to zero. Current in AB wire is 10 amp. 11*. V = E – ir

from graph V = 10 – 5i r = 5, E = 10V

imax = E 10

r 5 = 2 amp

12*. for short circuited, = E

r

V = E – r = E – E

r. r = 0

When current flow from negative terminal to positive terminal

V = E – r which is less than E When current flow from positive terminal to negative terminal

V = E + r which is greater than E.

13*.

Current i = R r

Cell generating power =i

Heat produced in R at the rate = i2R = iR. R r

= i.

R

R r

Heat produced in r at the rate i2r = ir

R r.

14*.

Current in the circuit

i = 12

3 2 1 = 2 amp VA = 0 VA – VM = – 2 × 3 VM = 6V and

VN – VA = – 2 × 2 VN = – 4 V Current in wire AG is zero.



Current Electricity




15.

imax = 20 1

5 75 120 10

amp

Vmax = imax R75 = 1

10 × 75 = 7.5 V

Range of potentiometer 0 to 7.5 V

16. For non ideal ammeter and voltmeter, ammeter have low resistance and voltmeter have high resistance. Therefore the main current in the circuit will be very low and almost all current will flow through the ammeter. It emf of cell is very high then current in ammeter is very high result of this current the devices may get damaged. If devices are ideal that means resistance of voltmeter is infinity. so that current in the circuit is zero. Therefore ammeter will read zero reading and voltmeter will read the emf of cell.

17. For 50 V, RV = –6

50

50 10 = 1000 K in series

For 10 V, RV = –6

10

50 10 = 200 K in series

For 5 mA, Rs= –6

–3

100 50 101

5 10

in parallel

For 10 mA, Rs = –6

–3

100 50 10

10 10

=

1

2 in parallel

18. VA – VB = Eeq =

1 2

1 2

1 2

r r

1 1

r r

= 1 2 2 1

1 2

r r

r r

If 1 > 2 source 1 act as a source and 2 act as a load.

and V1 = 1 – ir1

V2 = 2 + ir2

V1 = V2 as i = 1 2

1 2r r

for 1 > 2

19*. In parallel combination potential difference are same So V1 + V2 = V3 Here V1 and V2 are in series but their resistances are different So V1 = iR1 V2 = iR2

V1 V2

20. Potential gradient x = E

r R ×

R

100

Where R = resistance of potentiometer wire.

E

2 = x

E

2 =

E

r R ×

R

100

= 50(r R)

R

> 50 cm. Ans.

Balance length should be less than or equal to 100 cm

100

50(r R)

R

100 R r Ans.



Current Electricity




21. It is easier to start a car engine on a warm day than on a chilly cold day because the internal resistance of battery decreases with rise in temperature and so current increases.

Power Loss = 2R, Power loss I2

Also P = V. = P

V

Since for given power & line P & R are constant

Power loss = 2R =2

2

P R

V Power loss

2

1

V

mica is good conductor of heat but bad conductor of electricity

PART - IV 1 - 3. As E2 is increasing it's current also increases, So, increasing graph is of i2. i1 = 0.1A, E2 = 4V, i2 = 0

As ; +

+–

– E1

0.1A

0.1A

E2

4VR2

R1

0.1 R1 + 0.1 R2 – E1 = 0 0.1 R2 – 4 V = 0

R2 = 40 Now ; i2 = 0.3A, i1 = – 0.1 A, E2 = 8V

0.1A

0.2A

8V

0.3A

40R1

Now ; 0.1 R1 +E1 – 8 = 0 When E2 = 6V, current in E1 is i1 = 0 (from graph) E1 = 6V

R1 = 4

0.2 = 20

5. (i)

Energy kirchoffs first law at first junction

0V

0 K

1

V

R

=

0 02

V V

K K

1R

+

0V

K

2

0

R

11

K

0

1

V

R = 0

1

V

KR

11

K

+ 0

2

V

KR

1

1

R 2

1 1 11

K K K

=

2

1

KR.

2

2

K 2K 1

K

= 1

2

R

KR

2

1

R

R =

2

K

(K 1) .........(1)

O 0n 1 n

1

V V

K K

R

=

0n

3

V

K

R, 1

3

R

R= (K – 1) .........(2)



Current Electricity




(1) & (2) 2

3

R

R =

K

K – 1

(ii) I = 01

2 3

V /KV

R R (K /K – 1) = 0

23

V (K – 1)

K R I = 0

23

V(K – 1)

RK

6.

A B

E r

100 Cm 50

Current is AB = 450

E

Potential gradient

x = 100

50

450

E

For balancing length

v = x

1.5 = 402

1

450

E

E = 4.05

7.

A B

10 k

50

1.5 10 div

G

RG = 50

10100

5.1I

Current sensitivity = Current

Deflection

101005.1

10

5.1

010100.010 =

A

Div067.0

8. Current through G is Case – I

I0 = GR

E

Current through G is Case – II

I1 = )GS(

S

RGS

GS

E

R >> G 2

10I

I



Current Electricity




9 to 12. Case-1

MP

R 100

i

10 i0 G = 12 × 10–3

10 i0 (100) = 16 × 10–3 103 i0 = 16 × 10–3 i0 = 16 × 10–6 A

G 3

100 4 G = 75

Case-2

G R

30 i0 (R + G) = 12

R + G = 6

–6

0

12 2 2 10

30i 805 16 10

= 2.5 × 104 = 25k

R = 25k – 75 Case-3

30 i0 (R + G) = 12

ni0 ( R + G) = 5

2425

n = 12

V

24

5 1

EXERCISE-3 PART - I

1. R = l

A

R = L

tL

=

t

Independent of L.

2. 100 = 2

100

V

R'

100

1

R' =

2

100

V

where R’100 is resistance at any temperature corresponds to 100 W

60 = 2

60

V

R'

60

1

R' =

2

60

V

40 = 2

40

V

R'

40

1

R' =

2

40

V

From above equations we can say

100

1

R' >

60

1

R' >

40

1

R' .

So, most appropriate answer is option (D).



Current Electricity




3. To verify Ohm's law one galvaometer is used as ammeter and other galvanometer is used as voltameter. Voltameter should have high resistance and ammeter should have low resistance as voltameter is used in parallel and ammeter in series that is in option (C).

4.

i = 2

2 R

J1 =

22

2 R

R

eq = 1 11 1

1 1

=

req = 1

2 i =

1R

2

= 2

2R 1

J2 =

22

1 2R

R

Given J1 = 9

4 J2

22

2 R

R =

9

4

22

1 2R

R

2

2 R =

3

1 2R

2 + 4R = 6 + 3R R = 4.

5.

1 2

1 2

1 2

E E 6 3r r 1 21 1 1 1

r r 1 2

= 15

3 = 5 volt Ans.

6. Due to input and output symmetry P and Q and S and T have same potential.



Current Electricity




Req = 6 12

18

= 4

1 = 12

4 = 3A

2 = 12

36 12

2 = 2 A VA – VS = 2 × 4 = 8V VA – VT = 1 × 8 = 8V

VP = VQ Current through PQ = 0 (A)

VP = VQ VQ > VS (C) I1 = 3A (B) I2 = 2A (D)

7. In given Kettle R = 2

L

d

2

= 2

4 L

d

, P =

2V

R

In second Kettle R1 =2

L

d

R2 = 2

L

d

So R1 = R2 = R/4

If wires are in parallel equivalent resistance RP = 2

L

d

Then power PP= 8P So it will take 0.5 minute If wires are in series equivalent resistance RS = R/2 Then power PS = 2P So it will take 2 minutes

8.

Potential of Junction O

1 2

1 30

1 2 3

V V0

R RV

1 1 1

R R R

Current through R2 will be zero if

V0 = 0 1

2

V

V = 1

3

R

R

9. 6

1000 (G + 4990) = 30

G + 4990 = 30,000

50006

G = 10



Current Electricity




6

1000 ×10 =

61.5 S

1000

S = 60 2n

1494 249

n = 249 30 2490

1494 498

= 5

10. For balanced meter bridge

X

R =

100 ,

X

40 =

90

60 X 60

X = R 100

X 0.1 0.1

X 100 40 60

X = 0.25

so X = (60 + 0.25)

11. RA and RFe are Parallel to each other

eq Al Fe

1 1 1

R R R

Al

1

R =

8

2 2

2.7 10 (.05)

(.007) (.002)

;

RAl= 8 4

6

2.7 10 5 10

10 45 100

= 3 × 10–5

RFe = 7

2

1.0 10 (.05)

(.002)

RFe = 7

2

6

105 10

4 10

= 35

104

eq

1

R =

5 3

1 4

3 10 5 10

=

5 310 4 10

3 5

eq

1

R =

5 3

15

5 10 12 10 =

5

15

5.12 10

eq

1

R =

150

5.12 =

1875

64

12.



Current Electricity




I6.5

1amp6.5

13. Suppose current for full deflection (i.e., For maximum range) through galvanometer is g.

(A)

G G g

R R

Total potential difference V1 = 2Ig(R + RC)

(B)

2g

R R G

G g

g

Total potential difference V2 = 2g

R2

2

RC = g (RC + 4R)

Now since 2RC < R

So V1 < 3gR

while V2 > 4Rg

So V2 > V1 (C)

1

G

g

G

g

gRC = R

= R

RCg

1 =

R

R12 C

g



Current Electricity




(D)

2

G g

G

R

2gRC = R = R

R2 C

g

2 = g

R

R41 C We can be see 1 > 2S

j(t)

(0, 0) t

15.

2q

02VE

h

–V0

+V0 +q

–q

2q

2q just before collision just after collision

Kq

r = Vo = q =

V ro

K

1

2

qE 2t

m

= h ; 1

2

0V r

K 02V

hmt2 = h ; t2 =

2

2

0

h

V

mk

r

t = 0

h mk

V r

During every collision 2q charge will flow from circuit.

Average current Iavg = 2q

t =

2

02V r r

h mk k

2

avg 0I V

14. The balls will bounce back to the bottom plate carrying the opposite charge they went up with.

15. 2

avg 0I V

16. N = 50

A = 2 × 10–4 m2 C = 10–4, R = 50 , B = 0.02 T, = 0.2 rad

Ni AB = C

ig = 4

4

C 10 0.2

N AB 50 2 10 0.02

= 0. 1 A

ig × G = (i – ig) S 0.1 × 50 = (1 – 0.1) × S

5 = 0.9 ×S ; S = 50

9



Current Electricity




17.

G Rg g

v = 100 × 10-3v v = Ig(Rg + Rv)

6

1

102

10

= Rg +Rv

5 × 104 Rv (Rv < 105 )

A

1000

RA

RV

V

i

i '

IgRg = (I – Ig)S

G

S

S = 63

6

10210

10102

S = 2 × 10—5 ×103

2 × 10—2

20m

RA = 10

101020 3

= 20 × 10–3

i = 0R105

51

1051

10501000A4

3

3

10001051

Ri'i

3

v

measured resistance Rm =

4.980

51

105105

51i

1000'i 44

PART - II 1. Let R be their individual resistance at 0ºC. Their resistance at any other temperature t is

R1 = R (1 + 1 t) and R2 = R (1 + 2 t). In series

Rseries = R1 + R2 = R [2 +(1 + 2) .t]

= 2R. 1 21 t2

Series = 1 2

2

In Parallel RParallel = 1 2

1 2

R .R

R R = 1 2

1 2

R(1 t) R(1 t)

R(2 ) t)

2

1 2

1 2

R (1 ( )t)

2R(1 t)2

1 21 t2

Parallel = 1 2

2

.



Current Electricity




2. R = A

( V = A const)

V = A

By differentiation 0 = dA + Ad ....(1)

By differentiation dR = 2

(Ad dA)

A

....(2)

dR = 2

2Ad

A , dR =

2 d

A

or dR d

2.R

So , dR d

% 2. %R

= 2 × 0.1%

dR

% 0.2%R

Ans.

3. x = V

= R

= R

A

= A

x = 7

7

0.2 4 10

8 10

=

0.8

8 = 0.1 V/m.

4.

As R1 = 220

22025

and R2 = 220

220100

R = R1 + R2

= 220 × 220 1 1

25 100

= 220 × 220 1

20

live = 440 40

A220 220 220

20

1st bulb (25 W) will fuse only

5.

120V

240

2VP

R

120 120

R60

= 240

Req. = 240 + 6 = 246

V1 = 240

120246

= 117.073 volt



Current Electricity




120V

60

2

48V 120 106.66 Volt

54

120Volt

48

V1 – V2 = 10.04 Volt

6. Statements I is false and Statement II is true 7. Total power (P) = (15 × 40) + (5 × 100) + (5 × 80) + (1 × 1000) = 2500W

P = VI

I = 2500

220 A

= 125

11

= 11.3 A Minimum capacity should be 12 A

8. = 0.1 m

V = 5v+

vd = 2.5 × 10–4 m/s n = 8 × 1028/m3 I = ne A vd

VA

= ne A vd

= d

V

nev =

28 19 4

5

8 10 1.6 10 2.5 10 0.1

= 1.6 × 10–5 m

9. S = g

g

iI

Gi

here ig = 10–3 A G = 102, = 10A

S ~ 10–2

10.

0V

1 1 1

6V 2V 4V 2V 2V 2V

0V 2V 2V 2V 2V 4V

p.d. across each resistance is zero so current is also zero. 11. In a balanced Wheatstone bridge if the position of cell and galvanometer is exchanged the null point

remains same.



Current Electricity




12. Full deflection current, g = 5mA

Resistance of galvanometer, G = 15.

R = g

VG

= 3

1015

5 10

= 2000 – 15

= 1985

= 1.985 × 103

13.

10

1 12V

2 13V

Eeq =

12 13

1 21 1

1 2

=

37372

3 3

2

req. = 2

3

I =

37 37373 3

2 32 3210

3 3

Voltage across load = IR = 37

1032

= 11.56V

14. Using formula r = 1

2

R 1

= 52

5 140

=

125 1.5

40

15.

R 1000–R

100–

G

Say resistances are R and 1000 – R

For case-I R 1000 R

100

For case-II 1000 R R

10 110

Multiplying both equation



Current Electricity




R(1000 R) (1000 R)R

( 10) (100 )(110 )

2 – 10= 11000 + 2 –210

200= 11000

= 55 cm

Putting in first equation

R 1000 R

55 45

45R = 55000 – 55R

R = 550 16.

Color Codes Values Multiplier Tolerance (%)

Black 0 1

Brown 1 10

Red 2 100

Orange 3 1K

Yellow 4 10K

Green 5 100K

Blue 6 1M

Violet 7 10M

Grey 8 100M

White 9 1G

Gold 5

Silver 10

17. In circuit, when Rh = 2

i1 = A124

6

, =

AB

41× AJ

AJ = 4

AB …..(1)

when Rh = 6

i2 = A6.064

6

2 = AJAB

46.0

AJ = 46.0

AB2

…..(2)

From (1) & (2)

4

AB

46.0

AB2

2 = 0.6 × = 0.6 × 0.5 = 0.3V



Current Electricity




18. dR =

cd

Acording to Question

1

0

dc

dc

10 22

222

24

4

1 = 0.25 m.

19. Q0 iG Q0C = iG

I-case CQ0 = v

220 R ….(1)

II-case 0 v 5C

5R5 5 R220

5 R

….(2)

from (1) and (2) R = 22 20. Total power is (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000) = 4325 W

So current is = 4325

19.66220

A

Ans is 20 Amp.

21.

VP

5V 20

1000 cm L = 1200 cm

G

= 60 mA

Potential gradient = 1200

V

1000

5 P

VP = 6V

and RP = 3

P

1060

6V

= 100

22. Vg = ig Rg = 0.1 V V = 10 V

R = gg

VR 1

V

= 100 × 99 = 9.9 K



Current Electricity




HIGH LEVEL PROBLEMS (HLP)

1. G

S

25I=1A I =20×10 ×30g

-6

Ig = 20 × 10–6 × 30 = 0.6 × 10–3 A

As we know

IgRg = S(I – Ig)

25 × 0.6 × 10–3 = S × (1 – 0.6 × 10–3)

S = 3

3

15.0 10

1 0.6 10

0.015 Ans.

For voltameter

V = (RA + R) i

Resistance of ammeter is

RA = g

g

S R

S R

RA ~ S = 0.015

V = (RA + R) i

1 = (0.015 + R) × 1

R = 0.985 Ans.

2. 4 = i × 10 × 103.

i = 4

4

10 =

4

120

X 10

X + 104 = 30 × 104

X = 29 × 104 Ans.

3. (a) P = 40 W h = 10 m V = 200 litre t = ?

= 90 %

mgh

t × = P

m

t =

40

10 10 0.9 =

4

9 kg/s. Ans.

(b) m = 200 × 103 × 10–3 kg

= 4

9 t

t = 1800

4 =

900

2 = 450 sec. Ans.



Current Electricity




4. E = 3.4 volt , r = 3 , RA = 2 , R = 100 .

i1 = 0.04 A , V = ? , RV = ?

0.04 = V

V

3.4

100 R3 2

100 R

5 + V

V

100 R

100 R =

3.4

0.04 = 85

RV = 400 Ans.

V = 0.04 × 100 400

500

V = 3.20 V Ans. For ideal voltmeter

Rv

i = 3.4

3 2 100 =

3.4

105

V = i × 100 = 3.4

105 × 100 =

68

21V Ans.

5. I1 =E

r R r + R =

1

E

I

I2 = 2E

2r R 2r + R =

2

E

I

I3 = E

rR

2

r

R2 =

3

E

I

To show that 3 I2 I3 = 2I1 (I2 + I3)

L.H.S = 3 × 2E

2r R ×

E

rR

2

= 2E

r R

2E E

r2r RR

2

= R.H.S

3E = 1 E

r R

r2 R 2r R

2

3r + 3R = r + 2R + 2r + R

Hence its prove.

6. nr

m = R I =

nE

2R

I’ = mE

mrR

n

= mE

m mR R

n n

= 2

2 2

mE n

R(m n )

=

2

2 2

mn

(m n ) ×

2I

n



Current Electricity




7.

FE

CD

A B

By symmetry

Current in branches FD and CE are zero. Because potential defference across them is zero

rr

r

rr

r

A B

r rF E

CD

Req =3r

5 Ans.

8. Rg = 100

Ig = 1 mA , S = ? , I = 10 A , P = 1W

100 × 10–3 = (10 – 10–3) S

S = 3

0.1

10 10 0.01 Ans.

P = (10 – 10–3)2 × S = (10 – 10–3) × 0.1

10 × 0.1 = 1 Watt

Power dissipated is slightly less than 1W so its use is safe Ans.

4 = E

100 Sr

100 S

...........(i), 4 = E

100 0.01r

100

1 = E

100 sr 1.5

100 s

...........(ii) , 1

100 + r =

2

4

From (i) & (ii)

1 = E

E1.5

4

, r = 0.49 Ans.

E = 2V Ans.

9. (a) VA = 6 V, VC = 2V Ans.

(b) E = x 4 = × 6

100 =

200

3 cm Ans.

(c) In secondary circuit current is zero Ans.

(d) 6 V, 6 – 7.5 = – 1.5 V, no such point D exists Ans.



Current Electricity




10. (a)

Req= 1 + 2 +50 200

250

= 43

i = 4.3

43 = 0.1 A Ans.

V = 40 × 0.1 = 4V Ans.

(b) Req= 1 + 52 200

252

= 252 10400

252

i = 4.3 252

252 10400

=

1083.6

10652 iA =

1083.6 200

10652 200 52

= 0.08 A Ans.

V= 4.3 – i × 1 = 4.2 V Ans.

11.

i = i1 + i2

0 0 0

2 1 1

V V V

R 2R R

2 1 1

1 1 1

R R 2R 1

2

R 1

R 2

R2 = 2R1

RAB = Req = R1 + 1 eq

1 eq

2R R

2R R

On solving R2

eq – R1Req –2R12 – 0 ; Req = 2R1

12. It follows from symmetry considerations that the initial circuit can be replaced by an equivalent one (as shown).

A B

R/2

R/2

R /2x

R/2

R/2

R

We replace the inner triangle consisting of an infinite number of elements by a resistor of resistance RA

B / 2, where the resistance RAB is such that RAB = Rx and RAB = a. After simplification, the circuit becomes a system of series and parallel connected conductors. In order to find Rx, we write the equation

Rx = R x

x

RR / 2R

R R / 2

1

x

x

RR / 2R R

R R / 2

Solving the equation, we obtain RAB = Rx = R( 7 1)

3

=

a ( 7 1)

3



Current Electricity




13. As, R = A

i.e., resistance of wire is proportional to its lengths other factors being constant. So, resistance of wire. AB or CD = 4 unit Resistances of arm BC or AD = 3 unit and resistance of diagonal AC = 5 unit (Taking proportionality constant to be unity) Resistance R1 between AC is parallel combination of RADC ,

RAC and RABC .

So, 1

1 1 1 1

R 7 5 7 i.e., R1 =

35

17 unit

Applying Kirchoff's laws for closed meshes BCAB and ADCA,

3( – 1) + 52 – 41 = 0

3 – 71 + 52 = 0 ........(i)

and 3(1 + 2) – 4( – 1 – 2) + 52 = 0

– 4 + 71 + 122 = 0 ........(ii) Adding Eqs.(i) and (ii),

2 = 17

Substituting for 2 in Eq.(i)

1 = 8

17

If R2 is the effective resistance between points B and D,

RBD = 41 + 3(1 + 2) = 71 + 32

Substituting for 1 and 2 ,

RBD = 7. 8 3

17 17

, i.e. RBD =

59

17units

BD

AC

R

R =

59

35 Ans.

2nd method

4

3

4

3

D

I

(I–I )1

I1

(I–I )1

B

(2I –I)1

5

V

I

I1

– 4I1 – 5 (2 I1 – I) + 3 (I – I1) = 0 ........(i) – V + 3 (I – I1) + 4 I1 = 0 ........(ii)

(i) and (ii) V = 59

17 I

RBD = 59

17 units



Current Electricity




14. (a) Let thermal energy QA generated when coil A is switched on

QA = 2

AA

Vt

R =

2

A

V

R10

Let thermal energy QB generated when coil B is switched on

QB = 2

BB

Vt

R =

2

B

V

R20

since QA = QB

we have 2

A

V

R10 =

2

B

V

R20

or 2RA = RB

when A & B are connected in series thermal energy QS = 2

Seq

Vt

R =

2

A B

Vt

(R R )

QS = QP = QA = QB

2

S

A B

V t

R R =

2

A

V

R10

or tS = 10 A B

A

(R R )

R

= A

A

3R

R10 = 30 minutes

(b) when A & B are in parallel

QP = 2

Peq

Vt

R' =

2P A B

A B

V t (R R )

(R R )

QP = QA

or 2

P A B

A B

V t (R R )

(R R )

=

2

A

V

R10

or tp A2A

3R

2R =

A

10

R

or tp = 20

3 minutes

15. (a) There are no positive and negative terminals on the galvanometer because only zero deflection is

needed.

(b) G

J

X 12

A B C D

(c) AJ = 60 cm

BJ = 40 cm If no deflection is taking place. Then the Wheatstone bridge is be said to be balanced. Hence

X

12 = BJ

AJ

R

R

or X

12 =

40

60 =

2

3

or X = 8 Ans.



Current Electricity




16.

(i) i = 2

1 2r R R 2 / a

Vmax = i (R1 + R2) =

2 2

3

2 / a 2 a

=

3

4 Ans

(ii) From A to B potential drop is = irAB = 2 22 / a 2 a

=

4

distance from B point = 2 2

'.

42 / a a

=

2

total length = 2 + 5

2 2 Ans

(iii)

Let R = 22 a

apply KVL in loop ABCDA – 3i1R = (i1 + i2)R ....(1)

applying KVL in loop EFDLE 2

– i2R = (i1 + i2) R ...(2)

from eq (1) and (2)

5

2 = 7 (i1 + i2) R i2 =

7R

where R =

22 a

Ans.

17. (a) R =0

Lx

L

0

e

dx

A = 0

A

Lx

– L

o

L e

= 0L

A

1–I e

R = 0L

A

e 1

e

, i = 0V

R.

(b) Resistance upto x = Rx

Rx = 0

A

xx–

L

o

L e

= 0L

A

x–L1 e

V0 – V = i Rx = 0V

R Rx =

xL

–

01

V (1 e )

(1 e )

V =

xL

– –10

1

V e – e

1 e

.

18.

Here I = E

GSR r

G S

and Ig =S

G S



Current Electricity




Ig = E S ES

.GS G S (R r) (G S) GS

R rG S

for S = 10, Ig = 9

50 × 200 × 10–3 = 36 × 10–3 amp

36 × 10–3 = 10E

100(10 G) 10G ....(1)

for S = 50, Ig = 30

50 × 200 × 10–3 = 120 × 10–3 amp

120 × 10–3 = 50 E

100 (50 G) 50G ....(2)

from (1) and (2)

36 100(50 G) 50G

24 100(10 G) 10G

G =

700

3 Ans.

from eq (1) E = 96 V Ans.

19.

RAB =5

6 R

Power = 2

AB

V

R =

2V

5R

6

=6

5P =

3xP

5 x = 2

20. (A)

R0 = 2 , I1 = 2.5A I2=1.5A VP = 19 – R0I = 19 –(2)×(2.5) VP= 14V Similarly VQ = 16V

I3 = Q P

0

V V

R

I3 = 1A

(B) I4 = P CV – V

R = I1+ I3

I4 = 14 – 0

R = I1+ I3

14

R = 2.5+1

R = 4



Current Electricity




(C) Applying KVL in FEQPF – I1 R0 – I3 R0+ (I– I1) ×R0 = 0 I = I3 + 2I1 ... (i) Applying KVL in AFEDCBA ; V–I1R1 – (I1– I3) RX = 0

I1 = 3 X

0 X

V I R

R R

.........(ii)

Applying KVL in AFCBA ; V – (I–I1) R0 – 3 X

0 X

V I R

R R

(I–I1+I3) R = 0

V – I (R0+R)+I1( R0 +R) –I3R = 0 V– (I3+2I1) (R0+R1) +I1 ( R0+R) – I3R = 0 Using (i) V – I3 (R0+2R) –I1( R0+R) = 0

V– I3 (R0+2R) – 3 X

0 X

V I R

R R

(R0+R) = 0 Using (ii)

x

3

0 x 0 0

V(R R)

(2R 3R)R R (R 2R)

The graph of the function is a hyperbola. Its special points are: at Rx = 0 I3= –3.8 A ; at Rx ; = 4I3 = 0; at

Rx = 32 I3 = 1A ; If Rx , I3 tends to 19/16 = 1.1875A.

21. = 2x

RAB =

1

0

dx

RAB = 1

RAP= x2

I = P

AB

E

R R = P

2

E

1 t

Now VAP = IRAP = ES

ES = 2

P2

E x

(1 t ) at t = 1sec, x = 1/2 ; ES =

P 2

1E

2

1 1

= PE

8 = 8



Current Electricity




22. (a) A Ammeter

V

Voltmeter

Variable Power supply

Black Box

A

B C

A

A

(b) See graphs for the calculations of slopes.

RAB = 0.98 k, RBC = 4.60 k, RCA = 5.67 k

(c)



Current Electricity






solutions of current electricity exercise-1...(ii) = 1 r dr dt 273 0 dt = 20 10 dr r .273 = n 20 10...

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