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CURRENT ELECTRICITY-III
¾
ELECTRIC CURRENT
¾
OHM’S LAW AND
¾
KIRCHHOFF’S LAW
1
1. Assume that each atom of copper contributes
one electron
electron. If the current flowing through a
copper wire of 1 mm diameter is 1.1 A, the drift
velocity of electrons will be
(density of Cu =9 gcm-3 , atomic wt. of Cu =63)
a)) 0.3 mm/s
c) 0.1 mm/s
b)) 0.5 mm/s
d) 0.2 mm/s
2
1 Solution:
1.
No.
=
of
atoms
in
63g
of
Volume of 63 g of copper =
= 7 cm3 =
n=
copper
=
m3
=
A=πr2= 3.142X(
X
m-3
)2 m2
D ift velocity,
Drift
l
it
=
=
= 0.1X 10-3 ms-1 =0.1mms-1
Ans : (c)
3
02.The potential difference applied to an X-ray
tube is 5 kV and the current through
g it is
3.2 m A. Then the number of electrons
striking the target per second is
a) 2 x 1016
b) 5 x 106
c) 1 x 1017
d) 4x 1015
4
2. Solution:
As = =
=
=
=
Ans: (a)
5
3.The current in a conductor varies with time
t as I = 2t + 3t2, where I is in ampere and t in
seconds. Electric charge flowing through a
section of the conductor during t=2
t 2 sec to
t = 3 sec is
a) 10 C
b) 24 C
c) 33 C
d) 44 C
6
3. Solution:
As I=
I
dq = I dt =
dt
Charge passed from t=2s to t= 3s is
q=
=
2
3
=
3
q= t )2 + t ) 2
+
)23
3
=(9-4)
(
) + ((27-8))
= 5+ 19 = 24C
Ans : b)
7
4.The current I and voltage V curves for a
given metallic wire at two different
temperatures T1 and T2 are shown in the
T1
fi
figure.
Then,
Th
a) T1 > T2
c) T1= T2
b) T1 < T2
d) T1 = 2T2
I
T2
V
8
4. Solution: for the same potential v0 ,
T1
I
At temperature T1, R1 = , At
temperature T2, R2= ,
As
,
T2
I1
I2
R2 R1.
V0
V
Since resistance
Si
i
off metall increases
i
with temperature,
So, T2 T1 or
Ans : (b)
9
5.
The
length
of
a
given
cylindrical wire is increased by
100%. Due to the consequent
decrease
in
diameter
diameter,
the
change in the resistance of the
wire will be
a) 200%
b) 100%
c) 50%
d) 300%
10
5. Solution: When the length is
increased by 100%, it becomes 2l and
area of cross section decreases to
N
New
resistance,
i t
=
=
=
= 4R
Ch
Change
iin resistance
i t
100%= (
Ans: (d)
)100 = 300%
11
6. To verify Ohm’s law, a student is
provided with a test resistor RT, a
high
resistance
R1,
a
small
resistance
R
identical
,two
galvanometers G1 and G2 and a
variable voltage source V. The
y out the
correct circuit to carry
experiment is
12
13
6.Solution:G1 behaves as voltmeter byy connecting
g it in series
with high resistance R1 and applying it across RT in
parallel G2 behaves as ammeter by shunting it with
parallel.
small resistance R2 and connecting in series with RT.
Ans : (c)
14
7. By increasing the temperature, the
specific resistance of a conductor and a
semiconductor
(a) Increases for both
(b) decreases for both
(c) Increases, decreases respectively
(d) d
decreases, iincreases respectively
ti l
15
Solution(7):On increasing the temperature , the specific
resistance of a conductor increases and that of a
semiconductor decreases
Ans . (c)
16
8. The resistance of a wire is 5Ω at 50 ˚C and
6 at 100 ˚C
6Ω
˚C. Th
The resistance
i
off the
h wire
i at
0 ˚C will be
(a) 3 Ω
(b) 2 Ω
(c) 1Ω
(d) 4Ω
17
8. Solution: Rt-Ro =
R50-R
Ro =
t
- ------- (1)
And R100 - Ro =
-----(2)
=
Or
=
Or 10-2Ro = 6-Ro ; R0 = 4
Ans: (d)
9. I-V characteristics of a copper wire of length L and area of
9
cross-section A is shown in figure. The slope of the curve
becomes
(a) More if the experiment is performed at higher temperature
(b) More if a wire of steel of same dimension is used
(c) More if the length of the wire is increased
(d) Less if the length of the wire is increased
I
V
19
9.Solution:
9
Solution: Slope of II-V
V graph = =
I
If the experiment is performed at a
higher temperature, the resistance R of
copper increases and hence slope
d
decreases,
so th
the option
ti ((a)) iis wrong. IIn
option (b) and (c), the resistance more
and
d so slope
l
b
becomes lless. IIn option
ti
(d), the resistance R increase and so
slope
l
d
decrease. H
Hence only
l option
ti (d) iis
correct
Ans: (d)
V
I
V
20
10.The ratio of masses of three wires is 1 : 2 : 3
and that of their lengths is 3 : 2 : 1. If the wires
are made of same material, the ratio of their
resistances will be
( )1:1:1
(a)
(b) 1 : 2 : 3
(c) 9 : 4 : 1
(d) 27 : 6 : 1
21
10 Solution:
10.
Mass= m= volume X density = AlXd
Or A =
R =
=
=
i.e; R
given m1: m2: m3: = 1:2 :3 and
l1 : l2 : l3 = 3 : 2: 1
R1 : R2 : R3 =
:
:
=9 : 2 :
= 27: 6 : 1
Ans: (d)
22
11. The colour sequence in a carbon resistor is
red brown,
red,
brown orange and silver.
silver The resistance of
the resistor is
(a) 21X103 Ω ± 10%
(b) 23 X 101Ω ± 10%
(c) 21 X 103Ω ± 5%
(d) 12 X 103 Ω ± 5%
23
11 Solution:
11.
S l ti
Red
2
Brown
Orange
1
3
R =21 X 10
3
Silver
0%
0%
Ans: (a)
24
12. The three resistances of equal value are arranged in
12
the different combinations shown below. Arrange them
in increasing order of power dissipation.
(a) III < II < IV < I
( ) I < IV < III < II
(c)
(b) II < III < IV < I
(d) I < III < II < IV
25
12 Solution:
12.
I
Req = R + R + R = 3R ; PI = I2 (3R)
II Req =
=
; PII = I2
PIII = I2
III Req = ;
IV Req =
Clearly PIII
+R =
PII
PIV = I2
;
PIV
PI
Ans: (a)
26
13. A 100 W bulb
13
b lb B1, and
d ttwo 60 W
bulbs B2 and B3, are connected to
a 250 V source as shown in the
figure. Now W1, W2 and W3 are
the output powers of the bulbs B1,
B2 and
d B3 respectively.
i
l
Then
(a)
(
) W1 > W2 = W3
(b) W1 > W2 > W3
(c)W1 < W2 = W3
(d) W1 < W2 < W3
27
13. S
13
Solution:
l i V
Voltage
l
across B3 is
i maximum,
i
so
B3 gglows with maximum brightness.
g
In the series combination of B1 and B2 , the
bulb B2 of lesser wattage glows brighter than
B1, i.e.
i e w2 w1.
Hence w1
w2
w 3.
Ans: (d)
28
14. Figure shows three resistor configurations R1,
R2 and R3 connected to 3 V battery. If the power
dissipated by the configuration R1, R2 and R3 is
P1, P2 and P3, respectively, then
(a)P1 > P2 > P3
(c)P2 > P1 > P3
(b) P1 > P3 > P2
(d) P3 > P2 > P1
29
14. S
14
Solution:
l ti
E
Equivalent
i l t resistance
i t
off the
th three
th
combinations are
R1 = 1
, R2 = 1/2
Power p=
As R2
P2
R1
P1
, R3 = 2
; for same V, p
1/R
R3
P3
Ans: (c)
30
15. Six identical resistors are connected
as shown in the figure. The equivalent
resistance will be
(a)Maximum between P and R
(b) Maximum between Q and R
(c) Maximum between P and Q
(d) all are equal
31
15. Solution:
= R;
= R ; RPR = R
is maximum
Ans: (c)
32
16 Th
16.
The resistance
i t
off the
th series
i combination
bi ti
of two resistances is S.
When they are joined in parallel, the total
resistance is P. If S = nP,, the minimum
possible value of n is
(a) 4
(b) 3
(c) 8/9
(d) 2
33
16 Solution: In series ,
16.
R1 + R2 = S
In parallel,
=P
But S = nP
R1 + R2 = n
For n to be minimum, R1 = R2 = R (say)
R+R=n
;
Or 4 R2 = n R2 or n = 4
Ans: ((a))
34
17.In
17
I the
th circuit
i
it shown
h
iin th
the fi
figure,
the current through the
(a) 3 Ω resistor is 0.50 A.
(b) 3 Ω resistor is 0.25 A.
(c) 4 Ω resistor is 0
0.50
50 A
A.
(d) 4 Ω resistor is 0.25 A.
35
17.Solution: Resistance between B and C in the
RHS of the circuit =
=4
Resistance between A and B in the RHS of the
circuit =
=4
Equivalent
q
resistance of the circuit,,
R = 3+4+2 = 9
C
Current
td
drawn ffrom th
the b
battery
tt
I=
= 1A
36
At the junction A, the current of 1A is divided
equally
q y between the 8 resistance and the
remaining circuit of resistance 8 .
At the junction B
B, the current of 0.5A
0 5A is divided
equally between the 8 resistance and the
remaining
i i circuit
i it off resistance
i t
8 .
current through the 4
resistor = 0.25A
Ans: (d)
37
18.The total current supplied to the circuit
b th
by
the b
battery
tt
wire
i iis
(a)
1A
((b))
2A
(c)
4A
(d)
6A
38
18.Solution:
18
Solution: The equivalent circuit is as
shown in the fig.
Net resistance in the upper arm = 1.5 +
2 6
= 1.5 + 1.5 = 3 Ω
2
6
Now we have 3 Ω and 3 Ω resistance in
parallel; the net resistance is
2
R =
3
3
3
3
2
= 1.5 Ω
1.5
1 .5
6
6
3
3
6 V
Current I =
Ans: (c)
6
1.5
= 4A
19 For a cell
19.
cell, the terminal difference is 2
2.2
2V
V,
when circuit is open and reduces to 1.8 V,
when cell is connected to a resistance
R=5Ω
The internal resistance of cell r is
(a)
((c))
10
9
Ω
11
Ω
9
(b)
(d)
9
10
Ω
5
Ω
9
40
19. Solution: the internal resistance of
the cell is
r=(
)R=(
)5=(
)
Ans: (a)
41
20. If two identical cells, when
p
,
connected in series or in parallel,
supply same amount of current
through an external resistance of
2Ω. The internal resistance of each
cellll iis
((a)) 8 Ω
((b)) 2Ω
(c) 4Ω
(d) 1Ω
42
20 Solution:
20.
Current for series combination = current
in parallel combination
=
2(
) =2r + 2 or r = 2
Ans: (b)
43
21. An ideal gas is filled in a closed rigid
and thermallyy insulated container. A
coil of 100 Ω resister carrying current
1 A for 5 minutes supplies heat to the
gas. The change in internal energy of
th gas iis
the
((a)) 10 kJ
((b)) 30 kJ
(c) 20 kJ
(d) 0 kJ
44
21 .Solution:
S l ti
The change in internal energy of the gas
= Heat produced in the coil
2
2
= I Rt = 1 X 100 X 5 X 60 J = 30,000J
= 30 kJ.
Ans: (b)
45
22. When
22
Wh
two
t
identical
id ti l b
batteries
tt i
off
internal resistance 1Ω each are
connected in series across a
resistor R, the rate of heat produced
in R is J1. When the same batteries
are connected in parallel across R,
the rate is J2,. If J1 = 2.25
2 25 J2 , then
what is the value of R in Ω?
(a) 2Ω
(b) 1Ω
(c) 4Ω
(d) 0.5Ω
46
22 .Solution
S l ti
I case; J1=
=
II case; J2 =
=
Or
r= 1
==
= 2.25 =
=
R=4
Ans: ((c))
47
23.A steady current flows in a metallic
conductor of non-uniform
non uniform cross
cross-section
section.
The quantity/quantities constant along the
length of conductor is/are
a) Current, electric field and drift speed
b) Drift speed only
c) Current and drift speed
d) Current only.
only
48
23. Solution:- When a steady current flows through a
metallic conductor of non-uniform cross section,
=
=
or
and
or E
(Note: j=
(Note:-
;j=
=
)
That is both
and E change
g with A .
Only current I remains constant.
Ans: (d)
49
24 A material B has twice the specific
24.
resistance of the material A. A
circular wire made of B has twice the
diameter of the wire made of A.
Then for the two wires to have the
Then,
same resistance, the ratio l A / l B of
th i respective
their
ti llengths
th mustt b
be
1
((a)) 4
((b)) 2
(c) 1
(d) 21
50
24. Solution:-
given RA = RB or
or
=
=
=
2
=
Ans: (d)
51
25. An electric current is passed through a
circuit containing two wires of the same
material, connected in parallel. If the lengths
and radii of the wires are in the ratio of 4/3 and
2/3 then the ratio of the currents passing
through the wire will be
(a) 3
(b) 1/3
(d) 2
(c) 8/9
52
25 S
25.
Solution:l i
= X
=
=
= X
=3
For wires connected in parallel
parallel,
V1 =V2 or I1R1 = I2R2 or
=
=
Ans: (b)
53
26. A 25 W-220 V bulb and a 100 W- 220 V bulb are
joined in series and connected to the mains.
mains
Which bulb will glow brighter?
(a) 25 W bulb
(b) 100 W bulb
(c) First 25 W bulb and then 100 W bulb.
(d) Both will glow with same brightness.
54
26. Solution:- In the series circuit, same current
g each bulb. But the 25W bulb has
flows through
higher resistance (R = V2 / P) . It produces
more heat per second (P = I2 R) and hence
glows brighter than 100 W bulb.
Ans: (a)
55
27. A 30 V, 90W lamp is to be operated on a 120
V DC line
line. For proper glow
glow, a resistor of ....Ω
Ω
should be connected in series with the lamp.
(a) 40
(b) 10
(c) 20
(d) 30
56
2 Solution:27.
S l i
Resistance of the lamp,
R =
=10
Safe current through the lamp,
I =
=
= 3A
When the lamp is operated on 120V line, the
current should not exceed 3A,
Then ,
=
=
=40
Required resistance to be put in series= R1-R =
40-10 = 30
Ans: (d)
57
28.The
28
Th equivalent
i l t resistance
i t
b
between
t
points
i t A
and B of an infinite network of resistances,
h off 1Ω connected
t d as shown,
h
iis
each
(a) infinite
(c)
1 √5
Ω
2
(b) 2 Ω
(d) zero
58
28.Solution:- Let X be the equivalent resistance
between A and B. the network
consists of infinite units of two resistors
of 1 1 . The addition of one more
such unit across AB will not affect total
resistance. The network will then appear
a shown below;
59
P
1
1
1 1
=
Or Or X
1
2
= X
Q
=X 1 √1 4 1 √5
– X – 1 = 0 or X = =
2
2
(since X cannot be ‐ ve) Ans: (C) 60
29.In the circuit shown the value of I in ampere is
(a) 1
(c) 0.4
(b) 0.60
(d) 1.5
61
29.Solution-:-the equivalent circuit is as shown in the fig.
Current I =
I
= 1A
i
4
2 .4
6
=
I
i
i
4V
1 .6
1 .6
Now I1 X 4 = 6 X I or I1 =
But I = I + I1 or 1 = I +
=+
Or I = +
Ans: (c)
62
30. Two cells, having the same emf are connected in
series through
g an external resistance R. Cells have
internal resistances r1 and r2 (r1 > r2) respectively.
When the circuit is closed, the potential difference
across the
th fifirstt cellll iis zero. Th
The value
l off R iis
63
30 .Solution:- given;
V1 = E – Ir1 =0
I
R
E 1 ,r1
I
E 2 ,r2
Or I =
Also I =
Or
=
Or R + r1 + r2 = 2r or R = r1 - r2
Ans: (b)
64
All the Best
Th k you
Thank
65
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