Practice Sheet WARRIOR
ELECTRICITY
1. Which of the following represents voltage? 7. Which one among a bar of an alloy of mass 2 kg
(A) Work done / Current × Time and a 3 kg iron bar of the same dimension has
(B) Work done × Charge greater resistivity?
(C) Work done × Time / Current (A) Iron bar because it has a higher mass
(D) Work done × Charge × Time (B) Alloy bar because it has a lower mass
(C) Iron bar because it has the same types of
2. A cooler of 1500 W, 200 volts and a fan of 500 atoms
W, 200 volts are to be used from a household (D) Alloy bar because it has different types of
supply. The rating of the fuse to be used is atoms
(A) 2.5 A (B) 5.0 A
8. Two resistors connected in series give an
(C) 7.5 A (D) 10 A
equivalent resistance of 10 Ω. When connected in
parallel, give 2.4 Ω. Then the individual resistance
3. Which combination of a 2 Ω resistor and 4 Ω
is
resistor offers the least resistance to current in the
(A) each of 5 Ω
circuit?
(B) 6 Ω and 4 Ω
(A) Series combination, which results in a net (C) 7 Ω and 4 Ω
resistance of 2 Ω (D) 8 Ω and 2 Ω
(B) Parallel combination, which results in a net
resistance of 2 Ω 9. A battery of 10 volt carries 20,000 C of charge
(C) Series combination, which results in a net through a resistance of 20 Ω. The work done in
resistance of 1.5 Ω 10 seconds is
(D) Parallel combination, which results in a net (A) 2 × 103 joule (B) 2 × 105 joule
resistance of 7.5 Ω (C) 2 × 104 joule (D) 2 × 102 joule
4. In an electrical circuit, two resistors of 2 Ω and 4 10. Two bulbs are rated 40W, 220W and 60W, 220W.
Ω, respectively, are connected in series to a 6 V The ratio of their resistances will be
battery. The heat dissipated by the 4 Ω resistor in (A) 4 : 3 (B) 3 : 4
5 s will be (C) 2 : 3 (D) 3 : 2
(A) 5 J (B) 10 J
(C) 20 J (D) 30 J 11. A current of 10 A flows through a conductor for
two minutes.
5. In order to reduce electricity consumption at (i) Calculate the amount of charge passed
through any area of cross section of the
home, what kind of appliance should one
conductor.
purchase?
(ii) If the charge of an electron is 1.6 × 10–19C,
(A) One which draws low power
then calculate the total number of electrons
(B) One which produces less heat
flowing.
(C) One which operates at a higher voltage
(D) One which draws a high amount of current
12. (a) Define the term 'volt'.
(b) State the relation between work, charge and
6. If n resistors each of resistance R are connected in potential difference for an electric circuit.
parallel combination, then their equivalent Calculate the potential difference between the two
resistance is terminals of a battery if 100 J of work is required
(A) R/n2 (B) n2/R to transfer 20 C of charge from one terminal of the
(C) n/R (D) R/n battery to the other.
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13. Draw a schematic diagram of an electric circuit 17. A fuse wire melts at 5 A. If its is desired that the
consisting of a battery of two cells each of 1.5 V, fuse wire of the same material melt at 10 A, then
5, 10 and 15 resistors and a plug key, all should the new fuse wire be of smaller or larger
connected in series radius than the earlier one?
14. State Ohm's law. Draw a labelled circuit diagram 18. Two lamps, one rated 100 W; 220 V and the other
to verify this law in the laboratory. If you draw a 60 W; 220 V, are connected in parallel to electric
graph between the potential difference and current mains supply. find the current drawn by the two
flowing through a metallic conductor, what kind lamps from the line, if the supply voltage is 220V.
of curve will you get? Explain how you would use
this graph to determine the resistance of the 19. What is meant by overloading of an electrical
conductor. circuit? Explain two possible causes due to which
overloading may occur in household circuit?
15. The resistance of a wire of 0.01 cm radius is 10. Explain one precaution that should be taken to
If the resistivity of the material of the wire is avoid the overloading of domestic electric circuit.
50 × 10−8 m, find the length of the wire.
20. Explain two disadvantages of series arrangement
16. Deduce the expression for the equivalent for household circuit.
resistance of the parallel combination of three
resistors R1, R2, and R3.
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ANSWER KEY
1. (A) 11. (Q = 1200 C)
2. (D) 12. (5V)
3. (D) 13. (A4)
4. (C) 14. (Check Solution)
5. (A) 15. (62.8)
6. (D) 16. (Check Solution)
7. (D) 17. (Check Solution)
8. (B) 18. ( 0.73 A)
9. (B) 19. (Check Solution)
10. (D) 20. (Check Solution)
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HINTS AND SOLUTIONS
1. (A) 12. (5 V)
Work done / Current × Time (a) 'Volt' is the S.I. unit of potential difference.
The potential difference between two points
2. (D)
is 1 volt when 1 joule of work is required to
10 A
move 1 coulomb of charge from one point to
3. (D) the other.
Parallel combination, which results in a net (b) Work (W), charge (Q), and potential
resistance of 7.5 Ω difference (V) are related as,
W
4. (C) V=
Q
20 J
Given,
5. (A) W = 100 J,
One which draws low power Q = 20 C.
Potential difference between the two
6. (D) terminals, V, is
R/n W
V=
Q
7. (D)
100J
Alloy bar because it has different types of atoms =
20C
8. (B) = 5 J/C = 5 V
6 Ω and 4 Ω
13. (A4)
9. (B)
5 10 15
2 × 105 joule
10. (D) I
3:2
105V
11. Given,
I = 10 A 14. Ohm's law: The electric current flowing through
t = 2 mins = 120s.
a conducting wire, I, is directly proportional to the
(i) Amount of charge, Q, passing through a
cross-section is related to the current, I, as potential difference across its ends, V, provided its
Q temperature stays unchanged.
I= Mathematically, V I.
t
= 10 A × 120s Rheostat
+ –
= 1200 A. s = 1200 C
(ii) Given: B Rh
net charge, Q = 1200 C, –
Charge of electron, e = 1.6 × 10–19 C, K A
Number of electrons, n = ? +
+ –
We have, Q = ne. V
Q
n=
e R
1200C The graph obtained by plotting the potential
=
1.6 10−19 C difference against the current flowing in a
= 7.5 × 1021 conductor is a straight line.
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From Ohm's law, V I, or V = IR, Where R is the If the equivalent resistance is RP, we have using
resistance. Ohm's law.
V
The slope of the V – I graph gives the I=
RP
resistance of the conductor
The potential difference across each individual
resistor is also V. Applying Ohm's law for each
15. (62.8) resistor, we find,
Given: V V V
I1 = ,I2 = , and I3 =
R1 R2 R3
radius of wire, r = 0.01 cm = 10−4 m,
Tying it all up.
resistivity, = 50 × 10–8 m,
I = I1 + I2 + I3
resistance, R = 10.
V V V V
length of wire, l = ? = + +
R P R1 R 2 R 3
These quantities are related by,
1 1 1 1
l V = V + +
R = RP R1 R 2 R 3
A
1 1 1 1
Where A = r2 is the cross-sectional area of the = + +
R P R1 R 2 R 3
wire. We have, The equivalent resistance, RP, is related to the
RA individual resistors as
l=
1 1 1 1
= + +
R P R1 R 2 R 3
R r 2
=
17. The heat produced every second in a fuse wire of
10 (10−4 m)2 resistance, R, and with a current, I, in it, is given
=
50 10−8 m as, H = I2R.
= 0.628 the heat produced for I = 5A melts the wire.
It follows from the relation that for the same heat
= 62.8
to be produced for I = 10 A, the resistance must be
smaller.
16. Derivation of equivalent parallel resistance: Since,
The following diagram shows a parallel 1
R
combination of the three resistors R1, R2, and R3. cross- sectionalarea
Currents I1, I2, and I3 flow though the resistors R1, A smaller resistance has a larger cross-section.
A larger cross-section implies a larger radius. So,
R2, and R3. respectively. The total current, I, is the
the new fuse wire has a larger radius.
sum of the currents through the three branches:
I = I1 +I2 + I3 18. ( 0.73 A)
R1 Since the lamps are connected in parallel to the
mains supply, the voltage across each lamp
M R2 N is 220 V.
R3 power rating
The current drawn by a lamps =
voltageapplied
Current drawn by the 100 W lamp,
100W 100
I1 = = A
+ – 220V 220
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Current drawn by the 600 W lamp, household circuits, their copper wires get heated
60 up to a very high temperature and can cause a fire.
I2 = A
220 Precaution: Thus, overloading can be highly
Net current drawn from the mains supply, I, is damaging to electrical appliances and buildings.
I = I1 + I2 So, fuse of proper rating must be used to avoid
100 60 such damages. Such a fuse-wire will melt before
= A+ A the temperature of the heated circuit wire becomes
220 220
too high and causes the circuit to break.
160
= A
220
20. Disadvantages of series circuits for domestic
0.73 A
wiring:
1. In series circuit, if one electrical appliance
19. Overloading: The current flowing in domestic stops working due to some defect then all
wiring at a particular time depends on the power other appliances also stop working because
ratings of the appliances being used. If too many the whole circuit is broken.
electrical appliances of high power rating are 2. In series circuit, all the electrical appliances
switched on at the same time, they draw extremely have only one switch due to which they
large quantity of current from the circuit. This is cannot be turned off or turned on separately.
known as the overloading of the circuit. Due to
large current flowing through the wires of the
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