MULTIPLE-CHOICE QUESTIONS
Q.1.When electric current is passed, electrons move from:
(a) high potential to low potential.
(b) low potential to high potential.
(c) in the direction of the current.
(d) against the direction of the current.
ANS -b
Q.2. The unit of resistivity is:
(a) V A
(b) V A
(c) V m /A
(d) VA/m
ANS-c
Q.3. What is the commercial unit of electrical energy?
(a) Joules
(b) Kilojoules
(c) Kilowatt-hour
(d) Watt-hour
ANS -c
Q.4. The instrument used for measuring electric current is :
(a) Ammeter
(b) Galvanometer
(c) Voltmeter
(d) Potentiometer
ANS- a
Q.5. Electrical resistivity of any given metallic wire depends upon
(a) its thickness
(b) its shape
(c) nature of the material
(d) its length
ANS -c
Q.6. In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are
connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in
5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J
ANS -c
Q.7. The heating element of an electric iron is made up of:
(a) copper
(b) nichrome
(c) aluminium
(d) iron
ANS -b
Q.8. Work of 14 J is done to move 2 C charge between two points on a
conducting wire. What is the potential difference between the two points?
(a) 28 V
(b) 14 V
(c) 7 V
(d) 3.5 V
ANS -c
Q.9. While a cell is being charged, energy is converted into energy.
a. mechanical, electrical
b. electrical, chemical
c. heat, electrical
d. chemical, heat
ANS -b
Q.10. The unit of potential difference is :
(a) Volt
(b) Ohm
(c) Ampere
(d) Faraday
ANS -a
Q.11. Copper is not preferred to make fuse wire because it .
a. is a good conductor of electricity
b. has a low melting point
c. has a high melting point
d. is not easily available
Q.12. Coulomb is the SI unit of:
(a) Charge
(b) current
(c) potential difference
(d) resistance
ANS-a
ASSERTION-REASON TYPE QUESTIONS
Following questions consist of two statements – Assertion (A) and Reason (R).
Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Q.1.. Assertion (A): When resistances are connected between the same two points
they are said to be in series.
Reason(R): When resistors are connected in series the current through each resistor
is the same.
ANS – d
Q.2. Assertion (A) : Tungsten metal is used for making filaments of incandescent
lamps.
Reason (R) : The melting point of tungsten is very low.
ANS -c
Q.3. Assertion (A): If a graph is plotted between potential difference and current the
graph is a straight line passing through the origin.
Reason(R): current is directly proportional to the potential difference.
ANS – a
Q.4. Assertion (A) : Longer wires have greater resistance and the smaller wires have
lesser resistance.
Reason (R) : Resistance is inversely proportional to the length of the wire.-
ANS - c
Q.5. Assertion (A) : Alloys are commonly used in electrical heating devices, like
electrical iron, toasters etc.
Reason (R) : Alloys do not oxidise (burn) readily at high temperatures.
ANS -a
Q. 6. Assertion (A) : Bending a wire does not affect electrical resistance.
Reason (R) : Resistance of a wire is proportional to resistivity of material.
ANS – b
Q.7. Assertion (A): A cell converts chemical energy into electrical energy.
Reason(R): A cell maintains a potential difference across its terminals due to
chemical reactions.
ANS – b
CASE STUDY BASED QUESTION
Electrical resistivities of some substances at 20°C are given in the table. Based on
the info in the table, answer the given questions.
Silver 1.60 X 10-8 Ω𝑚
Copper 1.62 X 10-8 Ω𝑚
Tungsten 5.2 X 10-8 Ω𝑚
Mercury 94 X 10-8 Ω𝑚
Iron 10 X 10-8 Ω𝑚
Nichrome 100 X 10-8 Ω𝑚
1. Which is a better conductor of electric current ?
(A) Silver
(B) Copper
(C) Tungsten
(D) Mercury
Ans. Option (A) is correct. Explanation: Silver is a better conductor because it
has lower resistivity.
2. Which element will be used for electrical transmission lines ?
(A) Iron
(B) Copper
(C) Tungsten
(D) mercury
Ans. Option (B) is correct. Explanation: Copper, because it is economical, less
oxidative than other metals and has low resistivity.
3. Nichrome is used in the heating elements of electric heating device
because:
(A) It has high resistivity
(B) It does not oxidise readily at high temperature
(C) Both of the above
(D) None of the above U
Ans. Option (C) is correct. Explanation: Nichrome, as it has very high
resistivity / as it is an alloy, it does not oxidize readily at high temperature.
4. Series arrangement is not used for domestic circuits because:
(A) Current drawn is less
(B) Current drawn is more
(C) Neither of the above
(D) Both of the above
Ans. Option (A) is correct.
Explanation: In series arrangement, same current will flow through all the
appliances which is not required and the equivalent resistance becomes
higher, hence the current drawn becomes less.
TWO MARKS QUESTIONS
Q.1.Calculate the number of electrons that would flow per second through the cross-
section of a wire when 1 A current flows in it.
Ans : Given: I = 1A , t = 1 s , Q = It , Q =1Ax 1s = 1C
But Q = ne or n = Q/e = 1 /1.6x10-19 = 6.25 x 108 electrons
Q.2. Define the following terms:
(a) one ampere (b) 1 volt.
Ans: One Ampere: The SI unit of electric current is ampere (A). One ampere is
the electric current when one coulomb of charge flows through a conductor in
one second.
One Volt: The SI unit of potential difference is volt (V). One volt is the potential
difference between two points in an electric circuit when one joule of work is
done to move a charge of one coulomb from one point to the other.
Q.3. Keeping the potential difference constant, the resistance of a circuit is doubled.
By how much does the current change?
Ansr: V = IR or V/R= l ,
Since the resistance and the current are inversely proportional, the current will
become half.
Q.4. How much work is done in moving a charge of magnitude 3 C across two points
having a potential difference of 12 V?
Ans: Given : Q = 3 C, V = 12 V
To find: W , as V = W/Q or W = VQ = 12 × 3 = 36 J
Q.5. Define electric power. Write an expression relating electric power, potential
difference and resistance.
Ans. Electric power : It is the amount of electric energy consumed in a circuit
per unit time. Expression: P = V2/R Where, P = Electric Power, V = Potential
difference, R = Resistance
Q.6.Give reason for the following:
a. Tungsten used almost exclusively for filament of electric lamp.
b. Why do we use copper and aluminium wires for transmission of electric current?
Ans : a. Tungsten is used in making the filament of electric lamp because it
has high resistivity and high melting point.
b. The copper and aluminium have low resistivity and high conductivity.
Q.7. Distinguish between resistances in series and resistances in parallel.
Ans:
Resistances in series:
1. If a number of resistances are connected in such a way that the same
current flows through each resistance, then the arrangement is called
resistances in series.
2. The current across each resistance is same.
3. The equivalent resistance in series combination is greater than the
individual resistances.
4. This combination decreases the current in the circuit.
Resistances in parallel:
1. If a number of resistances are connected between two common points in
such a way that the potential differences across each of them is the same,
then the arrangement is called resistances in parallel.
2. The voltage across each resistance is same.
3. The equivalent resistance in parallel combination is smaller than each of the
individual resistances.
4. This combination increases the current in the circuit.
Q. 8. What is the better way of connecting lights and other electrical appliances in
domestic wiring? Why?
Ans: The better way of connecting lights and other electrical appliances in
domestic wiring is
parallel connection because of the following advantages:
• In parallel circuit, if one appliance stops working due to some defect, then all
other appliances keep working normally.
• In parallel circuit, each electrical appliance has its own switch due to which it
can be turned on or off, without affecting other appliances.
• In parallel circuit, each electrical appliance gets the same voltage (220 V) as
that of the power supply line.
• In parallel circuit, the overall resistance of the domestic circuit is reduced
due to which the current from the power supply is high.
THREE MARKS QUESTIONS
Q.1. (a) List the factors on which the resistance of a conductor in the shape of a wire
depends.
(b) Why are metals good conductors of electricity whereas glass is a bad conductor
of electricity? Give reasons..
(c) Why are alloys commonly used in electrical heating devices? Give reason.
Ans: a. Factors on which resistance of a wire depends:
i. Resistance is directly proportional to length (l)
ii. Resistance is inversely proportional to area of cross-section(A).
i.e. R α l, R α 1/A or R α l/A
or R = ρ l/A, here ρ is the resistivity of the material at a particular temperature
(ie, resistivity
depends on material and temperature)
b. Metals are good conductors due to having large number of free electrons
and their low
resistivity. Glass is a bad conductor because it has no free electrons and its
resistivity is higher.
c. Alloys are commonly used in electrical heating devices due to their high
resistivity and high
melting point.
Q. 2. A nichrome wire has a resistance of 10 Ω. Find the resistance of another
nichrome wire, whose length is three times and area of cross-section four times the
first wire.
Ans: we have resistance R = ρ l/A
For first wire length L1= l, Area of cross section A1= A
So, for first wire resistance R1= ρ l/A = 10 Ω
For second wire length L2 = 3l, Area of cross section A2 = 4A
So, for second wire resistance R2 = ρ 3l/4A
Q.3. State the formula co-relating the electric current flowing in a conductor and the
voltage applied across it. Also, show this relationship by drawing a graph. What
would be the resistance of a conductor, if the current flowing through it is 0.35
ampere when the potential difference across it is 1.4 volt?
Ans: potential difference V = IR where I is electric current and R, resistance of
conductor
ie, V α l
If we plot a graph b/w V and I, it is a straight line.
Graph b/w V and I:
Given current l = 0.35 A, potential difference V = 1.4 V
Resistance R = = V/I , R = 1.4 / .35 = 4 Ω
Q. 4.Calculate the total cost of running the following electrical devices in April month
if the rate of 1 unit of electricity is Rs. 6.00. (i) Electric heater of 1000 W for 5 hours
daily. (ii)
Electric refrigerator of 400 W for 10 hours daily
Ans. P1 = 1000 W = 1kW, t1= 5h,
P2 = 400 W = 400 /1000 kW = 0.4KW, t2 = 10h
No. of days in September, n = 30
E1 =P1 × t1 × n = 1 kW × 5h × 30 = 150 kWh
E2 =P2 × t2 × n = 0.4kW x 10h x 30 = 120kWh
؞Total energy = (150 + 120) kWh = 270 kWh ,so Total cost = 270 x 6 = Rs. 1620/-
Q. 5. (i) Consider a conductor of resistance ‘R’, length ‘L’, thickness ‘d’ and resistivity
‘ρ’. Now this conductor is cut into four equal parts. What will be the new resistivity of
each of these parts? Why?
(ii) Find the resistance if all of these parts are connected in:
(a) Parallel (b) Series
(iii) Out of the combinations of resistors mentioned above in the previous part, for a
given voltage which combination will consume more power and why?
Ans. (i) Resistivity will not change as it do not depend on the dimensions of
the conductor. It depends on the nature of material of the conductor.
(ii) The length of each part become L/4, ρ is constant and R = ρL/A
(iii) We know that Power P given as P=V.I =V2/R ( V= IR)
For given voltage parallel connection consume more power because it have
low equivalent resistance.