CBSE Class 10 Science Electricity Assignment Set F

Very Short Answer Type Questions :

Question. Why do we use copper and aluminium wire for transmission of electric current?
Answer: Copper and aluminium wires are used for electric transmission due to their low resistivity.

Question. If the charge on an electron be 1.6#10-19C, find the approximate number of electrons in 1 C.
Answer: 1.6×10-19C charge is of = 1 electron and 1 C charge is of = 1/1.6 x 10^-19 electron No. of electrons = 6.25×10¹⁸

Question. List any two factors on which resistance of a conductor depends.
Answer: Resistance of a conductor:
a. is directly proportional to its length R ? r …(1)
b. is inversely proportional to its area of cross section.
R ∝ 1/A
Combining (1) and (2), we get R ∝ P/A

Question. What is the SI unit of electric potential?
Answer: Volt is the SI unit of electric potential.

Question. Give an example of a metal which is the best conductor of heat.
Answer: Gold, Silver, Copper etc. metals are good conductor of heat.

Question. Define electric circuit. Distinguish between open and closed circuit.
Answer: Electric circuit is the arrangement in which electric current can flow when circuit is switched on. In open circuit there is no flow of current as the switch is off.
In closed circuit a current flows in the circuit when switch is on.

Question. Nichrome is used to make the element of electric heater. Why?
Answer: Nichrome is used to make element of electric heater because nichrome is an alloy which has high melting point and high resistances.

Question. happens to the resistance of a conductor when the length of the conductor is reduced to half?
Answer: Resistance is directly proportional to the length of the conductor. If length becomes half the resistance also become half of its initial value.

Question. What happens to resistance of a conductor when its area of cross-section is increased?
Answer: Resistance decreases as R x 1/A

Question. State difference between the wire used in the element of an electric heater and in a fuse wire.
Answer: The wire used in the element of electric heater has a high resistivity and have a high melting point, i.e. even at a high temperature element do not burn while fuse wire have a low melting point and high resistivity.

Question. How is an ammeter connected in a circuit to measure current flowing through it?
Answer: In series

Question. Write S.I. unit of resistivity.
Answer: Ohm-metre (Ωm).

Short Answer Type Questions :

 Question. Draw a schematic diagram of a circuit consisting of a cell of 1.5 V, 10 ohm resistor and 15 ohm resistor and a plug key all connected in series.
Answer: Schematic diagram is shown below.

Question.The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if it operates on a 6 V battery (Neglect the change in resistance due to unequal heating of the filament in the two cases).
Answer: 

Question.The charge possessed by an electron is 1.6 X 10-19 coulombs. Find the number of electrons that will flow per second to constitute a current of 1 ampere.
Answer: 

Question. A piece of wire of resistance 20 Ω is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new Situation.
Answer: 

 Question. A current of 10A exists in a conductor.Assuming that this current is entirely due to the flow of electrons (a) find the number of electrons crossing the area of cross section per second, (b) if such a current is maintained for one hour, find the net flow of charge.
Answer: Current, I = 10 A
Charge flowing through the circuit
in one second, 

(a) We know, Charge on an electron
= 1.6 × 10^–19C
So, No. of electrons crossing per second
=10C/( 1.6× 10^-19) = 6.25 × 10¹⁹
(b) Net flow of charge in one hour
= Current × Time
= 10 A × 1 h
10 A × (1 × 60 × 60 s) = 36000 C

Question. A current of 5.0 A flows through a circuit for 15 min. Calculate the amount of electric charge that flows through the circuit during this time.
Answer: Given : Current, I = 5.0 A
Time, t = 15 min. = 15 × 60 s = 900 s
Then, Charge that flows through the circuit,
Q = Current × Time
= 5.0A × 900 s
= 4500 A.s = 4500 C

 Question. The value of current (I) flowing through a given resistor of resistance (R), for the corresponding values of potential difference (V) across the resistor are as given below :

Plot a graph between current (I) and potential difference (V) and determine the resistance (R) of the resistor.
Answer: 

Question. If four resistances each of values 1 ohm are connected in series. Calculate equivalent resistance.
Answer: In series,
R₁ = R₂ = R₃ = R₄ = 1ohm
putting values, we get,
R_s = 1 + 1 + 1 + 1 = 4 

Question. A piece of wire is redrawn by pulling it until its length is doubled. Compare the new resistance with the original value.
Answer: Volume of the material of wire remains same.
So, when length is doubled, its area of crosssection will get halved. So, if l and a are the original length and area of cross-section of wire,
Original value of the resistance, R = Ρ×λ/a
and,
New value of the resistance, 

Question. Study the following electric circuit and find (i) the current flowing in the circuit and (ii) the potential difference across 10 Ω resistor.

Answer: 

Question. Find the current drawn from the battery by the network of four resistors Shown in the figure.

Answer: 

Question. In an experiment to study the relation between the potential difference across a resistor and the current through it, a student recorded the following observations:

On examine the above observations, the teacher asked the student to reject one set of readings as the values were out of agreement with the rest. Which one of the above sets of readings can be rejected? Calculate the mean value of resistance of the resistor based on the remaining four sets of readings.
Answer: The third reading for V = 3.0 volt and I — 0.6 A will be rejected as it has larger deviation from the rest of the readings.
The value of resistance in the other four observations will be I (using R = V/I) 10Ω, 11 Ω, 10 Ω and 10.67 Ω.
So, the mean value of resistance = 41.67/4 = 10.417 = 10.42 Ω

Question. Find the number of electrons transferred between two points kept at a potential difference of 20 V if 40 J of work is done.
Answer: Given: V = 20 Volt
W = 40 J
W = P x t
= Vx I x t
= V X Q/t x t
or W = VxQ
40 = 20xQ
or Q = 2C
ne = Q
n = Q/e = 2/1.6 x 10^-19
= 1.25 x 10¹⁹

Question. An electric kettle of 2 kW works for 2 h daily. Calculate the (a) energy consumed in SI and commercial units (b) cost of running it in the month of June at the rate of `3.00 per unit.
Answer: (a) Given: P = 2 kW = 2000W
t = 2 h
Electric energy, E = P#t = 2×2 = 4 kWh
(b) Total energy consumed in month of June (having 30 days)
Electric kettle = Ω 4X30h kWh = 120 kWh
= 120 units.
Cost of running electric kettle:
= `120 X 3 = `360

Question. (a) Name an instrument that measures electric current in a circuit. Define unit of electric current.
(b) What are the following symbols mean in an electric circuit?

Answer: a. Ammeter.
If IC charge flows in an electric circuit is 1 s then the current is said to be 1 A.
b. (i) Rheostat (ii) Closed key 99

Long Answer Type Questions :

 Question. Two lamps, one is rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to a 220 V supply. Find the current drawn from the supply line.
Answer: Given: Lamp one 100 W, 220 V, Lamp 2 60 W, 220 V.
Let their resistances are R1 and R2

Question. (a) Name an instrument that measures potential difference between two points in a circuit. Define the unit of potential difference in terms of SI unit
of charge and work. Draw the circuit symbols for (i) variable resistor, (ii) a plug key which is closed one.
(b) Two electric circuits I and II are shown below “

(i) Which of the two circuits has more resistance?
(ii) Through which circuit more current passes?
(iii) In which circuit, the potential difference across each resistor is equal?
(iv) If R1 > R2 > R3 in which circuit more heat will be produced in R1 as compared to other two resistors?
Answer: a. Voltmeter
The amount of work done in bringing a unit positive charge from one point to another in an electric field is said to be potential difference

b. (i) In series combination resistance is more than parallel combination.
(ii) Lesser the resistance more the current in circuit i.e., in parallel (II) current is max.
(iii) In parallel combination (II)
(iv) More heat in (I) across R1

Question. (a) Name and state the law that gives relationship between the current through a conductor and the potential difference across its two terminals. Also, express this law mathematically.
(b) Draw the V-I graph for this law. Justify your answer.
(c) Write the name and use of the circuit components whose symbols are given below.

Answer: 
a. The law is Ohm’s law.
If the physical conditions of a conductor is kept constant then current through it is directly proportional to the potential difference applied across it.
V ∝ I or V = RI
b. Since V ∝ I so a graph b/w V and I is a straight line.

c. (i) Symbol is of variable resistor and it is used to regulate the current.
(ii) Plug key is closed. When plug key is closed current flows through the circuit.

 Question. A piece of wire of resistance 6 W is connected to battery of 12 V. Find the amount of current flowing through it. Now, the same wire is redrawn by stretching it to double its length. Find the resistance of the new (redrawn) wire.
Answer: Given: R = 6 Ω
V = 12 Volt
I = ?

Question. Draw a schematic diagram of an electric circuit (in the “on” position) consisting of a battery of five cells of 2 V each, a 5 W resistor, a 8 W resistor, a 12 W resistor and a plug key, all connected in series. An ammeter is put in the circuit to measure the electric current through the resistors and a voltmeter is connected so as to measure the potential difference across the 12 W resistor.
Calculate the reading shown by the: (a) ammeter (b) voltmeter in the below electric circuit.
Answer: Resistors of 5 W , 8 W , 12 W all the connected in series.
Hence,
RS = R1+ R2+ R3
= ^5 + 8 + 12hW
RS = 25 W
As V = IR

 Question. (i) Draw a closed circuit diagram consisting of a 0.5 m long nichrome wire XY, an ammeter, a voltmeter, four cells of 1.5 V each and a plug key.
(ii) Following graph was plotted between V and I values :

What would be the values of V /I ratios when the potential difference is 0.8 V, 1.2 V and 1.6 V respectively? What conclusion do you draw from these values?
Answer: 

Question. (a) Calculate the resistance of the wire using graph.

(b) How many 176 W resistors in parallel are required to carry 5 A on a 220 V line?
(c) Define electric power, Derive relation between power, potential difference and resistance.
Answer: 

b. Resistance of the circuit to carry a current of 5A on 220V.

Question. (a) Three resistors R1, R2 and R3 are connected in parallel and the combination is connected to battery, ammeter, voltmeter and key. Draw suitable circuit diagram. Obtain an expression for the effective resistance of the combination of resistors in parallel.
(b) Why are electric bulbs filled with chemically inactive nitrogen or argon?
(c) What is meant by the statement that the rating of a fuse in a circuit is 5 A?
Answer: a. Let equivalent (effective) resistance is R then I = V/R

1/R = 1/R1 + 1/R2 + 1/R3
b. To prevent oxidising the filament due to high temperature.
c. The current in the fuse wire must not exceed 5 A otherwise it will melt.

Question. A TV set shoots out a beam of electrons. The beam current is 10μA.
(a) How many electrons strike the TV screen in each second ?
(b) How much charge strikes the screen in a minute?
Answer: Beam current, I = 10 μA = 10 ×10^–6A
Time, t = 1 s
So,
(a) Charge flowing per second,
Q = I × t = 10 × 10^–6A × 1s = 10 × 10^–6C
We known,
Charge on an electron = 1.6 × 10^–19C
So, No. of electrons striking the TV screen

(b) Charge striking the screen per min
= (6.25 × 10¹⁴ × 60) × 1.6 × 10^–19C
= 6.0 × 10^–3C

Question. A resistance of 6 ohms is connected in series with another resistance of 4 ohms. A potential difference of 20 volts is applied across the combination. Calculate the current through the circuit and potential difference across the 6 ohm resistance.
Answer: For better understanding we must drawn a proper circuit diagram. It is shown in fig.

We use proper symbols for electrical components.Resistances are shown connected in series, with 20 V battery across its positive and negative terminals. Direction of current
flow is also shows from positive terminal of the battery towards its negative terminal.
Potential difference, V = 20 V
Potential difference across 6Ω
V₁ = ? (to be calculated)
Total circuit resistance = 10Ω
From Ohm’s law, R_s =V/R_S
Circuit current, I = 2 ampere or (2A)
Putting values, we get, V₁ = 2 × 6 = 12 volts
Potential difference across 6Ω resistance = 12 V 

Question. Resistors R₁, R₂ and R₃ having values 5Ω,10Ω, and 30Ωrespectively are connected in parallel across a battery of 12 volt. Calculate
(a) the current through each resistor (b) the total current in the circuit and (c) the total circuit resistance.
Answer: Here,
R₁ = 5Ω R₂ = 10Ω, R₃ = 30Ω, V = 12 V
(a) I₁ = ? I₂ = ? I₃ = ?
(b) I = I₁ + I₂ + I₃ = ?
(c) R_p = ?

Question. Calculate the resistance of 100 m long copper wire. The diameter of the wire is 1 mm.
Answer: Using the relationship,

Question. For the circuit shown in the following diagram what is the value of

(i) current through 6 Ω resistor
(ii) potential difference (p.d.) across 12 Ω
Answer: (i) For current through 6Ω Current from 4 V battery flows through first parallel branch having 6 Ω and 3 Ω in series.
Current in this branch

Question. In the circuit diagram given below. find

(i) total resistance of the circuit
(ii) total current flowing in the circuit
(iii) potential difference across R1
Answer: (i) For total resistance
8Ω and 12Ω are connected in parallel.
Their equivalent resistance comes in series with 7.2Ω resistance as shown in fig.

With 7.2Ω and 4.8Ω in series
R_s = 7.2 + 4.8 = 12Ω
Total circuit resistance = 12 ohms.

(ii) For total current
Total circuit resistance, R = 12 ohm
Potential difference applied, V = 6 V
                                         I = ?
From Ohm’s law                 R =V/I
                                        I =V/R
                                        I =6/12
                                          = 0.5
Total circuit current = 0.5 A Ans.

(iii) For potential difference across R₁
R =V/I
V = IR
V₁ = IR₁
V₁ = 0.5 × 7.2
= 3.6 V
Potential difference across, V₁ = 3.6 V. Ans

Question. Two resistances are connected in series as shown in the diagram.

(i) What is the current through the 5 ohm resistance ?
(ii) What is the current through R ?
(iii) What is the value of R ?
(iv) What is the value of V ?
Answer: First resistance, R₁ = 5 Ω
(i) Current through 5 ohm resistance, I = ?
(ii) Current through R, I = ?
(iii) Value of second resistance, R = ?
(iv) Potential difference applied by the battery,
 V = ?

Current through 5Ω resistance = 2 ampere (2A).
(ii) Since R is in series with 5Ω same
current will flow through it,
Current through R = 2 A.

(iv) From relation, V = V₁ + V₂
V = 10 + 6 = 16 volts
V = 16 volts

Question. Three resistances are connected as shown in diagram through the resistance 5 ohms, a current of 1 ampere is flowing :

(i) What is the current through the other two resistors?
(ii) What is the potential difference (p.d.) across AB and across AC ?
(iii) What is the total resistance.
Answer: (i) For current in parallel resistors
For same potential difference across two parallel resistors,

Current is 0.6 A through 10 Ω
(ii) For p.d. across AB
From Ohm’s law, R =V/I, V = IR
V = 1 × 5 = 5V
P.D. across AB = 5 V. Ans
For parallel combination of 10Ω and 15Ω P.D. across BC, V = I1R1 = 0.6 × 10 = 6 V
P.D. across AC = P.D. across AB + P.D. across BC.
= 5 + 6 = 11 V
(iii) For total circuit resistance
For 10Ω and 15Ω in parallel

Question. In the circuit diagram.

Find (i) total resistance
(ii) current shown by the ammeter A.
Answer: 3 Ω and 2 Ω in series become 5Ω.Equivalent circuit is shown in fig.

 Question. Electric current flows through three lamps when arranged in (a) a series (b) a parallel. If the filament of one lamp breaks. Explain what happens to the other two lamps in both the cases.
Answer: a. In series combination if the filament of one lamp breaks then the circuit will be broken and hence other lamps stops glowing.

Question. Study the V-I graph for a resistor as shown in the figure and prepare a table showing the values of I (in amperes) corresponding to four different values of V (in volts). Find the value of current for V = 10 volts.
How can we determine the resistance of the resistor from this graph?

Answer: When V = 10 volt from the graph

Question. Resistors R₁ = 10 ohms, R₂ = 40 ohms,R₃ = 30 ohms, R₄ = 20 ohms, R₅ = 60 ohms and a 12 volt battery is connected as shown.
Calculate :
(a) the total resistance and (b) the total current flowing in the circuit.
Answer: The situation is shown in (figure).