CBSE Class 10 Physics Electricity Worksheet Set A

Question. Calculate the number of electrons that would flow per second through the cross- section of a wire when 1 A current flows in it.
Answer :
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

Question. Define the following terms:
(a) one ampere (b) 1 volt.
Answer :
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.

Question. Keeping the potential difference constant, the resistance of a circuit is doubled. By how much does the current change?
Answer :
V = IR or V/R= l ,
Since the resistance and the current are inversely proportional, the current will become half.

Question. How much work is done in moving a charge of magnitude 3 C across two points having a potential difference of 12 V?
Answer:
Given : Q = 3 C, V = 12 V
To find: W , as V = W/Q or W = VQ = 12 × 3 = 36 J

Question. Define electric power. Write an expression relating electric power, potential difference and resistance.
Answer : 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

Question. 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?
Answer :
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.

Question. List in a tabular form two differences between a voltmeter and an ammeter.

Question. Write the factors on which heat produced in a resistor depends
Answer : heat produced in a resistor is directly proportional to
• Square of current (I2)
• Resistance of the resistor (R) and
• Time for which the current flows through the resistor. (t)
H = I2Rt joules , hence By Ohm’s law, we get H = VIt joules = V2 t/R joules

Question. Distinguish between resistances in series and resistances in parallel.
Answer :
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.

Question. What is the better way of connecting lights and other electrical appliances in domestic wiring? Why?
Answer : 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.

Question. Two students perform experiments on series and parallel combinations of two given resistors R1 and R2 and plot the following V-I graphs.

Which of the graphs is (are) correctly labelled in terms of the words ‘Series and parallel’? justify your answer.
Answer : In case of series combination, the effective resistance = R1 + R2 is more, hence slope of V – I graph will be more. It is otherwise in case of I – V graph. So, series and parallel are correctly marked in graph (ii).

Question. A bulb is rated at 5.0 V, 100 mA. Calculate its (a) power and (b) resistance.
Answer : Rating of bulb, V = 5 0. Volt, I = 100 mA = 100 x10^-3 A = 0.1A
a. Power of bulb P = V x l or P = 5 .0 x 0 .1 = 0.5W
b. V = IR or R =V/l = 5/0.1 = 50 Ω

Question. (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 reason.
(c) Why are alloys commonly used in electrical heating devices? Give reason.
Answer : 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.

Question. 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.
Answer : 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

Question. 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?
Answer : 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

Question. Calculate the total cost of running the following electrical devices in the month of September, 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
Answer : 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/-

Question. (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?
Answer : (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

(ii) 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.

Question. Two bulbs A and B are rated as 90W–120V and 60W–120V respectively. They are connected in parallel across a 120V source. Find the current in each bulb. Which bulb will consume more energy?
Answer : First Bulb: 90 W–120 V.
Resistance of first bulb R1= V2 /P1 = 120 𝑋 120/90 = 160 Ω
Current in first bulb I1 = V/R1 = 120 /160 = .75 A
Resistance of second bulb R2 = V2 /P2 = 120 𝑋 120/60 = 240 Ω
Current in second bulb I2 = V/R2 = 120 /240 = .50 A
Power of first bulb is more than second bulb, so first bulb will consume more energy.

Question. In the given circuit, A, B, C and D are four lamps connected with a battery of 60 V.

Analyse the circuit to answer the following questions.
(i) What kind of combination are the lamps arranged in (series or parallel)?
(ii) Explain with reference to your above answer, what are the advantages (any two) of this combination of lamps?
(iii) Explain with proper calculations which lamp glows the brightest?
(iv) Find out the total resistance of the circuit R
Answer : (i) The lamps are in parallel.
(ii) Advantages: If one lamp is faulty, it will not affect the working of the other lamps. They will also be using the full potential of the battery as they are connected in parallel. (ii) The lamp with the highest power will glow the brightest. Since P=VI and In this case, all the bulbs have the same voltage. But lamp C has the highest current.
Hence, for Lamp C, power P = 5 × 60 Watt = 300 W. (the maximum).
(iii) The total current in the circuit = 3+4+5+3 A = 15A , Voltage = 60V
R = V/I = 60/15 = 4 Ω

Read the passage carefully and answer the following questions from The electrical energy consumed by an electrical appliance is given by the product of its power rating and the duration for which it is used. SI unit of electrical energy is the joule. Where a large quantity of energy is involved, using a joule is not convenient as a unit. So, for commercial purposes, bigger units of electrical energy are involved. 1 kilowatt-hour is equal to 3.6 × 106 joules of electrical energy.

Question. Choose the correct statement: 
(a) 1 watt-hour = 3600 J
(b) 1 kWh = 36 × 106 J
(c) Energy in kWh = power in W(watt) × time in hour(h)
(d) Energy in kWh = V × I × T1000

Answer: A

Question. Choose the incorrect statement. 
(a) Higher the resistance, the lesser the power consumed.
(b) Lower the resistance, more the voltage drawn.
(c) Higher the resistance, the higher the current flown.
(d) Higher the resistance, the lesser the voltage drawn.

Answer: C

Question. The value of energy dissipated by a certain heater is E. If the duration of operation of the heater is doubled, the energy dissipated will be: 
(a) halved
(b) doubled
(c) four-times
(d) remains same

Answer: B

Question. 60 W is the power of a lamp. The energy dissipated in one minute is: 
(a) 360 J
(b) 36 J
(c) 3.6 J
(d) 3600 J

Answer: D

Question. Calculate the energy transformed by a 5 A current flowing through a resistor of 2 Ω for 30 minutes. 
(a) 90 kJ
(b) 80 kJ
(c) 60 kJ
(d) 40 kJ

Answer: A

Observe the figure carefully and answer the following questions from The following graphs represent the current versus voltage and voltage versus current for six conductors A, B, C, D, E, and F. 

Question. Which indicates the correct sum of least resistances of two graphs? 
(a) Curve C + Curve F
(b) Curve A + Curve D
(c) Curve A + Curve F
(d) Curve C + Curve D

Answer: C

Question. If resistances shown by curve A and curve E are added, the value will be: 
(a) 1.83 W
(b) 1.50 W
(c) 1.64 W
(d) 1.25 W

Answer: D

Question. Which is true for these graphs? 
 
(a) Both are ohmic conductors
(b) Curve A is ohmic and B is non-ohmic conductor
(c) Both are non-ohmic conductors
(d) Curve B is ohmic and A is non-ohmic conductor

Answer: C

Question. Among conductors A, B, C, D, E, F, the maximum resistance is shown by: 
(a) curve C
(b) curve A
(c) curve F
(d) curve D

Answer: A

Question. Which of the following does not indicate the resistance of curve B? 
(a) The slope of curve B
(b) The ratio of V-intercept to I-intercept
(c) The ratio of total grids on the y-axis to total grids on the xaxis
(d) 3/4 Ω

Answer: D