CBSE Class 10 Science Electricity Assignment Set B

Question. What is the current through a 5.0 ohm resistor if the voltage across it is 10 V
(a) zero
(b) 0.5 A
(c) 2.0 A
(d) 5.0 A
Answer : C

Question. 100 J of heat is produced each second in a 4 W resistance. The potential difference across the resistor is
(a) 20 V
(b) 10 V
(c) 5 V
(d) 15 V
Answer : A

Question. A cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A
Answer : C

Question : An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:
a. 100 W
b. 75 W
c. 50 W
d. 25 W

Answer : D

Very Short Answers :

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

Question : Define resistance. Give its S.I. unit.
Answer :  Resistance is the property of a conductor to ppose the flow of charges through it. SI unit of resistance is Ohm (Ω).

Question. State Joule’s law of heating.
Answer: Joule’s law of heating states that the amount of heat produced in a conductor is directly proportional to:
(a) Square of current (I2)
(b) Resistance of wire (R)
(b) Time (t), for which current is passed.

Question. Define electric power.
Answer: The electrical work done per unit time is called electric power.

Question. What does an electric circuit mean?
Answer: A continuous conducting path consisting of wires and other electrical components (like resistance or electric bulb, switch etc.) between the two terminals of a cell or battery, along which an electric current flows, is called an electric circuit.

Question. Distinguish between an open and a closed circuit.
Answer: An electric circuit is said to be an open circuit when the switch is in ‘off’ mode (or key is unplugged) and no current flows in the circuit.
The circuit is said to be a closed circuit when the switch is in ‘on’ mode (or key is plugged) and a current flows in the circuit.

Question. How many 220 Ω resistors (in parallel) are required to carry 5 A on a 220 V line ?
Answer: 5.

Question. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A ? 
Answer: 110.

Question. Which uses more energy :
(a) a 250 W T.V. set in 1 hour
(b) a 1200 W toaster in 10 minutes ? 
Answer: (a) 9 × 105 J, (b) 7.2 × 105 J (T.V. set uses more energy)

Question. Study the circuit diagram given in figure carefully and calculate :
(a) Current in main circuit.
(b) Current in each of the resistors in the parallel circuit.
Answer: (a) 1.80 A, (b) 1.2 A and 0.6 A, 1.2 A and 0.6 A.

Question. A wire of resistance 1.5 Ω is stretched to double its length. What will be its new resistance ?
Answer: 6 W.

Question. A wire of resistance 36 Ω and length 60 cm is tripled on itself. What is the new resistance ?
Answer: 4 W.

Question. How many electrons are flowing per second past a point in a circuit in which there is a current of 1 A ?
Answer: 6.25 × 1018.

Question. The graph between V and I for a conductor is a straight line passing through the origin.
(a) Name the law illustrated by such a graph ?
(b) What should remain constant in a statement of this law ?
Answer: (a) Ohm's law, (b) Temperature.

Question : The potential difference across the wire having fixed resistance is tripled. By how much does the electric power increase?
Answer :  The electric power will increase by nine times when the potential difference across the wire having fixed resistance is tripled. According to Ohm’s law, potential difference V is proportional to current, I. Therefore, when V is made 3 times, I will increase 3 times. As Power P = VI, therefore, Power will increase by 9 times.

Question : find the minimum rating of fuse that can be safely used on a line on which two 1.1 KW electric geysers are to run simultaneously.
The supply voltage is 220 V.
Answer :  Power P = VI. As the two geysers have power rating 1.1 kW or 1100 W and are connected in parallel, each geyser draws a current I = P/V = 1100/220 A = 5 A.

Question : Should the resistance of an ammeter be low or high? Give reason. 
Answer :  An ideal ammeter is one which has zero resistance. But that is not possible. Therefore, the resistance of an ammeter should be as close to zero as possible. If it is non-zero and
substantial, it will affect the current flowing through the circuit. This is because an ammeter is connected in series in the circuit for the measurement of electric current.

Question : The electric power consumed by a device may be calculated by either of the two expressions P = I2R or P = V2/R. The frist expression indicates that it is directly proportional to R whereas the second expression indicates inverse proportionality. How can the seemingly different dependence of P on R in these expressions be explained?
Answer :  The expression P = I2R is used for calculating electric power when only current I and resistance R are known, whereas P = V2/R is used for calculating power when voltage V and
resistance R are known.

Short Answers :

Question : Define the unit of electric current. 

Answer :  The SI unit of electric current is ampere (A). When 1 coulomb of electric charge flows through any cross-section of a conductor in 1 second, the electric current flowing through it is said to be 1 ampere.

 

 

Question : How many joule are there in 1 kWh? 

Answer : 

1kWh = 1,000 Ω × 1h = 1,000 J/S × 3600S

1kWh = 3,600,000 J = 3.6 ×106 J 

 

Question. The given figure represents V-I graph for a series combination and for a parallel combination of two resistors. Which of the two, A or B, represents the series combination. Give a reason for your answer. 

Answer: Since, the straight line A is steeper than B, so the straight line A represents a greater resistance. The equivalent resistance in a series combination is greater than in parallel combination. Hence, A represents series combination.

Question. You are given three resistors of resistance 2 W, 4 W and 6 W. Show by a diagram, how you can get a 3 W resistance with the help of these resistors.
Answer: Connect 2 W and 4 W in series then this combination is connected with the parallel 6 W resistance. By this arrangement we can get resistance 3 W as equivalent resistance.

Question. Derive the relation 1/R = 1/R1 + 1/R2 + 1/R3 when resistors are joined in parallel.
Answer: In parallel combination of three resistance R1, R2 and R3, the current in each of the resistance is different. If I is the current drawn from the cell then it is divided into branches as I1, I2 and I3. Thus,
I = I1 + I2 + I3
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
I1 = V /R1 , I2 = V/R2 , I3 = V/R3
If R is the equivalent resistance then, 

V/R = V/R1 + V/R2 + V/R3 
and 1/R = 1/R1 + 1/R2 + 1/R3 .

Question. How many 10 W resistors are required to get a 25 W resistor?
Answer: Four 10 W resistors are required and they are connected as shown. 

Question. You are given fifty 5 W resistors. What is (a) smallest and (b) largest resistance can be obtained by using these?
Answer: (a) To get the smallest resistance, all the 5 W resistors must be connected in parallel.
Smallest resistance = 5/50 = 0.1 W
(b) To get the largest resistance, the 5 W resistors must be
connected in series.
Largest resistance = 5 × 50 = 250 W.

Question. Why are fairy decorative lights always connected in parallel?
Answer: When the fairy lights are connected in series the resistance offered will be greater and brightness of the bulbs will be affected. But in parallel connection all the bulbs will glow with same intensity and if any bulb gets fused the other bulbs will continue to glow.

Question. Resistance of a conductor of length 80 cm is 4.0 W. Calculate the resistance of a similar conductor of length 400 cm.
Answer: Here, l1 = 80 cm, R1 = 4.0 W, l2 = 400 cm, R2 = ?
∴ R1/R2 = 1₁/1₂
or R2 = R1. 1/2 , 1/1
= 4.0 × 400/80 = 20.0 W.

Question. Out of 60 W and 40 W lamps, which one has a higher electrical resistance when in use.
Answer: Power (P) = V2/R .
From the above formula, P is inversely proportional to R (resistance) as voltage remaining the same. Hence, 40 We lamp has high resistance.

Question. Write the formula for current ‘I’ flowing through a conductor if ‘n’ electrons flow through the cross-section of a conductor in time ‘t’.
Answer: If ‘n’ electrons pass through the cross-section of a conductor in time ‘t’, the total charge ‘Q’ passing through the conductor is :
Q = ne (e is the charge on an electron = 1.6 ×10–19 C) The current ‘I’ in the conductor is :
l = Q/t = ne/t

Question. What is commercial unit of electrical energy ? Convert it into joules.
Answer: The commercial unit of electrical energy is kWh.
1 kWh = 1000 W × 1 hour
= 1000 j/s x 60 x 60s
= 3.6 × 106 J

Question : Heat is generated continuously in an electric heater, but the temperature of its element becomes constant after some time. Why? 

Answer :   When the temperature of the heater becomes greater than the temperature of the surroundings, some of the heat is lost to the surroundings in the form of thermal radiations and through convection. After some time,the rate at which heat is being produced becomes equal to the rate of which heat is lost. Hence, the temperature of the element becomes constant. This happens because an equilibrium or steady state is reached.

Question : Define 1 volt. Express it in terms of SI unit of work and charge calculate the amount of energy consumed in carrying a charge of 1 coulomb through a battery of 3 V.
Answer :  When 1 joule of work is done in carrying 1 coulomb of charge, from infinity to a point in the electric field, then potential at that point is called 1 volt. Potential difference between two points is

 

Question :  A heater coil is rated 100 W, 200 V. It is cut into two identical parts. Both parts are connected together in parallel to the same source of 200 V. Calculate the energy liberated per second in the new combination. 

Answer : Resistance of heater coil,

As the coil is cut into two identical parts, resistance of each part = 400/2  = 200 Ω

When connected in parallel, the net resistance is given by

 

Energy liberated per second in the new combination is 400J 

Question : Why is parallel arrangement used in domestic wiring? 
Answer :  Parallel arrangement is used in domestic wiring due to the following reasons:
• Each device will have the same voltage which is equal to the voltage of the supply.
• If two or more devices are used at the same time, then each appliance will be able to draw the required current.
• If one of the devices fails, then the other keeps working.

Answer :

(i) What happens to the glow of the other two bulbs when the bulb B1 gets fused?
(ii) What happens to the reading of A₁, A₂, A₃ and A when the bulb gets fused?
(iii) How much power is dissipated in the circuit when all the three bulbs glow together?
(i) Since B₁, B₂ and B₃ are in parallel, the potential difference across each of them will remain same. So when the bulb B₁ gets fused,B₁ ,B₂ and B₃ have the same potential and continues with the same energy dissipated per second, i.e. they will glow continuously as they were glowing before.
(ii) Resistance of the parallel combination when all the three bulbs are glowing

Since resistance of each arm is same and p.d. is also same, current divides them equally. So 1A current will pass through each bulb Bl and By 
Therefore, ammeter A₁ and A₃ reads l A current while A₂ will read zero and ammeter A read 2 A current.
(iii) In parallel, total power consumed
Peq=P₁ +P₂ +P₃
So, when all the three bulbs glow together

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

Answer :

Question : A 9Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.
Answer :

Question : a.Two resistors R₁ and R₂ may form (i) a series combination or (ii) a parallel combination, and the combination may be connected to a battery of 6 volts. In which combination, will the potential difference across R₁ and across R₂ be the same and in which combination, will the current through R₁ and through R₂ be the same?
b. For the circuit shown in this diagram, calculate

(i) the resultant resistance.
(ii) the total current.
(iii) the voltage across 7 Ω resistor.
Answer. a. Potential difference across R₁ and R₂ is same in parallel combination of R₁ and R₂ and the current through R₁ and R₂ will be same when they are connected in series.