Electricity
NCERT Solutions
NCERT Exercises
1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is _____.
(a) 1/25
(b) 1/5
(c) 5
(d) 25
(d) 25
2. Which of the following does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2/R
(b) IR2
3. 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
(d) 25 W
4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be _____.
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
(c) 1:4
5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
The voltmeter should be connected in parallel between the two points.
6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Given
Diameter = 0.5 mm
Resistivity = 1.6 × 10–8 Ω m
R=\rho \frac{l}{A}…………..1
The area of cross section of wire A=\pi (\frac{diameter^2}{4})……………….2
By substituting the equation 2 in 1, we get
l=\frac{R \pi d^2}{4 \rho}
By substituting the values in the above equation we get;
A=\pi (\frac{10 \times 3.14 \times (0.5 \times 10 ^-3)^2}{4 \times 1.6 \times10^-8}) = 122.7m
R \alpha \frac{1}{d^2}
In series connection, there is no division of current, so that the current flowing across all the resistors is the same. Total resistance of resistors connected in series is given by: R = R_1+R_2+R_3+R_4+R_5
R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω
By using Ohm’s Law
I=\frac{V}{R}=\frac{9}{13.4}=0.67A
10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Let us consider the number of resistors required as ‘x.’
The equivalent resistance of the parallel combination of resistor R is given by
\frac{1}{R}=x \times \frac{1}{176}
R =\frac{176}{x}
By using Ohm’s law
R = V/I
\frac{176}{x} = \frac{V}{I}
x=\frac{176 \times 5}{220} = 4
The number of resistors required is 4.
11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
(i) In order to get 9 Ω resistance using resistors having resistance of 6 Ω, we need to connect two 6 Ω resistors in parallel, then connect that parallel connection in series with a 6 Ω. The effective resistance of resistors connected in parallel is \frac{1}{R}=\frac{1}{6}+\frac{1}{6} =\frac{1}{3}.
R = 3Ω
The effective resistance of resistors connected in series is 3Ω + 6Ω = 9Ω
(ii) In order to get 4 Ω resistance connect two resistors in series, their equivalent resistance is given by R = 6 Ω + 6 Ω = 12 Ω.
The third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is: R=\frac{1}{6}+\frac{1}{12}= 4 \varOmega
12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. 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?
Given V= 220 V
P = 10W
I = 5A
R=\frac{V}{P} = \frac{220^2}{10} = 4840 \varOmega
Let the number of bulbs be x.
According to Ohm’s law V = IR
V/P = 220/5 = 44Ω
Let x be the number of bulbs of resistance of 176Ω the equivalent resistance of the resistors connected in parallel is 44Ω.
\frac{1}{44} = \frac{1}{4840} + \frac{1}{4840} +…….x -times
\frac{1}{44} = \frac{x}{4840}
x=\frac{4840}{44}= 110
Hence, 110 lamps can be connected in parallel.
13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Case (i) When coils are used separately
V = 220 V
R = 24 Ω
I =V/R = 220/ 24
I = 9.17A
Case (ii) When coils connected in series
V = 220 V
R = 48 Ω
I =V/R = 220/ 48
I = 4.58A
Case (iii) When coils connected in parallel
Resistance \frac{1}{R} = \frac{1}{24}+ \frac{1}{24} = \frac{2}{24} = \frac{1}{12}
R= 12Ω
I =V/R = 220/ 12
I = 18.34A
14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Given
P = 6V
(i) Equivalent resistance of the circuit when 1Ω and 2Ω are connected in series
R = 1+2 = 3Ω
We know that
V = IR
I = V/R = 6/3 = 2A
In series connection the current in the circuit remains constant
So, Power P = I^2 R = 2^2 \times 2 = 8W
(ii) 1Ω and 2Ω resistors are connected in parallel.
I= \frac{V}{R} = \frac{6}{3}=2A
In parallel connection voltage in circuit remain constant. Therefore power is
P= \frac{V^2}{R} = \frac{4^2}{2}=8W
In both case power is same.
15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Current drawn by the bulb of rating 100 W: I = P/V = 100 W/220 V = 100/220 A
Current drawn by the bulb of rating 60 W: I = 60 W/220 V = 60/220 A
Therefore, the current drawn from the line is
\frac{100}{220} + \frac{60}{220} = 0.727A
16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
By using the equation H = Pt, the energy consumed by electrical appliances.
The energy consumed by electrical appliances: H = 250 W × 3600 seconds = 9 × 105 J
The energy consumed by a toaster of power rating 1200 W: H = 1200 W × 600 s = 7.2 × 105 J
So the energy consumed by the TV is greater than the toaster.
17. An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
The rate at which the heat develops in the heater can be calculated by using the formula,
P = I2 R
P = (15A) 2 × 8 Ω = 1800 watt
18. Explain the following.
a. Why is the tungsten used almost exclusively for filament of electric lamps?
b. Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
c. Why is the series arrangement not used for domestic circuits?
d. How does the resistance of a wire vary with its area of cross-section?
e. Why copper and aluminium wires are usually employed for electricity transmission?
(a) The resistivity and melting point of tungsten is very high. When heated, it doesn’t burn.
(b) Alloys show high resistivity. Due to its high resistivity it produces large amount of heat.
(c) In series arrangement, the voltage gets divided. So that each component in the circuit receives a small voltage which in turn decreases amount of current.
(d) Resistance is inversely proportional to the area of cross section. When the area of cross section increases the resistance decreases and vice versa.
(e) Because copper and aluminium are good conductors of electricity and have low resistivity.
Intext Exercises
Pg. No 200
1. What does an electric circuit mean?
Electric circuit is a closed and continuous path for the movement electric current and consist of electric components.
2. Define the unit of current.
The unit of current is ampere. It is the flow of one coulomb of charge per second.
3. Calculate the number of electrons constituting one coulomb of charge.
We can find the number of electrons by using the equation n=\frac{Q}{q_e}
qe is the charge of an electron = 1.6 × 10-19 C
Q = 1C
n=\frac{1}{1.6 \times 10^-19}=6.25 \times 10^18
So the number of electrons constituting one coulomb of charge is 6. 25 × 1018.
Pg. No. 202
1. Name a device that helps to maintain a potential difference across a conductor.
Cell or battery
2. What is meant by saying that the potential difference between two points is 1 V?
Potential difference between two points is 1 V, when 1 J of work is done to move a charge of 1 C from one point to another.
3. How much energy is given to each coulomb of charge passing through a 6 V battery?
We know the equation V = W/Q
Form the above equation we can find W = V × Q
W = 6V × 1C = 6 J
So 6 J of energy is given to each coulomb of charge passing through a 6 battery.
Pg. No. 209
1. On what factors does the resistance of a conductor depend?
- Length of the conductor
- Temperature of the conductor
- Cross-sectional area of the conductor
- Nature of the material of the conductor
2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Resistance is influenced by the cross section area of the conductor. Resistance is given by the equation, R = ρ l/A. It is clear that resistance is inversely proportional to cross section are of the conductor. If area decreases ( thin wire) resistance increases and if area increases, resistance decreases. So current flows more easily through a thick wire than a thin wire.
3. Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
We know from Ohm’s law I = V/R
From question we find that the new potential difference across the two ends of the component decreases to half of its former value. Lets denote this new Potential difference as V’ = V/2. Let I’ be the new amount of current and new resistance be R’ = R .
The change in the current can be found by substituting the value in Ohm’s law
i.e. I’=\frac{V’}{R’}= \frac{V/2}{R} =\frac{1}{2} \frac{V}{R}
So the current flowing the electrical component is reduced by half.
4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Alloys show high resistivity and got a high melting point. Due to its high resistivity it produces large amount of heat.
5. Use the data in the table given below and answer the following questions.
a. Which among iron and mercury is a better conductor?
b. Which material is the best conductor?
(a) Iron
(b) Silver
Pg. No. 213
1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
The total resistance of the circuit: 5 Ω + 8 Ω +12 Ω = 25 Ω.
Given V= 6 V,
I = V/R = 6/25 = 0.24A
Let the potential difference across the 12 Ω resistor be V1.
From the obtained current V1 can be calculated as follows:
V1 = 0.24A × 12 Ω = 2.88 V
Therefore, the ammeter reading will be 0.24 A and the voltmeter reading be 2.88 V.
Pg. No. 216
1. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 106 Ω, (b) 1 Ω, 103 Ω, and 106 Ω.
(a) When 1 Ω and 106 are connected in parallel, the equivalent resistance is given by
\frac{1}{R}=\frac{1}{1}+\frac{1}{10^6}
R=\frac{10^6}{1+10^6}=1 \Omega
(b) When 1 Ω, 103 Ω, and 106 Ω are connected in parallel, the equivalent resistance is given by
\frac{1}{R}=\frac{1}{1}+\frac{1}{10^6}+\frac{1}{10^3}
R=\frac{10^6 + 10^3 + 1}{10^6}=0.99\Omega
2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
The equivalent resistance of the resistors
\frac{1}{R}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}
=\frac{5+10+ 1}{500}=16/500\Omega
R=\frac{500}{16}\Omega
I=\frac{V}{R}
I=\frac{220}{500/16}
I = 7.04A
Current drawn across all the appliances connected in parallel are same, here it is I = 7.04 A
Hence current drawn by electric iron is also the same.
Resistance of the iron box: R=\frac{V}{I}
R=\frac{220V}{7.04A} = 31.25 \Omega
3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
- The total resistance in the circuit is decreased.
- If any of the device in parallel connection fails, other devices working willnot be affected.
4. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
(a) Connect 3 Ω and 6 Ω parallel.
Their effective resistance R = 2Ω
The equivalent resistor 2 Ω is in series with the 2 Ω resistor. Now the new resistance R = 2Ω +2Ω =4Ω
(b) Connect all resistors in parallel. So their effective resistance R = 1 Ω.
5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
(i) Equivalent resistance of the resistors connected in series: 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.
(ii) Equivalent resistance of resistor connected in parallel: \frac{1}{R}= \frac{1}{4} + \frac{1}{8}+\frac{1}{12}+\frac{1}{24}
R = 2Ω
So the lowest total resistance is 2 Ω.
Pg. No. 218
1. Why does the cord of an electric heater not glow while the heating element does?
The heating element of an electric heater is made of alloy which has a high resistance. When the current flows through it the alloy becomes red hot and glows red. At the same time cord is made of copper or aluminum which has low resistance. Hence the cord doesn’t glow.
2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Given V = 50V
t = 1 hour = 3600 seconds
Amount of current = Amount of charge/ Time flow of charge
I = 96000/ 3600
I= 26.66A
The heat generated can be found by using Joule’s law :
H = VIt = 50 × 26.66 × 3600 = 4.8 \times 10^6 J
4.8 \times 10^6 J is 4.8 \times 10^6 J
3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Given
R = 20 Ω
I = 5A
t = 30 s
By using the formula H = VIt we get the for heat developed.
H = 100 × 5 × 30 = 1.5 × 104 J
The amount of heat developed by the electric iron in 30 s is 1.5 × 104 J.
Pg. No. 220
1. What determines the rate at which energy is delivered by a current?
The rate at which energy is delivered by a current is the power of the appliance.
2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Given
V = 220 v
I = 5 A
T = 2 hours
The power of the motor can be calculated by the equation,
P = VI = 220 V × 5 A = 1100 W
The energy consumed by the motor can be calculated using the equation,
E = P × T = 1100 W × 7200 = 7.92 × 106 J.