{"id":64611,"date":"2020-12-15T13:14:05","date_gmt":"2020-12-15T13:14:05","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=64611"},"modified":"2023-09-19T03:58:00","modified_gmt":"2023-09-19T03:58:00","slug":"ncert-solutions-for-class-10-science-chapter-12-electricity","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10-science-chapter-12-electricity\/","title":{"rendered":"NCERT Solutions for Class 10 Science: Chapter 12 – Electricity"},"content":{"rendered":"\n

Class 10: Science Chapter 12 solutions. Complete Class 10 Science Chapter 12 Notes.<\/p>\n\n\n\n

NCERT Solutions for Class 10 Science: Chapter 12 – Electricity<\/strong><\/h2>\n\n\n\n

NCERT 10th Science Chapter 12, class 10 Science Chapter 12 solutions<\/p>\n\n\n\n

Page No: 200<\/strong><\/p>\n\n\n\n

1. What does an electric circuit mean?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

A continuous and closed path of an electric current is called an electric circuit. An electric circuit consists of electric devices, source of electricity and wires that are connected with the help of a switch.<\/p>\n\n\n\n

2. Define the unit of current.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The unit of electric current is ampere (A). 1 A is defined as the flow of 1 C of charge through a wire in 1 s.<\/p>\n\n\n\n

3. Calculate the number of electrons constituting one coulomb of charge.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

One electron possesses a charge of 1.6 \u00d710-19<\/sup>C, i.e., 1.6 \u00d710-19<\/sup>C of charge is contained in 1 electron.<\/p>\n\n\n\n

\u2234 1 C of charge is contained in 1\/1.6 x 10-19<\/sup> = 6.25 x 1018<\/sup> = 6 x 1018<\/sup><\/p>\n\n\n\n

Therefore, 6 x 1018<\/sup> electrons constitute one coulomb of charge.<\/p>\n\n\n\n

Page No: 202<\/strong><\/p>\n\n\n\n

1. Name a device that helps to maintain a potential difference across a conductor.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Any source of electricity like battery, cell, power supply, etc. helps to maintain a potential difference across a conductor.<\/p>\n\n\n\n

2. What is meant by saying that the potential difference between two points is 1 V<\/em>?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

If 1 J of work is required to move a charge of amount 1 C from one point to another, then it is said that the potential difference between the two points is 1 V<\/em>.<\/p>\n\n\n\n

3. How much energy is given to each coulomb of charge passing through a 6 V<\/em> battery?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The energy given to each coulomb of charge is equal to the amount of work which is done in moving it.
Now we know that,
Potential difference = Work Done\/Charge
\u2234 Work done = Potential difference \u00d7 charge
Where, Charge = 1 C and Potential difference = 6 V<\/em>
\u2234 Work done = 6\u00d71
= 6 Joule.<\/p>\n\n\n\n

Page No: 209
<\/strong>
1. On what factors does the resistance of a conductor depend?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The resistance of a conductor depends upon the following factors:
\u2192 Length of the conductor
\u2192 Cross-sectional area of the conductor
\u2192 Material of the conductor
\u2192 Temperature of the conductor<\/p>\n\n\n\n

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?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The current will flow more easily through thick wire. It is because the resistance of a
conductor is inversely proportional to its area of cross – section. If thicker the wire, less is resistance and hence more easily the current flows.<\/p>\n\n\n\n

NCERT 10th Science Chapter 12, class 10 Science Chapter 12 solutions<\/p>\n\n\n\n

3. Let the resistance of an electrical component remains 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?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

According to Ohm\u2019s law
V<\/em> = IR<\/em>
\u21d2 I<\/em>=V\/R<\/em> …                   (1)
Now Potential difference is decreased to half
\u2234 New potential difference V\u02b9<\/em>=V<\/em>\/2
Resistance remains constant
So the new current I\u02b9<\/em> = V\u02b9\/R<\/em>
= (V<\/em>\/2)\/R<\/em>
= (1\/2) (V<\/em>\/R<\/em>)
= (1\/2) I<\/em> = I<\/em>\/2<\/p>\n\n\n\n

Therefore, the amount of current flowing through the electrical component is reduced by half.<\/p>\n\n\n\n

4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.<\/p>\n\n\n\n

5. Use the data in Table 12.2 to answer the following –
Table 12.2 Electrical resistivity of some substances at 20\u00b0C<\/p>\n\n\n\n

\u2212<\/td>Material<\/strong><\/td>Resistivity (\u03a9 m)<\/strong><\/td><\/tr>
Conductors<\/td>Silver<\/td>1.60 \u00d7 10\u22128<\/sup><\/td><\/tr>
Copper<\/td>1.62 \u00d7 10\u22128<\/sup><\/td><\/tr>
Aluminium<\/td>2.63 \u00d7 10\u22128<\/sup><\/td><\/tr>
Tungsten<\/td>5.20 \u00d7 10\u22128<\/sup><\/td><\/tr>
Nickel<\/td>6.84 \u00d7 10\u22128<\/sup><\/td><\/tr>
Iron<\/td>10.0 \u00d7 10\u22128<\/sup><\/td><\/tr>
Chromium<\/td>12.9 \u00d7 10\u22128<\/sup><\/td><\/tr>
Mercury<\/td>94.0 \u00d7 10\u22128<\/sup><\/td><\/tr>
Manganese<\/td>1.84 \u00d7 10\u22126<\/sup><\/td><\/tr>
Constantan
(alloy of Cu and Ni)<\/td>		49 \u00d7 10\u22126<\/sup><\/td><\/tr>
Alloys<\/td>Manganin
(alloy of Cu, Mn and Ni)<\/td>		44 \u00d7 10\u22126<\/sup><\/td><\/tr>
Nichrome
(alloy of Ni, Cr, Mn and Fe)<\/td>		100 \u00d7 10\u22126<\/sup><\/td><\/tr>
Glass<\/td>1010<\/sup> \u2212 1014<\/sup><\/td><\/tr>
Insulators<\/td>Hard rubber<\/td>1013<\/sup> \u2212 1016<\/sup><\/td><\/tr>
Ebonite<\/td>1015<\/sup> \u2212 1017<\/sup><\/td><\/tr>
Diamond<\/td>1012<\/sup> \u2212 1013<\/sup><\/td><\/tr>
Paper (dry)<\/td>1012<\/sup><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) Resistivity of iron = 10.0 x 10-8<\/sup> \u03a9
Resistivity of mercury = 94.0 x 10-8<\/sup> \u03a9
Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.<\/p>\n\n\n\n

(b) It can be observed from Table 12.2 that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.<\/p>\n\n\n\n

NCERT 10th Science Chapter 12, class 10 Science Chapter 12 solutions<\/p>\n\n\n\n

Page No: 213<\/strong><\/p>\n\n\n\n

1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V<\/em> each, a 5 \u03a9 resistor, an 8 \u03a9 resistor, and a 12 \u03a9 resistor, and a plug key, all connected in series.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Three cells of potential 2 V<\/em>, each connected in series therefore the potential difference of the battery will be 2 V<\/em> + 2 V<\/em> + 2 V <\/em>= 6V<\/em>. The following circuit diagram shows three resistors of resistances 5 \u03a9, 8 \u03a9 and 12 \u03a9 respectively connected in series and a battery of potential 6 V<\/em> and a plug key which is closed means the current is flowing in the circuit.<\/p>\n\n\n\n

\"NCERT<\/a><\/figure>\n\n\n\n

2. Redraw the circuit of question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure potential difference across the 12 \u03a9 resistor. What would be the readings in the ammeter and the voltmeter?<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, as shown in the following figure.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

The resistances are connected in series.
Ohm\u2019s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm\u2019s law,
V <\/em>= IR<\/em>,
Where,
Potential difference, V <\/em>= 6 V
Current flowing through the circuit\/resistors = I<\/em>
Resistance of the circuit, R<\/em> = 5 + 8 + 12 = 25\u03a9
I<\/em> = V\/R<\/em> = 6\/25 = 0.24 A
Potential difference across 12 \u03a9 resistor = V1<\/sub><\/em>
Current flowing through the 12 \u03a9 resistor, I <\/em>= 0.24 A
Therefore, using Ohm\u2019s law, we obtain
V1<\/sub><\/em> = IR<\/em> = 0.24 x 12 = 2.88 V<\/em>
Therefore, the reading of the ammeter will be 0.24 A.
The reading of the voltmeter will be 2.88 V<\/em>.<\/p>\n\n\n\n

NCERT 10th Science Chapter 12, class 10 Science Chapter 12 solutions<\/p>\n\n\n\n

Page No: 216<\/strong><\/p>\n\n\n\n

1. Judge the equivalent resistance when the following are connected in parallel \u2212 (a) 1 \u03a9 and 106<\/sup>\u03a9, (b) 1 \u03a9 and 103<\/sup>\u03a9 and 106<\/sup>\u03a9.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) When 1 \u03a9 and 106<\/sup> \u03a9 are connected in parallel:
Let R<\/em> be the equivalent resistance.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Therefore, equivalent resistance \u2248 1 \u03a9<\/p>\n\n\n\n

(b) When 1\u03a9, 103 \u03a9 and 106 \u03a9 are connected in parallel:<\/p>\n\n\n\n

Let R<\/em> be the equivalent resistance.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Therefore, equivalent resistance = 0.999 \u03a9<\/p>\n\n\n\n

2. An electric lamp of 100 \u03a9, a toaster of resistance 50 \u03a9, and a water filter of resistance 500 \u03a9 are connected in parallel to a 220 V<\/em> 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?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Resistance of electric lamp, R1<\/sub><\/em> = 100 \u03a9
Resistance of toaster, R2<\/sub><\/em> = 50 \u03a9
Resistance of water filter, R3<\/sub><\/em> = 500 \u03a9
Potential difference of the source, V<\/em> = 220 V<\/em>
These are connected in parallel, as shown in the following figure.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Let R<\/em> be the equivalent resistance of the circuit.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.
The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.<\/p>\n\n\n\n

4. How can three resistors of resistances 2 \u03a9, 3 \u03a9 and 6 \u03a9 be connected to give a total resistance of (a) 4 \u03a9, (b) 1 \u03a9?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

There are three resistors of resistances 2 \u03a9, 3 \u03a9, and 6 \u03a9 respectively.<\/p>\n\n\n\n

(a) The following circuit diagram shows the connection of the three resistors.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Here, 6 \u03a9 and 3 \u03a9 resistors are connected in parallel.
Therefore, their equivalent resistance will be given by<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

This equivalent resistor of resistance 2 \u03a9 is connected to a 2 \u03a9 resistor in series.
Therefore, the equivalent resistance of the circuit = 2 \u03a9 + 2 \u03a9 = 4 \u03a9
Hence the total resistance of the circuit is 4 \u03a9.<\/p>\n\n\n\n

(b) The following circuit diagram shows the connection of the three resistors.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

All the resistors are connected in series. Therefore, their equivalent resistance will be given as<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Therefore, the total resistance of the circuit is 1 \u03a9.<\/p>\n\n\n\n

5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 \u03a9, 8 \u03a9, 12 \u03a9, 24 \u03a9?<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

There are four coils of resistances 4 \u03a9, 8 \u03a9, 12 \u03a9 and 24 \u03a9 respectively.<\/p>\n\n\n\n

(a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24 = 48 \u03a9

(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Therefore, 2 \u03a9 is the lowest total resistance.<\/p>\n\n\n\n

NCERT 10th Science Chapter 12, class 10 Science Chapter 12 solutions<\/p>\n\n\n\n

Page No: 218<\/strong><\/p>\n\n\n\n

1. Why does the cord of an electric heater not glow while the heating element does?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminium is very law so it does not glow.<\/p>\n\n\n\n

2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V<\/em>.<\/strong><\/p>\n\n\n\n

Given Charge, Q = 96000C
Time, t<\/em>= 1hr = 60 x 60= 3600s
Potential difference, V<\/em>= 50volts
Now we know that H<\/em>= VIt
So we have to calculate I<\/em> first
As I<\/em>= Q\/t<\/em>
\u2234 I = 96000\/3600 = 80\/3 A<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Therefore, the heat generated is 4.8 x 106<\/sup> J.<\/p>\n\n\n\n

3. An electric iron of resistance 20 \u03a9 takes a current of 5 A. Calculate the heat developed in 30 s.
Answer<\/strong><\/p>\n\n\n\n

The amount of heat (H<\/em>) produced is given by the joule’s law of heating asH<\/em>= Vlt
Where,
Current, I<\/em> = 5 A
Time, t<\/em> = 30 s
Voltage, V<\/em> = Current x Resistance = 5 x 20 = 100 V<\/em><\/p>\n\n\n\n

H<\/em>= 100 x 5 x 30 = 1.5 x 104<\/sup> J.<\/p>\n\n\n\n

Therefore, the amount of heat developed in the electric iron is 1.5 x 104<\/sup> J.<\/p>\n\n\n\n

NCERT 10th Science Chapter 12, class 10 Science Chapter 12 solutions<\/p>\n\n\n\n

Page No: 220<\/strong><\/p>\n\n\n\n

1. What determines the rate at which energy is delivered by a current?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.<\/p>\n\n\n\n

2. An electric motor takes 5 A from a 220 V<\/em> line. Determine the power of the motor and the energy consumed in 2 h.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Power (P<\/em>) is given by the expression,P<\/em> = VI<\/em>
Where,
Voltage,V<\/em> = 220 V
Current, I<\/em> = 5 A
P<\/em>= 220 x 5 = 1100 W
Energy consumed by the motor = Pt
Where,
Time, t<\/em> = 2 h = 2 x 60 x 60 = 7200 s
\u2234 P<\/em> = 1100 x 7200 = 7.92 x 106<\/sup> J
Therefore, power of the motor = 1100 W
Energy consumed by the motor = 7.92 x 106<\/sup> J<\/p>\n\n\n\n

NCERT Solutions for Class 10 Science Chapter 12 – Electricity Exercise<\/h4>\n\n\n\n

Exercise<\/strong><\/h3>\n\n\n\n

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<\/strong>
(b) 1\/5
(c) 5
(d) 25
\u25ba (d) 25<\/p>\n\n\n\n

2. Which of the following terms does not represent electrical power in a circuit?
<\/strong>(a) I2<\/sup>R<\/em>
(b) IR2<\/sup><\/em>
(c) VI<\/em>
(d) V2<\/sup>\/R<\/em>
\u25ba (b) IR2<\/sup><\/em>
3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
<\/strong>(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
\u25ba (d) 25 W<\/p>\n\n\n\n

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 –<\/strong><\/p>\n\n\n\n

(a) 1:2<\/p>\n\n\n\n

(b) 2:1<\/p>\n\n\n\n

(c) 1:4<\/p>\n\n\n\n

(d) 4:1<\/p>\n\n\n\n

\u25ba (c) 1:4<\/p>\n\n\n\n

5. How is a voltmeter connected in the circuit to measure the potential difference between two points?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.<\/p>\n\n\n\n

NCERT 10th Science Chapter 12, class 10 Science Chapter 12 solutions<\/p>\n\n\n\n

6. A copper wire has diameter 0.5 mm and resistivity of 1.6 \u00d7 10\u22128<\/sup> \u03a9 m. What will be the length of this wire to make its resistance 10 \u03a9? How much does the resistance change if the diameter is doubled?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Area of cross-section of the wire, A =\u03c0 (d\/2) 2
Diameter= 0.5 mm = 0.0005 m
Resistance, R = 10 \u03a9
We know that<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 \u03a9.<\/p>\n\n\n\n

7. The values of current I<\/em> flowing in a given resistor for the corresponding values of potential difference V<\/em> across the resistor are given below \u2212<\/strong><\/p>\n\n\n\n

I <\/em>(amperes )<\/td>0.5<\/td>1.0<\/td>2.0<\/td>3.0<\/td>4.0<\/td><\/tr>
V<\/em> (volts)<\/td>1.6<\/td>3.4<\/td>6.7<\/td>10.2<\/td>13.2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Plot a graph between V<\/em> and I<\/em> and calculate the resistance of that resistor.<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The plot between voltage and current is called IV<\/em> characteristic. The voltage is plotted on x<\/em>-axis and current is plotted on y<\/em>-axis. The values of the current for different values of the voltage are shown in the given table.<\/p>\n\n\n\n

V<\/em> (volts)<\/td>1.6<\/td>3.4<\/td>6.7<\/td>10.2<\/td>13.2<\/td><\/tr>
I <\/em>(amperes )<\/td>0.5<\/td>1.0<\/td>2.0<\/td>3.0<\/td>4.0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

The IV<\/em> characteristic of the given resistor is plotted in the following figure.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

The slope of the line gives the value of resistance (R<\/em>) as,<\/p>\n\n\n\n

Slope = 1\/R <\/em>= BC\/AC = 2\/6.8<\/p>\n\n\n\n

R<\/em>= 6.8\/2 = 3.4 \u03a9<\/p>\n\n\n\n

Therefore, the resistance of the resistor is 3.4 \u03a9.<\/p>\n\n\n\n

8. When a 12 V<\/em> battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Resistance (R<\/em>) of a resistor is given by Ohm’s law as,V<\/em>= IR<\/em>
R<\/em>= V<\/em>\/I<\/em>
Where,
Potential difference, V<\/em>= 12 V<\/em>
Current in the circuit, I<\/em>= 2.5 mA = 2.5 x 10-3<\/sup> A<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Therefore, the resistance of the resistor is 4.8 k\u03a9<\/p>\n\n\n\n

9. A battery of 9 V<\/em> is connected in series with resistors of 0.2 \u03a9, 0.3 \u03a9, 0.4 \u03a9, 0.5 \u03a9 and 12 \u03a9, respectively. How much current would flow through the 12 \u03a9 resistor?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm\u2019s law as<\/p>\n\n\n\n

V<\/em>= IR<\/em><\/p>\n\n\n\n

I<\/em>= V<\/em>\/R<\/em><\/p>\n\n\n\n

Where,<\/p>\n\n\n\n

R<\/em> is the equivalent resistance of resistances 0.2 \u03a9, 0.3 \u03a9, 0.4 \u03a9, 0.5 \u03a9 and 12 \u03a9. These are connected in series. Hence, the sum of the resistances will give the value of R<\/em>.<\/p>\n\n\n\n

R<\/em>= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 \u03a9<\/p>\n\n\n\n

Potential difference, V<\/em>= 9 V<\/em><\/p>\n\n\n\n

I<\/em>= 9\/13.4 = 0.671 A
Therefore, the current that would flow through the 12 \u03a9 resistor is 0.671 A.<\/p>\n\n\n\n

10. How many 176 \u03a9 resistors (in parallel) are required to carry 5 A on a 220 V<\/em> line?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

For x<\/em> number of resistors of resistance 176 \u03a9, the equivalent resistance of the resistors connected in parallel is given by Ohm’s law asV<\/em>= IR<\/em>
R<\/em>= V<\/em>\/I<\/em>
Where,
Supply voltage, V<\/em>= 220 V<\/em>
Current, I<\/em> = 5 A
Equivalent resistance of the combination = R<\/em>,given as<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Therefore, four resistors of 176 \u03a9 are required to draw the given amount of current.<\/p>\n\n\n\n

11. Show how you would connect three resistors, each of resistance 6 \u03a9, so that the combination has a resistance of (i) 9 \u03a9, (ii) 4 \u03a9.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 \u03a9 + 6 \u03a9 + 6 \u03a9 = 18 \u03a9, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be 6\/2 = 3 \u03a9 is also not desired. Hence, we should either connect the two resistors in series or parallel.
(a) Two resistor in parallel<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Two 6 \u03a9 resistors are connected in parallel. Their equivalent resistance will be<\/p>\n\n\n\n

\"NCERT<\/a><\/figure>\n\n\n\n

The third 6 \u03a9 resistor is in series with 3 \u03a9. Hence, the equivalent resistance of the circuit is 6 \u03a9+ 3 \u03a9 = 9 \u03a9.
(b) Two resistor in series<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Two 6 \u03a9 resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 \u03a9.<\/p>\n\n\n\n

The third 6 \u03a9 resistor is in parallel with 12 \u03a9. Hence, equivalent resistance will be<\/p>\n\n\n\n

\"NCERT<\/a><\/figure>\n\n\n\n

Therefore, the total resistance is 4 \u03a9.<\/p>\n\n\n\n

12. Several electric bulbs designed to be used on a 220 V<\/em> 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<\/em> line if the maximum allowable current is 5 A?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Resistance R1<\/sub> of the bulb is given by the expression,
Supply voltage, V<\/em> = 220 V<\/em>
Maximum allowable current, I<\/em> = 5 A
Rating of an electric bulb P=10watts
Because R<\/em>=V2<\/sup><\/em>\/P<\/em><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

\u2234 Number of electric bulbs connected in parallel are 110.<\/p>\n\n\n\n

13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 \u03a9 resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Supply voltage, V<\/em>= 220 V<\/em>
Resistance of one coil, R<\/em>= 24 \u03a9<\/p>\n\n\n\n

(i) Coils are used separately<\/p>\n\n\n\n

According to Ohm’s law,
V<\/em>= I1<\/sub>R<\/em>1<\/em><\/sub>
Where,<\/p>\n\n\n\n

I1<\/sub> is the current flowing through the coil<\/p>\n\n\n\n

I1<\/sub> = V<\/em>\/R1<\/sub><\/em> = 220\/24 = 9.166 A
Therefore, 9.16 A current will flow through the coil when used separately.<\/p>\n\n\n\n

(ii) Coils are connected in series<\/p>\n\n\n\n

Total resistance, R2<\/sub><\/em> = 24 \u03a9 + 24 \u03a9 = 48 \u03a9<\/p>\n\n\n\n

According to Ohm’s law,V<\/em> = I2<\/sub>R2<\/sub><\/em>
Where,<\/p>\n\n\n\n

I<\/em>2<\/em> <\/sub>is the current flowing through the series circuit<\/p>\n\n\n\n

I<\/em>2<\/em> <\/sub>= V\/R2<\/sub><\/em> = 220\/48 = 4.58 A
Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel<\/p>\n\n\n\n

Total resistance, R3<\/sub><\/em> is given as =<\/p>\n\n\n\n

\"NCERT<\/a><\/figure>\n\n\n\n

According to Ohm’s law,<\/p>\n\n\n\n

V<\/em>= I3<\/sub>R3<\/sub><\/em>
Where,
I3<\/sub> is the current flowing through the circuit I3<\/sub> = V<\/em>\/R3<\/sub><\/em> = 220\/12 = 18.33 A
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.<\/p>\n\n\n\n

14. Compare the power used in the 2 \u03a9 resistor in each of the following circuits: (i) a 6 V battery in series with 1 \u03a9 and 2 \u03a9 resistors, and (ii) a 4 V battery in parallel with 12 \u03a9 and 2 \u03a9 resistors.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Potential difference, V<\/em> = 6 V
1 \u03a9 and 2 \u03a9 resistors are connected in series. Therefore, equivalent resistance of the circuit, R<\/em> = 1 + 2 = 3 \u03a9
According to Ohm\u2019s law,
V<\/em> = IR<\/em>
Where,
I<\/em> is the current through the circuit
I<\/em>= 6\/3 = 2 A
This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 \u03a9 resistor is 2 A. Power is given by the expression,
P<\/em>= (I<\/em>)2<\/sup>R<\/em> = (2)2<\/sup> x 2 = 8 W<\/p>\n\n\n\n

(ii) Potential difference, V<\/em> = 4 V
12 \u03a9 and 2 \u03a9 resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 \u03a9 resistor will be 4 V.
Power consumed by 2 \u03a9 resistor is given by
P<\/em>= V2<\/sup><\/em>\/R<\/em> = 42<\/sup>\/2 = 8 W
Therefore, the power used by 2 \u03a9 resistor is 8 W.<\/p>\n\n\n\n

15. Two lamps, one rated 100 W at 220 V<\/em>, and the other 60 W at 220 V<\/em>, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V<\/em>?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V<\/em>, because no division of voltage occurs in a parallel circuit.
Current drawn by the bulb of rating 100 W is given by,Power = Voltage x Current
Current =  Power\/Voltage = 60\/220 A
Hence, current drawn from the line = 100\/220 + 60\/220 = 0.727 A<\/p>\n\n\n\n

16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Energy consumed by an electrical appliance is given by the expression,H= Pt
Where,
Power of the appliance = P<\/em>
Time = t<\/em>
Energy consumed by a TV set of power 250 W in 1 h = 250 \u00d73600 = 9 \u00d7105<\/sup> J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 \u00d7600
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 \u00d7600
= 7.2\u00d7105<\/sup> J
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.<\/p>\n\n\n\n

17. An electric heater of resistance 8 \u03a9 draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Rate of heat produced by a device is given by the expression for power as, P<\/em>= I2<\/sup>R<\/em>
Where,
Resistance of the electric heater, R<\/em>= 8 \u03a9
Current drawn, I<\/em> = 15 A<\/p>\n\n\n\n

P<\/em>= (15)2<\/sup> x 8 = 1800 J\/s<\/p>\n\n\n\n

Therefore, heat is produced by the heater at the rate of 1800 J\/s.<\/p>\n\n\n\n

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 are copper and aluminium wires usually employed for electricity transmission?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) The melting point and of Tungsten is an alloy which has very high melting point and very high resistivity so does not burn easily at a high temperature.<\/p>\n\n\n\n

(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of metals which produces large amount of heat.<\/p>\n\n\n\n

(c) In series circuits voltage is divided. Each component of a series circuit receives a small voltage so the amount of current decreases and the device becomes hot and does not work properly. Hence, series arrangement is not used in domestic circuits.<\/p>\n\n\n\n

(d) Resistance (R<\/em>) of a wire is inversely proportional to its area of cross-section (A), i.e. when area of cross section increases the resistance decreases or vice versa.<\/p>\n\n\n\n

(e) Copper and aluminium are good conductors of electricity also they have low resistivity. So they are usually used for electricity transmission.<\/p>\n\n\n\n

NCERT 10th Science Chapter 12, class 10 Science Chapter 12 solutions<\/p>\n\n\n\n

NCERT Solutions for Class 10 Science: Chapter 12: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for Class 10 Science: Chapter 12 – Electricity<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for Class 10 Science: Chapter 12 – Electricity PDF<\/a><\/p>\n\n\n\n

Chapterwise NCERT Solutions for Class 10 Science :<\/strong><\/h2>\n\n\n\n