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Electricity

Madhya Pradesh Board · Class 10 · Science

NCERT Solutions for Electricity — Madhya Pradesh Board Class 10 Science.

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40 Questions Solved · 8 Sections

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Q U E S T I O N S

1What does an electric circuit mean?Show solution
An electric circuit means a continuous and closed path for the flow of electric charges (current). If the path is broken anywhere, current stops flowing.

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2Define the unit of current.Show solution
The SI unit of electric current is ampere (A). 1 ampere is the current when 1 coulomb of charge flows in 1 second; that is, 1 A=1 C s−11\,\text{A}=1\,\text{C s}^{-1}.

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3Calculate the number of electrons constituting one coulomb of charge.Show solution
Since 1 coulomb is equivalent to the charge contained in nearly **6×10186\times10^{18} electrons**, the number of electrons constituting 1 C of charge is about that value.

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Q U E S T I O N S

1Name a device that helps to maintain a potential difference across a conductor.Show solution
A battery (or a cell) produces and maintains the potential difference across a conductor, which sets charges in motion.

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2What is meant by saying that the potential difference between two points is 1 V?Show solution
Potential difference is V=W/QV=W/Q. So if the potential difference between two points is 1 V, then 1 J of work is required to move 1 C of charge from one point to the other.

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3How much energy is given to each coulomb of charge passing through a 6 V battery?Show solution
A potential difference of 6 V means 6 J of energy is given to each 1 C of charge, because 1 V=1 J C−11\,\text{V}=1\,\text{J C}^{-1}. So energy per coulomb = 6 J.

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Q U E S T I O N S

1On what factors does the resistance of a conductor depend?Show solution
The resistance of a conductor depends on:
- its length,
- its area of cross-section,
- the nature of its material.

The chapter also states that resistance and resistivity vary with temperature.

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2Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?Show solution
Current flows more easily through a thick wire of the same material. A thicker wire has a larger area of cross-section, so its resistance is less. Since R∝1AR \propto \frac{1}{A}, the thicker wire offers less opposition to current.

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3Let 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?Show solution
By Ohm's law, I=VRI=\frac{V}{R}. If resistance remains constant and the potential difference is halved, the current also becomes half of its former value.

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4Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?Show solution
Coils of electric toasters and irons are made of an alloy rather than a pure metal because alloys:
- have higher resistivity,
- can withstand high temperatures without oxidising/burning readily,
- are suitable for producing heat efficiently.

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5Use the data in Table 11.2 to answer the following -Show solution
From Table 11.2:
- Iron has resistivity 10.0×10−8 Ω m10.0\times10^{-8}\,\Omega\,\text{m}
- Mercury has resistivity 94.0×10−8 Ω m94.0\times10^{-8}\,\Omega\,\text{m}

A lower resistivity means a better conductor. So:
- Iron is the better conductor among the two.
- The best conductor in the table is silver because it has the lowest resistivity (1.60×10−8 Ω m1.60\times10^{-8}\,\Omega\,\text{m}).

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Q U E S T I O N S

1Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω\Omega resistor, an 8 Ω\Omega resistor, and a 12 Ω\Omega resistor, and a plug key, all connected in series.Show solution
The circuit should have the three cells in series with the 5 Ω, 8 Ω, and 12 Ω resistors also in series, along with a plug key in the same path.

A schematic form is:

Battery of 3 cells (2 V each) → key → 5 Ω resistor → 8 Ω resistor → 12 Ω resistor → back to battery

All components are connected end to end in one loop.

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2Redraw 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\Omega resistor. What would be the readings in the ammeter and the voltmeter?Show solution
First find the total resistance in series:
R=5+8+12=25 Ω R=5+8+12=25\,\Omega
The battery has three cells of 2 V each, so total voltage:
V=2+2+2=6 V V=2+2+2=6\,\text{V}
Current in the circuit:
I=VR=625=0.24 A I=\frac{V}{R}=\frac{6}{25}=0.24\,\text{A}
So the ammeter reading is 0.24 A.

Potential difference across the 12 Ω resistor:
V12=IR=0.24×12=2.88 V V_{12}=IR=0.24\times12=2.88\,\text{V}
So the voltmeter reading is 2.88 V.

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Q U E S T I O N S

1Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10⁶ Ω, (b) 1 Ω and 10³ Ω, and 10⁶ Ω.Show solution
For resistors in parallel, equivalent resistance is always less than the smallest resistance.

### (a) 1 Ω1\,\Omega and 106 Ω10^6\,\Omega
Since one resistor is very large, the equivalent resistance is **almost 1 Ω1\,\Omega**.

### (b) 1 Ω1\,\Omega, 103 Ω10^3\,\Omega, and 106 Ω10^6\,\Omega
Again, the 1 Ω resistor dominates, so the equivalent resistance is slightly less than 1 Ω, but approximately 1 Ω.

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2An 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?Show solution
In parallel, each appliance gets the same voltage, and currents add.

Current drawn by each appliance:
I1=220100=2.2 A I_1=\frac{220}{100}=2.2\,\text{A}
I2=22050=4.4 A I_2=\frac{220}{50}=4.4\,\text{A}
I3=220500=0.44 A I_3=\frac{220}{500}=0.44\,\text{A}
Total current:
I=2.2+4.4+0.44=7.04 A I=2.2+4.4+0.44=7.04\,\text{A}
If the iron takes the same current as all three together, then at 220 V:
R=VI=2207.04=31.25 Ω R=\frac{V}{I}=\frac{220}{7.04}=31.25\,\Omega
So the electric iron has resistance 31.25 Ω and current 7.04 A.

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3What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?Show solution
Advantages of parallel connection:
- Each device gets the same potential difference as the battery.
- Devices can be operated independently.
- If one device fails, the others continue working.
- The current is divided according to the resistance of each device, so each gadget can take the current it needs.

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4How 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?Show solution
We need to use 2 Ω, 3 Ω, and 6 Ω resistors.

### (a) Total resistance 4 Ω
Connect 3 Ω and 6 Ω in parallel:
1R=13+16=12⇒R=2 Ω \frac{1}{R}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2} \Rightarrow R=2\,\Omega
Then connect this 2 Ω in series with the 2 Ω resistor:
Rtotal=2+2=4 Ω R_{total}=2+2=4\,\Omega

### (b) Total resistance 1 Ω
Connect all three in parallel:
1R=12+13+16=1⇒R=1 Ω \frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1 \Rightarrow R=1\,\Omega

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5What 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?Show solution
- The highest resistance is obtained when all resistors are connected in series:
Rmax=4+8+12+24=48 Ω R_{max}=4+8+12+24=48\,\Omega
- The lowest resistance is obtained when all are connected in parallel:
1Rmin=14+18+112+124=6+3+2+124=1224=12 \frac{1}{R_{min}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24} =\frac{6+3+2+1}{24}=\frac{12}{24}=\frac{1}{2}
So,
Rmin=2 Ω R_{min}=2\,\Omega

Therefore, the answers are 48 Ω and 2 Ω.

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Q U E S T I O N S

1Why does the cord of an electric heater not glow while the heating element does?Show solution
The cord does not glow because it has very low resistance and is made thicker so that it does not get heated much. The heating element has high resistance and therefore produces a large amount of heat when current flows through it.

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2Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.Show solution
Use Joule’s law:
H=VQ H=VQ
Given V=50 VV=50\,\text{V} and Q=96000 CQ=96000\,\text{C},
H=50×96000=4800000 J H=50\times96000=4800000\,\text{J}
So the heat generated is **4.8×1064.8\times10^6 J**.

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3An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Q U E S T I O N S

1What determines the rate at which energy is delivered by a current?
2An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

E X E R C I S E S

1A piece of wire of resistance RR is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′R', then the ratio R/R′R/R' is –
2Which of the following terms does not represent electrical power in a circuit?
3An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
4Two 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 –
5How is a voltmeter connected in the circuit to measure the potential difference between two points?
6A copper wire has diameter 0.5 mm and resistivity of 1.6×10−8Ω1.6 \times 10^{-8} \Omega m. What will be the length of this wire to make its resistance 10 Ω\Omega? How much does the resistance change if the diameter is doubled?
7The values of current II flowing in a given resistor for the corresponding values of potential difference VV across the resistor are given below –
8When a 12 V 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.
9A battery of 9 V is connected in series with resistors of 0.2 Ω\Omega, 0.3 Ω\Omega, 0.4 Ω\Omega, 0.5 Ω\Omega and 12 Ω\Omega, respectively. How much current would flow through the 12 Ω\Omega resistor?
10How many 176 Ω\Omega resistors (in parallel) are required to carry 5 A on a 220 V line?
11Show how you would connect three resistors, each of resistance 6 Ω\Omega, so that the combination has a resistance of (i) 9 Ω\Omega, (ii) 4 Ω\Omega.
12Several 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?
13A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω\Omega resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
14Compare the power used in the 2 Ω\Omega resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω\Omega and 2 Ω\Omega resistors, and (ii) a 4 V battery in parallel with 12 Ω\Omega and 2 Ω\Omega resistors.
15Two 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?
16Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
17An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

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Frequently Asked Questions

What are the important topics in Electricity for Madhya Pradesh Board Class 10 Science?
Key topics in Electricity include Electricity Chapter Concept Overview, Electricity Chapter Overview, Electricity Chapter Concept Map. These are the concepts Madhya Pradesh Board Class 10 examiners draw on most — study them first, then practise related questions.
How to score full marks in Electricity — Madhya Pradesh Board Class 10 Science?
Understand the core concepts first, then work through the 178 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Electricity Class 10 Science?
This page has free step-by-step NCERT Solutions for every exercise question in Electricity (Madhya Pradesh Board Class 10 Science) — written the way examiners award marks: given, formula, working, answer.

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