In this article you will get CBSE Class 10 Science chapter 12, Electricity: NCERT Exemplar Problems and Solutions 

Multiple Choice Questions

Question 1. A cell, a resistor, a key and an ammeter are arranged as shown in the circuit diagrams of figure. The current recorded in the ammeter will be

(a) Maximum in (i)

(b) Maximum in (ii)

(c) Maximum in (iii)

(d) The same in all the cases

Answer. (d)

Explanation:

In series connections the order of elements in the circuit will not affect the amount of current flowing in the circuit.

Question 2. In the following circuits, heat produced in the resistor or combination of resistors connected to a 12 V battery will be

(a) Same in all the cases

(b) Maximum in case (i)

(c) Maximum in case (ii)

(d) Maximum in case (iii)

Answer. (d)

Explanation:

Question 3. Electrical resistivity of a given metallic wire depends upon

(a) Its length

(b) Its thickness

(c) Its shape

(d) Nature of the material

Answer. (d)

Explanation: The resistivity of a material is constant for a particular temperature at a constant temperature.

Resistivity of material does not depend on length, thickness and shape of the material. It only depends on the temperature.

Question 4. A current of 1 A is drawn by a filament of an electric bulb. Number of electron passing through a cross-section of the filament in 16 seconds would be roughly

(a) 10²⁰

(b) 10¹⁶

(c) 10¹⁸

(d) 10²³

Answer. (a)

Explanation:

Question 5. Identify the circuit in which the electrical components have been properly connected.

(a) (i)                          

(b) (ii)                         

(c) (iii)                        

(d) (iv)

Answer. (b)

Explanation:Essential conditions are necessary when electrical components are connected

• Voltmeter should be connected in parallel.
• Ammeter is always connected in series.
• Positive terminals of voltmeter and ammeter should be connected to positive terminal of the cell and their negative terminals should be joined to the negative terminal of the cell.

Thus, the above conditions are satisfied in case (ii).

Question 6. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

(a) 1/5 Ω

(b) 10 Ω

(c) 5 Ω

(d) 1 Ω

Answer. (d)

Explanation: The maximum resistance is obtained when resistors are connected in series combination.

Thus equivalent resistance obtained by connecting five resistors of resistance 1/5 Ω each, in series = (1/5 + 1/5 + 1/5 + 1/5) = 1 Ω

Question 7. What is the minimum resistance which can be made using five resistors each of 1/5 Ω?

(a) 1/5

(b) 1/25

(c) 1/10

(d) 25

Answer. (b)

Explanation:The minimum resistance is obtained when resistors are connected in parallel combination.

Thus equivalent resistance obtained by connecting five resistors of resistance 1/5 Ω each, parallel to each other =

Question 8. The proper representation of series combination of cells obtaining maximum potential is

(a) (i) 

(b) (ii)                         

(c) (iii)                        

(d) (iv)

Answer. (a)

Explanation:Maximum potential is obtained when cells are connected in series such that, negative terminal of the cell is connected to the positive terminal of the second cell and so on, as shown in the following diagram.

Question 9. Which of the following represents voltage?

Question 10. 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 samematerial has area of cross-section.

(a) A/2

(b) 3A/2

(c) 2A

(d) 3A

Answer. (c)

Explanation:

Question 11. A student carries out an experiment and plots the V ‒ I graph of three samples of nichrome wire with resistances R₁, R₂ and R₃ respectively as shown in figure. Which of the following is true?

(a) R₁ = R₂ = R3

(b) R₁ > R₂ > R₃

(c) R₃ > R₂ > R₁

(d) R₂ > R₃ > R₁

Answer. (c)

Explanation:

Slope of VI graph is proportional to 1/ Resistance.

It means when slope will be maximum, then resistance will be minimum.

From the figure, we can see that, slope of R₁ is maximum; hence its resistance will be minimum.

As, slope of R₃ is minimum  so, its resistance will be maximum.

Therefore, R₃ > R₂ > R₁

Question 12. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

(a) 100%

(b) 200%

(c) 300%

(d) 400%

Answer. (c)

Explanation: If I is current and R is resistance then,

Power, P = I²R

Power in first case, P₁ = I²R

100% increase in current means that current becomes 2I

Power in second case, P₂ = (2I)²R = 4I²R

Now, increase in dissipated power = P₂ – P₁ = 4I²R – I²R = 3I²R

Percentage increase in dissipated power = 3P₁/ P₁ × 100 = 300%

Question 13. The resistivity does not change if

(a) the material is changed

(b) the temperature is changed

(c) the shape of the resistor is changed

(d) both material and temperature are changed

Answer. (c)

Explanation: The resistivity depends on the nature of the material and the temperature.

It does not depends on dimension of resistor.

Question 14. In an electrical circuit three incandescent bulbs. A, B and C of rating 40W, 60 W and 100 W, respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

(a) Brightness of all the bulbs will be the same

(b) Brightness of bulb A will be the maximum

(c) Brightness of bulb B will be more than that of A

(d) Brightness of bulb C will be less than that of B

Answer. (c)

Explanation: We know that power is defined as rate of doing work. A bulb consumes electric energy and produces heat and light. Now, bulb with more power rating will produce more heat and light or we can say that

Power rating of bulb is directly proportional to the brightness produced by bulb.

Therefore, brightness of bulb B with power rating 60 W will be more than the brightness of bulb A having power rating as 40W.

Question 15. In an electrical circuit, two resistors of 2 and 4 respectively are connected in series to a 6V battery. The heat dissipated by the 4 resistor in 5s will be

(a) 5J

(b) 10J

(c) 20J

(d) 30J

Answer. (c)

Explanation:

Here, firstresistor, R₁ = 2Ω

And second resistor, R₂ = 4Ω

Voltage of cell, V = 6V

Time taken = t = 5s

Total resistance of the circuit = R = R₁ + R₂ = 2 + 4 = 6 Ω

Current, I = V/R = 6/6 = 1A

Heat dissipated by the 4Ω resistor in 5s is given as,

            H = I2Rt

⟹       H = 1 × 4 × 5 = 20J

Question 16. An Electric kettle Consumes 1 KW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

(a) 1 A

(b) 2 A

(c) 4 A

(d) 5 A

Answer. (d)

Explanation: Here, power = P = 1 KW = 1000 W

Voltage = V = 220 V

Current = I = ?

Now, I = P/V = 1000/220 = 4.5 A

Now rating of fuse wire must be slightly greater than 4.5 A i.e., 5 A.

Question 17. Two resistors of resistance 2 and 4when connected to a battery will have

(a) same current flowing through them when connected in parallel

(b) same current flowing through them when connected in series

(c)same potential difference across them when connected in series

(d) different potential difference across them when connected in parallel

Answer. (b)

Explanation: In series combination of resistor, the current through both the resistor are same but potential difference across each will be different.

In parallel combination current across each resistor will be different but the potential difference will be same.

Question 18. Unit of electric power may also be expressed as

(a) volt ampere

(b) kilowatt hour

(c) watt second

(d) joule second

Answer. (a)

Explanation: Electric power = voltage × current

SI Unit of voltage = Volt

SI Unit of current = Ampere

So, unit of electric power is also given by, volt ampere.

Short Answer Type Questions

Question 19. A child has drawn the electric circuit to study Ohm’s law as shown in figure. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

Answer. 

Errors	Corrections
Ammeter is connected in parallel with R.	Ammeter should be connected in series.
Signs of the terminals of voltmeter are not mentioned and that of ammeter are not correct.	Signs of the terminals of voltmeter need to be mentioned and that of ammeter must be corrected.
Voltmeter connected in series with R.	Voltmeter connected in Parallel with R.
Current is coming out from negative terminal and going into negative terminal again.	Direction of current and polarity of cell needs to be corrected.
Cells are not connected in series properly.	Cells must be connected in series and that too properly

The correct diagram is as shown below:

Question 20. Three 2resistors, A, B and C are connected as shown in figure. Each of them dissipates energy and can with stand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors?

Answer.

Here, first resistor in the path of current = R = 2

Maximum power 2 resistor can withstand = P_max = 18 W

Let maximum current through 2 Ω resistance be I_max

Now, power = P = I²R

The maximum current which can pass through 2 Ω resistance is 3 A.

As resistances of B and C are equal, therefore, 3 ampere current (or maximum current passing through resistor 2Ω) after passing through the node will divide equally.

Therefore, maximum current from B and C will be 3/2 = 1.5 A.

Question 21. Should the resistance of an ammeter be low or high? Give reason.

Answer.

The resistance of an ammeter should be low. This is because an ammeter is connected in series with the circuit for the measurement of electric current. In case, its resistance is high, then some amount of current may lost in heating it leading to the inaccurate reading. An ideal ammeter is one which has zero resistance.

Question 22. Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 in series with a combination of two resistors (4 each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 resistors be the same as that across the parallel combination of 4 resistors? Give reason.

Answer. 

The required figure is as shown below:

The equivalent resistance of two 4 Ω resistances is equal to 2Ω. This combination is connected to Therefore, 2Ω resistance in series.

Hence, the total voltage drop is same across 2resistor and the parallel combination of two resistors, cell of 4.

Question 23. How does use of a fuse wire protect electrical appliances?

Answer. 

An electric fuse is a safety device used to protect circuits and appliances by stopping the flow of any unduly high electric current. It works on the principle of the heating effect of electric current.

It is made up of material having low melting point and connected in series with the circuit. When current passing through the fuse exceeds a certain limit then it melts. Due to which the circuit breaks and current stops flowing. 

Question 24. What is electrical Resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Answer.

If l is the length of the conductor, A its area of cross section and R its total resistance then,

Where, ρ is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is ohm meter. It is a characteristic property of the material. The resistivity of a material varies with temperature.

Let, l = initial length of the conductor, ρ = electrical resistivity of the material, A = cross-sectional area, then,

So, if initial reading of ammeter is I then after changes it will become I/2.

Question 25. What is the commercial unit of electrical energy? Represent it in terms of joules.

Answer.

The commercial unit of electrical energy is kilowatt hour. It is written as kWh

1kWh = 1kW × h = 1000W × 3600s = 3.6 × 10⁶ J.

Question 26. A current of 1 A flow in a series circuit containing an electric lamp and a conductor of 5when connected to a 10V battery. Calculate the resistance of the electric lamp.

Now, if a resistance of 10is connected in parallel with this series combination, what change (if any) in current flowing through 5conductor and potential difference across the lamp will take place? Give reason.

Answer. 

The situation given in question is shown in the figure given below:

Here, current, i = 1A

Resistance of conductor, R = 5

Voltage, V = 10V

Resistance of lamp, R_L =?

Total resistance in the circuit, R_T = V/ i = 10/1 = 10 Ω

Now,   R_T = R_L + R

⟹       R_L = R_T – R = 10 – 5 = 5Ω

Potential difference across the lamp = IR_L= 1 × 5 = 5V

Now, 10Ω resistances is connected in parallel with total resistance R_T, then R_Total will be the total resistance and is given as:

Now, both the branches have 10 Ω resistance, so the incoming current from the battery will divide equally in both the branches.

Hence, current through 5Ω resistance and lamp = 2/2 = 1 ampere.

Now, current through lamp is 1 V therefore, potential difference across the lamp = (current through lamp) × (Potential difference across lamp) = 1 × 5 = 5 V.

Question 27. 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 other keeps working.

Question 28. B1, B2 and B3 are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

(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 bulbs B₂ gets fused?

(iii) How much power is dissipated in the circuit when all the three bulbs glow together?

Answer. 

Let R_eq is the net resistance of combination of three bulbs in parallel then,

R_eq = V/I = 4.5/3 = 1.5 Ω

All bulbs are identical so they must have same resistance.

Let R is the resistance of each bulb and all bulbs are connected in parallel hence,

Current through each bulb,

(i) If B₁ gets fused, the current in B₂ and B₃ will remain unaffected as voltage across bulb B₂ and B₃ bulb remains same. Hence, glow of bulb will not be affected.

(ii) When bulb B₂ gets fused, the current through B₂ will be zero and current in B₁ and B₃ will remain 1A.

Now net Current, I = I₁ + I₂ + I₃ = 1 + 0 + 1= 2A

Thus, current in ammeter, A₁ = 1 ampere

Current in ammeter, A₂ = 0

Current in ammeter, A₃ = 1 ampere

Current in ammeter, A = 2 ampere.

Long Answer Type Questions

Question 29. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another set of three bulbs of the same wattage are connected in parallel to the same source.

(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.

(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

Answer.

(a) The two situations given in questions is shown in the figure given below:

Let us assume that the resistance of each bulb is R and potential difference is V

Equivalent resistance in series combination = R_eq = R + R + R = 3R.

Let current through each bulb in series combination be I₁.

So, brightness of each bulb in parallel combination will increase. Each bulb will glow 3 times brighter to that of each bulb in series combination.

(b) If one bulb gets fused in series combination then, circuit gets broken and current stops flowing and remaining bulb don’t glow.

If one bulb gets fused in series combination then, same voltage continue to act on the remaining voltage and hence, other bulbs continue to glow with same brightness.

Question 30. State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

Answer. 

According to Ohm’s law the electric current flowing through a conductor is directly proportional to the potential difference applied across its ends provided the physical conditions such as temperature etc remains unchanged.

Let V be the potential difference applied across the ends of conductor through which current I flows, then according to Ohm’s law.

Ohm’s law can be verified experimentally by the activity given below:

(i) Firstly, Set up a circuit as shown in figure given below, consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel,chromium, manganese, and iron metals.)

(ii) First use only one cell as the source in the circuit. Note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given.

(iii) Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.

(iv) Repeat the above steps using three cells and then four cells in the circuit separately.

(v) Calculate the ratio of V to I for each pair of potential difference V and current I.

(vi) Plot a graph between V and I. The graph will be a straight line as shown below.

(vii) This verifies the Ohm’s law.

Ohm’s does not hold under all conditions as it is basically not a fundamental law which means it has exceptions. Following are the conditions when Ohm’s law does not hold:

(i) It is not obeyed when physical conditions of conductors like temperature keep on changing.

(ii) It is not obeyed by a lamp filament,  junction diode, thermistor etc.
(iii) It is not obeyed in case of superconductors whose resistance is equal to zero.

Question 31.What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends.

Answer. 

If l is the length of the conductor, A its area of cross section and R its total resistance then,

Where, ρ is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is ohm meter.

Experiment to study the factors on which the resistance of conducting wires depend

• Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length l [say, marked (1)] and a plug key, as shown in figure given below.

• Now, plug the key. Note the current in the ammeter
• Replace Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 2l [marked (2) in the Fig. above]
• Note the ammeter reading
• Now replace the wire by a thicker nichrome wire, of the same length l [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
• Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig. above] in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)]. Note the value of the current.
• Note the difference in the current in all cases
• We will find that the current depend on the length of the conductor
• We will also find that the current depend on the area of cross-section of the wire used

Question 32. How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?

Answer. 

Suppose 3 resistances are R₁, R₂ and R₃. We will connect these resistances in series with an ammeter, key and a battery of known voltage as shown in the figure given below.

Now, we will switch the position of ammeter i.e., between R₁ and R₂, between R₂ and R₃, after R₃ and note the reading in each case by closing and opening the key. We will find that the reading in each case will be same in all cases.

Question 33. How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?

Ans.  Suppose 3 resistances are R₁, R₂ and R₃. We will connect these resistances in parallel with

an ammeter (A), a voltmeter (V), a plug key (K)and a battery of known voltageas shown in figure given below.

We will close the key K and record the ammeter and voltmeter readings.

Now we will open the key and switch the position of voltmeter and ammeter as show in the figure given below.

We will close the switch and note the readings. Similarly, we will switch the position of voltmeter to resistances R₂ and R₃ and observe the readings.

We will find that readings of ammeter keeps on changing but readings of voltmeter in all cases almost remain same in all cases.

Question 34. What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

Ans.

When an electric current is passed through a high resistance wire, like nichrome wire, the resistance wire becomes very hot and produces heat. This effect is known as heating effect of current or Joule’s law of heating.

It states that the heat H produced by a resistor of resistance R due to current flowing through it for time t isH = I²Rt

A simple experiment to demonstrate heating effect of current is that if we switch on the bulb for a long period of time then it will become hot.

Four applications of Joule’s heating effect in daily life are:

(i) Electric fuse is a safety circuit devices work on this principle.

(ii) Electric iron we use to iron our clothes works on this principle.

(iii) Electric kettle used to boil water also works on this principle.

(iv) Electric toaster to make bake breads works on the same principle.

Question 35. Find out the following in the electric circuit given in figure:

(a) Effective resistance of two 8Ω resistors in the combination

(b) Current flowing through 4Ω resistors

(c) Potential difference across 4Ω resistances

(d) Power dissipated in 4 Ω resistors

(e) Difference in ammeter readings, if any

Answer.

(a) Here, two 8 Ω resistances are connected in parallel so their effective resistances are     

Here, current through resistor will be same as current from battery = 1 Ampere

(c) Potential difference (V’) across 4resistors (R’) is given by V’ = IR’ = 1 × 4 = 4 V

(d) Power dissipated across 4Ω resistor, P = I²R =1² × 4 = 4 W

(e) There is no difference in the readings of ammeters A₁ and A₂ as same amount of current flows through the ammeter.