Chapter 11: Electricity

Multiple Choice Questions

1. A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure12.1. 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) the same in all the cases .

2. In the following circuits (Figure 12.2), heat produced in the resistor or combination of resistors connected to a 12 V battery will be

(a) same in all the cases      (b) minimum in case (i)

(c) maximum in case (ii)       (d) maximum in case (iii)

Answer : (c) maximum in case (ii)

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) nature of the material.

[ Electrical resistivity (ρ) is an intrinsic property of the material and is independent of the wire's length, thickness, or shape. It is a measure of how strongly a given material opposes the flow of electric current. Different materials have different resistivities, and this property is fundamental to the material itself. ]

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

(a)        (b)        (c)      (d) 

Answer: (a)

[ Here , I = 1 A , t = 16 s

We have, 

and  ]

5. Identify the circuit (Figure 12.3) in which the electrical components have been properly connected.

  

(a) (i)           (b) (ii)            (c) (iii)          (d) (iv)

Answer: (b) (ii)

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) 1 Ω

[ Here ,   

We have,

So, the maximum resistance that can be made using five resistors, each of 15 Ω, is 1 Ω. Therefore, the correct answer is (d) 1 Ω. ]

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

(a) 1/5 Ω        (b) 1/25 Ω        (c) 1/10 Ω      (d) 25 Ω

Answer  : (b)  1/25 Ω

 [  Here,

We have,

 

So, the minimum resistance that can be made using five resistors, each of 15 Ω, is 125 Ω. Therefore, the correct answer is (b) 125 Ω ]

8. The proper representation of series combination of cells (Figure 12.4) obtaining maximum potential is

(a) (i)               (b) (ii)             (c) (iii)              (d) (iv)

Answer: (a)  i

9. Which of the following represents voltage?

(a)     

 (b) Work done × Charge

(c)

(d) Work done × Charge × Time

Answer:  (a) 

10. A cylindrical conductor of length and uniform area of cross- section A has resistance . Another conductor of length  and resistance R of the same material has area of cross section

(a)         (b)            (c) 2A          (d) 3A

Answer : (c) 2A .

[We have,

and    ]

11. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances  and  respectively (Figure.12.5). Which of the following is true?

           

(a) 

(b) 

(c) 

(d) 

Answer :  (c)  .

[ In a V-I (Voltage-Current) graph for resistors, the resistance (R) is given by the slope of the graph. The steeper the slope, the smaller the resistance. From the given options, the correct order of resistances based on the slopes of the V-I graphs is  .]

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) 300%.

[ We have, 

If the current (I) is increased by 100%, the new current  will be .

The increase in power () is  

So, the increase in power is 300% of the original power . ]

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) the shape of the resistor is changed.

[ Resistivity () is an intrinsic property of a material and is independent of the material's shape. It also remains constant as long as the temperature remains constant. Therefore, if the shape of the resistor is changed, the resistivity of the material used to make the resistor will still stay the same.]

14. In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 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) Brightness of bulb B will be more than that of A.

[ In a parallel circuit, each bulb gets the full voltage of the source. The brightness of an incandescent bulb is directly proportional to its power. Therefore, the bulb with higher power (wattage) will be brighter. In this case, bulb B has a higher wattage (60 W) compared to bulb A (40 W), so the brightness of bulb B will be more than that of A.]

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

(a) 5 J       (b) 10 J        (c) 20 J          (d) 30 J

Answer:  (c) 20 J.

[ Here,     volt

 

We have, 

   

 J

Therefore, the heat dissipated by the 4 Ω resistor in 5 s is 20 J. ]

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) 5 A.

[ We have , 

 

and considering safety margins, a 5 A fuse is suitable.]

17. Two resistors of resistance 2 Ω and 4 Ω when 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) same current flowing through them when connected in series.

[ In a series circuit, the current remains constant and resistances add up. ]

18. Unit of electric power may also be expressed as

(a) volt ampere      (b) kilowatt hour        (c) watt second     (d) joule second

Answer: (b) kilowatt hour 

[ The commercial unit of electric power consumption is kilowatt hour (kWh), commonly called one unit of electricity.]

Short Answer Questions

19. A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

        

Answer:

20. Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors?

   

Answer: Here, P = 18 W ,  

We have,

  

Therefore, the maximum current that can flow through the three resistors is 3A .

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

Answer : The resistance of an ammeter should be very low. An ammeter is connected in series in a circuit to measure current. If its resistance is high, it will oppose the flow of current and change the actual current in the circuit. A low resistance allows current to pass through easily without affecting the circuit significantly, so the measurement remains accurate.

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

Solution: Here,  

We have, 

    

The potential difference across the 2 Ω resistor will be the same as that across the parallel combination of 4 Ω resistors because they are connected in series, sharing the same current.

Photo

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

Answer: A fuse wire protects electrical appliances by melting when excessive current flows through the circuit. This breaks the circuit and stops the flow of current, preventing overheating, short circuits and damage to appliances. The fuse works because it has a low melting point and high resistance.

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

Answer: Electrical resistivity is the property of a material that measures how strongly it opposes the flow of electric current. Its SI unit is ohm-meter (Ω·m).

Resistance of a wire is given by . When the length  is doubled, resistance also doubles (since  are constant). In a series circuit, current . If resistance doubles and voltage remains the same, current becomes half. Hence, the ammeter reading decreases from 5 A to 2.5 A.

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

Answer: The commercial unit of electrical energy is the kilowatt-hour (kWh).

We have, 1kWh = 1 kW × h = 1000 W × 60×60 second

= 1000×3600 W second

= 3600000 joule = Joule (J)

26. A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.

Solution: Given ,  Current (I) = 1 A , Voltage (V) = 10 V and 

Using Ohm’s law:

 

Resistance of an electric lamp  

So, the resistance of the electric lamp is 5 Ω.

Next Part : Here, Series resistance = 5 Ω (conductor) + 5 Ω (lamp) = 10 Ω.

New resistance  

We have ,

 In parallel, each branch gets full battery voltage, so current in the original branch does not change.

27. Why is parallel arrangement used in domestic wiring?

Answer: Parallel arrangement is used in domestic wiring because each appliance gets the same voltage (equal to the mains supply). Also, appliances can be operated independently, and if one device fails, others continue to work normally.

28.  and  are three identical bulbs connected as shown in Figure 12.8. 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  gets fused?

(ii) What happens to the reading of ,  ,  and A when the bulb  gets fused?

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

Answer: (i) The other two bulbs ( and ) will continue to glow normally. In a parallel circuit, each bulb gets the full voltage and is unaffected if one bulb fails.

(ii) When  fuses, the current through  and  remains unchanged, while A decreases slightly due to reduced total current from two working bulbs.

(iii) Here,  ,  volt

Using ohm’s law ,

We have, 

  

Long Answer Questions

29. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit 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) No, the bulbs in the two circuits will not glow with the same brightness.

In the series circuit, the supply voltage gets divided among the three bulbs, so each bulb receives less voltage and glows dimly.

In the parallel circuit, each bulb gets the full supply voltage. Hence, the bulbs glow with normal and brighter intensity.

(b) In series circuit: If one bulb gets fused, the circuit breaks and current stops flowing. So, the remaining bulbs will also stop glowing.

In parallel circuit: If one bulb gets fused, the other bulbs still have separate paths for current. Therefore, the remaining bulbs continue to glow normally.

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

Answer:  Ohm’s Law states that the current  flowing through a conductor between two points is directly proportional to the voltage  across the two points, provided the temperature and other physical conditions remain constant. It is mathematically expressed as: , where  is the voltage,  is the current and  is the resistance.

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No, Ohm’s law does not hold good under all conditions. It is valid only when the physical conditions such as temperature remain constant. It is not obeyed by devices like electric bulbs, semiconductors and electrolytes where resistance changes with temperature or other factors.

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: Electrical resistivity is the property of a material that measures how strongly it opposes the flow of electric current. Its SI unit is ohm-meter (Ω·m).

 Electrical resistivity is a measure of how much a material resists the flow of electric current. It indicates how strongly a material opposes the movement of electric charges through it.

The unit of electrical resistivity is the ohm-meter (Ω⋅m).

 

To study the factors affecting the resistance of a conducting wire, follow these steps:

 

(i) Set up a circuit with a cell, ammeter, a nichrome wire of length l (marked (1)), and a plug key. Close the circuit and record the current.

(ii) Replace the nichrome wire with one of twice the length 2l (marked (2)), and measure the current.

(iii) Replace the wire with a thicker nichrome wire of length l (marked (3)) and record the current.

(iv) Substitute the nichrome wire with a copper wire of the same length and cross-sectional area as the first nichrome wire (marked (4)), and note the current.

The current in a wire is influenced by both its length and cross-sectional area. Longer wires have higher resistance, which reduces the current, while thicker wires have lower resistance, allowing more current to flow. The material of the wire also impacts current due to differences in resistivity between materials, such as copper and nichrome. Applying Ohm's Law reveals that the resistance of a conductor is determined by (i) its length, (ii) its cross-sectional area, and (iii) the nature of its material. Precise measurements show that resistance is directly proportional to length (l) and inversely proportional to the cross-sectional area (A).

R ∝ l  and

 R ∝1A 

We get ,

 

R ∝l×1A=lA  

R ∝lA

R=ρlA

where ρ (rho) is a constant of proportionality.

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:  To infer that the same current flows through every part of a circuit containing three resistances connected in series to a battery, you can conduct the following experiment:

(i) Connect three resistors, R1, R2 , and R3, in series with a battery. Label the ends of the series combination of resistors as X and Y.

(ii) Insert a voltmeter across the ends X and Y of the series combination. Plug the key into the circuit and note the voltmeter reading, which gives the potential difference across the series combination of resistors. Let this value be V.

(iii) Measure the potential difference across the two terminals of the battery using the voltmeter. Compare this value with V.

(iv) Disconnect the voltmeter from the circuit and insert it across the ends X and P of the first resistor (R1). Plug the key back into the circuit and measure the potential difference across R1. Let this value be  V1.

(v) Similarly, measure the potential difference across the second resistor (R2) and the third resistor (R3). Let these values be V2 and V3, respectively.

(vi) According to the principle of series circuits, the total potential difference (V) across the series combination of resistors should equal the sum of the potential differences across each resistor:

V=V1+V2+V3

Using ohm’s law : We have

V=IRs

On using ohm’s law to the three resistors separately , we have

V1=IR1

V2=IR2

V3=IR3

Now ,

IRs=IR1+IR2+IR3

Rs=R1+R2+R3

Since the same current flows through each resistor in series, the potential differences are directly proportional to their respective resistances. This confirms that the same current flows through every part of the circuit.

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

Answer :

To conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery, you can perform the following steps:

(i) Connect three resistors R1 , R2 , and R3 in parallel between points X and Y. Connect this parallel combination to a battery along with a plug key and an ammeter. Attach a voltmeter in parallel with the combination of resistors to measure the potential difference.

(ii) Plug the key into the circuit and note the reading on the ammeter, which gives the total current (I) in the circuit. Then, read the voltmeter to determine the potential difference (V) across the parallel combination of resistors.

(iii) The potential difference across each resistor in a parallel circuit should be the same. Verify this by connecting the voltmeter across each individual resistor (R1 , R2 and R3 ). The voltmeter reading should indicate the same potential difference (V) for each resistor.

(iv) Disconnect the ammeter and voltmeter from the overall circuit and insert the ammeter in series with each resistor individually. Measure the current through each resistor, let the readings be I1 for R1 ,  I2  for R2 and I3 for R3 .

(v) According to the principle of parallel circuits, the total current (I) is the sum of the individual currents flowing through each resistor:

I=I1+I2+I3

Using ohm’s law , we have

I=VRp

Using ohm’s law to each resistor , we get

I1=VR1

I2=VR2

I3=VR3

Now ,
I=I1+I2+I3

VRp=VR1+VR2+VR3

VRp=V1R1+1R2+1R3

1Rp=1R1+1R2+1R3

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

Q. How can Joule’s heating effect be demonstrated experimentally?

Answer : Joule's heating effect states that when an electric current passes through a conductor, the electrical energy is converted into heat energy. This effect is directly proportional to the square of the current (I), the resistance of the conductor (R), and the time (t) for which the current flows. Mathematically, it is expressed as:

H=I²×R×t

where:

H is the heat produced (in joules),

I is the current (in amperes),

R is the resistance of the conductor (in ohms),

t is the time the current flows (in seconds).

 

To experimentally demonstrate Joule’s heating effect :

Materials Required:

(i) A resistor with known resistance R

(ii) A power supply (DC source) capable of providing a known current I

(iii) An ammeter to measure the current I through the resistor

(iv) A voltmeter to measure the potential difference V across the resistor

(v) A stopwatch to measure the time t during which the current flows

(vi) A thermometer or a calorimeter to measure the heat generated (optional for precise measurements)

Experimental Setup:

Circuit Assembly:

(i) Connect the resistor R in series with the ammeter and the power supply. This will ensure that the current III flowing through the circuit can be measured.

(ii) Connect the voltmeter across the resistor to measure the potential difference V across it.

Initial Measurements:

(i) Set the power supply to provide a specific voltage V.

(ii) Close the circuit to allow current I to flow through the resistor.

(iii) Measure and record the current I using the ammeter and the voltage V across the resistor using the voltmeter.

(iv) Record the initial temperature of the resistor or the surroundings (if using a thermometer or calorimeter).

Time Measurement:

(i) Start the stopwatch to measure the time t during which the current flows through the resistor.

(ii) Allow the current to flow for a specific duration, t (say for a few minutes).

Observation of Heating:

(i) After the time t has elapsed, stop the stopwatch and disconnect the circuit.

(ii) If using a thermometer or calorimeter, measure the final temperature of the resistor or surroundings to determine the amount of heat generated.

Calculation of Heat Generated:

(i) According to Joule's law of heating, the heat HHH generated in the resistor can be calculated using the formula:

H=I²Rt

Where, I is the current through the resistor, R is the resistance of the resistor, t is the time for which the current flows.

The experimental results should confirm that the heat produced in the resistor is directly proportional to the square of the current, the resistance of the resistor, and the time for which the current flows. This confirms Joule's law of heating.

35. Find out the following in the electric circuit given in Figure 12.9

    

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

(b) Current flowing through 4 Ω resistor

(c) Potential difference across 4 Ω resistance

(d) Power dissipated in 4 Ω resistor

(e) Difference in ammeter readings, if any.

Answer: (a) Here,  ,

We have, 

 

(ii) Here, V = 8 volt , and  

 

Using ohm’s law , we have

(iii) Here, ,

We have,

(iv) Here, R = 4 Ω , V = 4 volt

We have,

(v) Here,

 

and 

There is no difference in the ammeter readings in this case.