NCERT Exemplar Class 10 Science Solutions Chapter 12 Electricity

NCERT Exemplar Class 10 Science Solutions Chapter 12-MCQ

1. A cell, a resistor, a key, and an ammeter are arranged as shown in the circuit diagrams of Figure 12.1. 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)
Solution:
A cell, a resistor, a key and an ammeter are arranged in series as shown in the circuit diagrams.
Current is given as: I = V/R
V and R are equal in all the three circuits, so current remains constant.
In series, the order of elements in the circuit does not matter to the amount of current flowing through it.
Therefore, option (d) is correct.

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: (d)
Solution:
(i) Net resistance, $R_1=2\mathrm{\Omega }$
(ii) Net resistance, $R_2=(2+2)\mathrm{\Omega }$ (In series)
$R_2=4\mathrm{\Omega }$
(iii) Net resistance, $1/R_3=(1/2+1/2)\mathrm{\Omega }$ (In parallel)
$R_3=1\mathrm{\Omega }$
As we know that heat produced is inversely proportional to the resistance.
So, heat produced will be maximum in (iii) and minimum in (ii)
Therefore, option (d) is correct.

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)
Solution:
The resistivity of material does not depend on length, thickness and shape of the material.
The electrical resistivity of a given metallic wire depends on the number density of free electrons in the conductor,
which is the nature of the material.
$p\alpha \frac{1}{n}$
where, n = number of free electrons per unit volume.
Electrical resistivity also depends on the temperature of the conductor.
Therefore, option (d) is correct.

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) 10²⁰
(b) 10¹⁶
(c) 10¹⁸
(d) 10²³

Answer: (a)
Solution:
Given Current:
I = 1A
Time (t) = 16s
$I=\frac{Q}{t}=\frac{ne}{t}$
$n=\frac{It}{e}$

Where e = charge of electron = $1.6\times {10}^{-19}C$
$n=\frac{1\times 16}{1.6\times {10}^{-19}}={10}^{20}$

Therefore, option (a) is correct.

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)
Essential conditions are necessary when electrical components are connected
1) In a circuit ammeter should be connected in series.
2) In a circuit voltmeter should be connected in parallel
3) Positive terminal of Voltmeter and Ammeter should be connected to positive terminal of battery. Same
goes with negative terminal. All the three conditions are valid only in circuit diagram 2.
Therefore, option (b) is correct.

6. What is the maximum resistance which can be made using five resistors each of $\frac{1}{5}\mathrm{\Omega }$ ?
(a) $1/5\mathrm{\Omega }$
(b) $10\mathrm{\Omega }$
(c) $5\mathrm{\Omega }$
(d) $1\mathrm{\Omega }$

Answer: (d)
Maximum resistance is obtained by connecting all the resistors in a series.
Series combination of n equivalent resistors of $R\mathrm{\Omega }$ is $(R\times n)\mathrm{\Omega }$
$R_s=R\times n=(5\times 1/5)\mathrm{\Omega }=1\mathrm{\Omega }$
Therefore, option (d) is correct.

7. What is the minimum resistance which can be made using five resistors each of 1/5 ?
(a) 1/5 $\mathrm{\Omega }$
(b) 1/25 $\mathrm{\Omega }$
(c) 1/10 $\mathrm{\Omega }$
(d) 25 $\mathrm{\Omega }$

Answer: (b)
Minimum resistance is obtained by connecting all the resistors in parallel.
Parallel combination of n equivalent resistors of $R\mathrm{\Omega }$ is $=(R/n)\mathrm{\Omega }$
$R_p=R/n=\frac{\frac{1}{5}}{5}\mathrm{\Omega }=\frac{1}{25}\mathrm{\Omega }$

Therefore, option (b) is correct.

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)
Maximum potential is obtained when cells are connected in series
Negative terminal of the cell is connected to the positive terminal of the second cell and so on
Series combination of cells representation.
​
Therefore, option (a) is correct

9. Which of the following represents voltage?
(a) $\frac{Workdone}{Current\times time}$
(b) $Workdone\times Charge$
(c) $\frac{Workdone\times Time}{Current}$
(d) $Workdone\times Time\times Charge$

Answer: (a)
Work done = $Voltage\times charge$
Now, Charge = $Current\times Time(q=It)$
Work done = $Voltage\times Current\times Time$
Voltage = $\frac{Workdone}{Current\times time}$
Therefore, option (a) is correct

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 same material has area of cross section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A

Answer: (c)
The resistivity of 1st conductor
$p=\frac{RA}{l}$
Resistivity of 2nd conductor should be the same as they are made of same material
Let area of 2nd conductor = A₁

$p=\frac{R{A}_1}{2l}$
Now,
$p=\frac{RA}{l}=\frac{R{A}_1}{2l}$
Which gives,
A₁ =2A

Therefore, option (c) is correct.

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 (Figure.12.5). Which of the following is true?
​
(a) R₁ = R₂ = R₃
(b) R₁ > R₂ > R₃
(c) R₃ > R₂ > R₁
(d) R₂ > R₃ > R₁

Answer:

Slope of Current vs Voltage graph is proportional to 1/R
(where R is resistance)
So, line with highest slope has least resistance due to the inverse proportional relation. Hence,
R₃>R₂>R₁
Therefore, option (c) is correct.

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)
Power, P=I²R
Case 1, P₁=I²R
Case 2, P₂=(2I)²R=(4I)²R
Power dissipated = P₂-P₁
=4I²R-I²R
=3I²R
So percentage increase from initial power
$\frac{3I^{2}R}{I^{2}R}\times 100=300$%
Therefore, option (c) is correct.

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)
Resistivity of material does not depend on length, thickness and shape of the material.
Electrical resistivity of a given metallic wire depends on number density of free electrons in the conductor, which is the nature of the material.
$p\alpha \frac{1}{n}$
where, n = number of free electrons per unit volume.
Electrical resistivity also depends on the temperature of conductor.
Therefore, option (c) is correct.

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:

Given that bulb A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source.
In parallel, the voltage applied across all the bulbs would be the same.
Power of a bulb is directly proportional to the brightness produced by it.
Hence the brightness of C > B > A.
Therefore, option (c) is correct.

15. In an electrical circuit two resistors of 2 $\mathrm{\Omega }$ and 4 $\mathrm{\Omega }$ respectively are connected in series to a 6 V battery. The heat dissipated by the 4 $\mathrm{\Omega }$ resistor in 5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J

Answer: (c)
Given resistors are $4\mathrm{\Omega },2\mathrm{\Omega }$ are connected in series
Equivalent Resistance = $4\mathrm{\Omega }+2\mathrm{\Omega }=6\mathrm{\Omega }$
Voltage = 6V
Current in the resistor = V/R = 6/6 = 1 A
Heat dissipated in resistor = I²Rt=1²×4×5=20J
Therefore, option (c) is correct.

16. An electric kettle consumes 1 kW of electric power when operated at 220V. A fuse wire of what rating must be used for it?
(a) 1 A
(b) 2 A
(c) 4 A
(d) 5 A

Answer:

Given Power: 1 kW = 1000 Watt
Power (P) = VI
Voltage = 220V
Current, I = P/V
= 1000/220 = 4.5 A
The rating should be greater than 4.5 A in order for the kettle to work. So, 5A is the only option that satisfies the condition.
Therefore, option (d) is correct.

17. Two resistors of resistance 2 $\mathrm{\Omega }$ and 4 $\mathrm{\Omega }$ 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)
In series combination, same current flows in both the resistors but voltage gets split according to resistance value.
In parallel combination, current splits and voltage remains same across the resistors.
Therefore, option (b) is correct.

18. Unit of electric power may also be expressed as
(a) volt ampere
(b) kilowatt hour
(c) watt second
(d) joule second

Answer: (a)
Electric power, P = VI
Unit of Voltage = Volt
Unit of current = Ampere
Unit of power = Volt ampere.
Therefore, option (a) is correct.

NCERT Exemplar Class 10 Science Solutions Chapter 12-Short Answer

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:

Ammeter is connected in parallel with R – Should be in series.

Voltmeter is connected in series with R – Should be in parallel.

Signs of the terminals of voltmeter are not mentioned.

Signs of the terminals of ammeter are incorrect.

Flow of current is shown from negative to negative terminal.

Cells are not connected in series properly.

The correct diagram is as follows:
​

20. Three 2 $\mathrm{\Omega }$ 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:

Given
Maximum Power a Resistor can handle (P): 18$\omega$
Resistance Value of A: 2 $\mathrm{\Omega }$
Resistance Value of B: 2 $\mathrm{\Omega }$
Resistance Value of C: 2 $\mathrm{\Omega }$
$$\begin{gathered} P=I^{2}R \\ {I}^2=\frac{P}{R} \\ I=\sqrt{\frac{P}{R}} \\ I=\sqrt{\frac{18}{2}}=\sqrt{9}=3A \end{gathered}$$
After substituting the values in the above equation 1 we get maximum current a resistor can take is 3 amperes.
I_A=I_B+I_c …(2)
So, if you note from equation 2, whatever current that passes through resistor A gets split and passes through resistor B and C.
Therefore, Resistor A has the maximum current of 3A passing through it due to an upper limit of power 18W.
Hence, I_A=3A
As the resistors B and C are parallel to each other, voltage across them is going to be same. Let us assume it to be V
$$\begin{gathered} I_B=\frac{V}{2} \\ {I}_C=\frac{V}{2} \end{gathered}$$
So we can see that current in Resistor B and C are going to be same and from equation 2 we can see that sum of current in both resistor to be 3A.
$$\begin{gathered} I_B=\frac{3}{2}=1.5A \\ {I}_C=\frac{3}{2}=1.5A \end{gathered}$$
So, the maximum current that can flow through each resistor is
I_A=3A
I_B=1.5A
I_C=1.5 A

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

Answer:

In order to measure the current through an element in a circuit, an ammeter is connected in series. In order to have the most accurate reading the resistor of the ammeter should be as low as possible so that it won’t tamper the current in the circuit.
If resistance is high, then some amount of current may be lost in heating it, leading to the inaccurate reading. An ideal ammeter is one which has zero resistance.

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

Answer:

​
Yes, the potential across the 2$\mathrm{\Omega }$ and the parallel combination of two 4$\mathrm{\Omega }$ resistors is the same. As the combination of two parallel 4$\mathrm{\Omega }$ resistor turns out to be 2$\mathrm{\Omega }$ and the potential across two same resistors in series combination comes to be the same.

Note: In series combination, the voltage gets divided according to the value of the resistor but the current remains the same but in parallel combination, the voltage remains constant but the current varies according to the voltage value.

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.
A fuse wire is nothing more than a resistor; every resistor produces heat when current passes through it. If the current in the resistor is high then heat produced in the resistor is also high and the fuse is made with materials which melt when the heat produce is high. So, if the current more than the preferred value passes through it, fuse wire melts and protects the rest of the circuit.

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 it is the constant of proportionality for each material which gives the resistance of a resistor when multiplied with length and divided with the area of cross section of the resistor.
$$\begin{gathered} V=IR \\ ,R=\rho \frac{l}{A} \end{gathered}$$
Where l is the length and A is the area of cross section
ρ is called the electrical resistivity of the material of the conductor.
The SI unit of resistivity is W-m. The resistivity of a material varies with temperature.
When the length of the wire is doubled the current in the circuits gets halved. The resistance is doubled, which in turn halves the current for the same voltage.

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

Answer:

The commercial unit of current is the kilowatt hour (kWh).
In one hour we have = 60×60 sec=3600 sec
So total joules spent in 3600 seconds = 3600×1000 J
= 3.6×10⁶ J
Therefore, 1kilowatt hour = 3.6×10⁶ J

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

Answer:

The situation can be depicted as:
​
Current, I = 1A
Resistance of conductor, R_C = 5$\mathrm{\Omega }$
Voltage, V = 10V
Resistance of lamp, R_L = ?
Total resistance in the circuit, RT = V/ I = (10/1) $\mathrm{\Omega }$ = 10 $\mathrm{\Omega }$
As the resistances are in series, R_T = R_C + R_L
RL = 10$\mathrm{\Omega }$ – 5$\mathrm{\Omega }$= 5$\mathrm{\Omega }$
Potential difference across the lamp = I R_L = (1 × 5) = 5V
Now, 10Ω resistances is connected in parallel with total resistance RT
1/R_new = (1/10) + (1/10) = 2/10 = 1/5
R_new = 5 $\mathrm{\Omega }$
New current, I_new = V/R_new = (10/5) A = 2 A
As both branches have equal resistance, current will divide equally.
Hence, current through 5Ω resistance and lamp = 2/2 = 1 A.
Potential difference across the lamp = 1 × 5 = 5 V.
There will be no change in the current flowing through 5 ohm conductor and potential difference across the lamp.
As voltage remains the same in both the branches the potential difference across the lamp will also remain the same.

27. Why is parallel arrangement used in domestic wiring?

Answer:

Parallel arrangement is used in domestic circuits so that when short circuit happens, few damages happen and all the electric equipment won’t get effected.
In parallel combination equal power is supplied to all the equipment in a house but in series connection the power received by each equipment changes according to the equipment connected in the house which will lead to a lot of problems.
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.

28. B₁, B₂ and B₃ 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 B1 gets fused?
(ii) What happens to the reading of A₁, A₂, A₃ and A when the bulb B2 gets fused?
(iii) How much power is dissipated in the circuit when all the three bulbs glow together?

Answer:

(i)Nothing happens to the glow of bulbs B₂ and B₃ when B₁ gets fused because all the bulbs are connected parallel therefore the power output won’t get changed.
$P=\frac{V^2}{r}$
In parallel connection voltage across the bulb does not change and resistance is independent of the connections in the circuit as it is an intrinsic property of wire based on its dimensions
(ii) Given all the three bulbs are identical therefore all the bulbs have same resistance R
All the three bulbs are in Parallel combination therefore
$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R} \\ \frac{1}{R_{eq}}=\frac{3}{R} \\ {R}_{eq}=\frac{R}{3} \end{gathered}$$
Current in the ammeter A is 3A
$$\begin{gathered} 3A=\frac{4.5}{\frac{R}{3}} \\ R=\frac{4.5}{\frac{3}{3}} \\ R=4.5\mathrm{\Omega } \end{gathered}$$
Resistance of each individual bulb is 4.5$\mathrm{\Omega }$
So reading in A₁, A₂, A₃ when the bulb A₂ gets short circuited is:
Reading in A₁
$A_1=\frac{4.5}{4.5}=1A$
So reading in A₁ when A₂ is short circuited is 1A.
Reading in A₂ is zero as it is short circuited
Reading in A₃
$A_3=\frac{4.5}{4.5}=1A$
So reading in A₃ when A₂ is short circuited is 1A.
(iii) Power dissipated when all the bulbs glow together: As they are in parallel combination, voltage across each bulb is same. Also, all of them are identical therefore, we can find the power consumption of one bulb and multiplying thrice will give us the total power.
Tripling will give you total power consumed
$Powerinonebulb=\frac{V^2}{R}=\frac{{4.5}^2}{4.5}=4.5W$
Total power consumed=4.5×3=13.5 W

NCERT Exemplar Class 10 Science Solutions Chapter 12-Long Answer

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)Let, resistance of each bulb is R
Case (1): connected in series
​
Net resistance = R + R + R = 3R
Current in each bulb = V/3R

(b)Case (2): connected in parallel
​
1/Net resistance = 1/R + 1/R + 1/R = 3/R
Net resistance = R/3
Net Current = 3V/R
In parallel, the current will be equally divided among all the bulbs.
Current in each bulb = V/R
As the current is more in case (2), the bulbs will glow brighter
Blubs will glow brighter in parallel combinations.

(b) If one bulb is fused in a series, all bulbs will not glow as the circuit gets broken and the current stops flowing.
In parallel combination, other bulbs will glow as short circuiting of one bulb will not affect the voltage of other bulbs.

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 electric current flowing through a conductor is directly proportional to the potential difference across the conductor as long as the temperature of wire doesn’t change and state of wire doesn’t change.
$\begin{array}{rl}& V\propto I\\ & V=IR\\ & I=\frac{V}{R}\end{array}$
R is the proportionality constant called resistance at a given temperature.
Experimental setup for Ohm’s law verification:
​
1. Set up a circuit as shown in the figure. This is made up of a nichrome wire XY of length, say 0.5m, an ammeter, a voltmeter and four cells of 1.5 V each.
2. First use only one cell as the source in the circuit. Note the reading in the ammeter A, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Note the readings.
3. 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.
4. Repeat the above steps using three cells and then four cells in the circuit separately.
5. Calculate the ratio of V to I for each pair of potential difference V and current I.
6. Plot a graph between V and I, and observe the nature of the graph
Once you plot the graph between V and I it will be a straight line with slope equivalent to resistance of wire. This is how we prove Ohm’s law experimentally. Make sure that the temperature in the room is same throughout the experiment.
Ohms law doesn’t obey all the time, it works only when temperature is constant.

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 of a material is the resistance of the conductor made of a material for unit length and unit cross section. It is represented with letter ρ
$P=\frac{RA}{l}$
Set an electric circuit consisting of a cell, an ammeter, a nichrome wire of length l and a plug key, as shown in the figure below.
​
1. Now, plug the key. Note the current in the ammeter.
2. Replace the nichrome wire by another nichrome wire of same thickness but twice the length, i.e., 2l
Note the ammeter reading.
3. Now replace the wire by a thicker nichrome wire, of the same length l. A thicker wire has a larger cross-sectional area. Again, note down the current through the circuit.
4. Instead of taking a nichrome wire, connect a copper wire in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire. Note the value of the current. Notice the difference in the current in all cases.
You can notice resistance gets doubled once length is doubled and halved when area is doubled

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:

Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in Figure. Let us assume that the resistors have values 1$\mathrm{\Omega }$, 2$\mathrm{\Omega }$, 3$\mathrm{\Omega }$ etc., and a battery of 6 V is used.
Plug the key. Note the ammeter reading. Change the position of ammeter to anywhere in between the resistors. Note the ammeter reading each time.
We can notice that where ever we connect the ammeter, the current is always the same.
​

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

Answer:

Make a parallel combination, XY, of three resistors having resistances R₁, R₂, and R₃, respectively. Connect it with a battery, a plug key and an ammeter, as shown in figure.
Now, connect a voltmeter in parallel with the combination of resistors. Plug the key and note the ammeter reading. Let the current be I. Also take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor.
​
Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R1, as shown in figure:

So, if you note all the voltmeter readings in each parallel resistor, they will be the same.

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

Answer:

The heat H produced by a resistor of resistance R due to current flowing through it for time t is H= I²Rt. It is also called ohmic heating and resistive heating.
In a conductor when voltage is applied across its ends, the free electrons available in it start drifting opposite to the direction of the positive terminal. These electrons collide with the positive ions (atoms which have lost the electrons). As a result of these collisions some energy of the electrons is transferred to the positive ions which
vibrate violently as they gain energy. Thus, heat is developed in the conductor.
Lesser the value of resistance higher the value of heat produced.
$H=\frac{V^2}{R}$
The four applications using heating effect of current are as follows
(i) Heating Rod
(ii) electric fuse
(iii) Electric iron
(iv) Incandescent Bulb

35. Find out the following in the electric circuit given in Figure 12.9
(a) Effective resistance of two 8$\mathrm{\Omega }$ resistors in the combination
(b) Current flowing through 4$\mathrm{\Omega }$ resistor
(c) Potential difference across 4$\mathrm{\Omega }$ resistance
(d) Power dissipated in 4$\mathrm{\Omega }$ resistor
(e) Difference in ammeter readings, if any.
​

Answer:

(a) Parallel combination of two 8-ohm resistors
Equivalent resistor in parallel combination:
$\begin{array}{rl}& \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\ & \frac{1}{R_{eq}}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}\\ & \frac{1}{R_{eq}}=\frac{1}{4}\\ & R_{eq}=4\mathrm{\Omega }\end{array}$
(b) Total resistance in the circuit is: 4 $\mathrm{\Omega }$ + 4 $\mathrm{\Omega }$ = 8 $\mathrm{\Omega }$
Given voltage in the circuit = 8 V
Current through the circuit: I = $\frac{V}{R}=\frac{8}{8}=1A$
(c) Potential difference across 4 $\mathrm{\Omega }$ = IR = (4 × 1) V = 4V
(d) Energy dissipated in 4 $\mathrm{\Omega }$ is I²R = 1² × 4 = 4W
(e) No, there is no difference in ammeter A₁ and A₂ reading as the same current passes through the ammeter.