Please refer to Electricity Class 10 Science Exam Questions provided below. These questions and answers for Class 10 Science have been designed based on the past trend of questions and important topics in your class 10 Science books. You should go through all Class 10 Science Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

**Class 10 Science Exam Questions Electricity**

Class 10 Science students should read and understand the important questions and answers provided below for Electricity which will help them to understand all important and difficult topics.

**Very Short Answer:**

**Question: Calculate the resistance of an electric bulb which allows a 10 A current when connected to a 220 V power source. Answer:** Let V be the potential difference created by power sources. I be the current in the circuit and R be the resistance of the bulb.

Given, V = 220 V, I = 10 A

Using Ohm’s law, V = IR

or R= V/I or R=220/10 Ω

R=22Ω

Resistance of given electric bulb comes out to be 22 Ω .

**Question**. The current flowing through a resistor connected in a circuit and the potential difference developed across its ends are as shown in the diagram by milliammeter and voltmeter readings respectively :**(a) What are the least counts of these meters ?****(b) What is the resistance of the resistor ?**

**Answer :**

(a) 10 mA and 0.1 V

(b) V = 2·4 volt, I = 250 mA = 0·25 A

From Ohm’s law,

**Question**. Why does the cord of an electric oven not glow while its heating element does ?**Answer : **The resistance of heating element of an electric oven is very high. As the current flows through the heating element it becomes too hot and glows red. On the other hand chord of an electric oven low resistance hence, it does not become red during the flow of current.

**Question**. What would you suggest to a student if while performing an experiment he finds that the pointer/ needle of the ammeter and voltmeter do not coincide with the zero marks on the scales when circuit is open ? No extra ammeter/voltmeter is available in the laboratory.**Answer : **This is called the zero error of the scale of ammeter or voltmeter. If there is a zero error then this error is subtracted from the value that depicts when the circuit is closed otherwise accurate readings will not be recorded.

**Question**. Name and define the SI unit of current.**Answer : **‘Ampere‘ is the SI unit of current.

1 Ampere current can be defined as a unit charge flowing per second in the circuit.

**Question**. Write the function of voltmeter in an electric circuit.**Answer : **Voltmeter measures the potential difference across two points in a circuit. It is always connected in parallel in the circuit.

**Question**. The following table gives the value of electrical resistivity of some materials :

**Which one is the best conductor of electricity out of ****them ?****Answer : **Silver, because its electrical resistivity is least out of the given materials.

**Question**. Which substance is used for making resistance coil of electric heater and why ?**Answer : **Nichrome, due to its high resistivity.

**Question**. Why is an ammeter connected in series in an electric circuit ?**Answer : **It is connected in series so that whole of electric current, which it has to measure, passes through it.

**Question**. Name the alloy which is used for making the filament of bulbs.**Answer : **Tungsten is used for making the filament of bulbs.

**Question**. Why closed path is required for the flow of current ?**Answer : **It makes possible to move the electrons in a particular direction, so closed path is necessary for the flow of current.

** Question**.

**State some properties of charge.**

**Answer :**Properties of Charge :

(a) Electric charge can neither be created nor be destroyed but it can transfer from one body to another i.e., total electric charge in an isolated system is conserved.

(b) Total charge on a body is equal to the algebraic sum of all the charges located on that body.

**Question**. When two ends of a metallic wire are connected across the terminals of a cell, some potential difference is set up between its ends. In which direction, electrons are flowing through the conductors ?**Answer : **From a lower potential end of a metallic conductor to its higher potential end.

**Question**. Electric current flows through a metallic conductor from its one end A to other end B. Which end of the conductor is at higher potential ? Why ?**Answer :** Current always flow from a higher potential to a lower potential end of the conductor. So end ‘A’ of the conductor is at a higher potential.

**Question**. Define current.**Answer :** Current is defined as the rate of flow of charge. If a charge ‘Q’ flows through the cross-section of a conductor in time ‘t’, then the current ‘I’ through it is given as :

**Question**. State Ohm’s law.**Answer : **According to the Ohm’s law, the current flowing in a conductor is directly proportional to the potential difference applied across its ends, provided the temperature and other physical conditions of the conductor remain constant.

**Question**. While studying the dependence of potential difference (V) across a resistor on the current (I) passing through it, in order to determine the resistance of the resistor, a student took 5 readings for different values of current and plotted a graph between V and I. He got a straight line graph passing through the origin. What does the straight line signify ? Write the method of determining resistance of the resistor using this graph.**Answer : **The straight line in the graph signify that potential difference and current are directly proportional to each other.

The method of determining resistance of resistor using the graph is by Ohm’s law, V = IR and by calculating the slope from the points mentioned on the graph

**Question**. (a) Write Joule’s law of heating.**(b) Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V.****Answer :** (a) According to Joule’s law of heating, the heat produced in a wire is directly proportional to

1. square of current (I2),

2. resistance of wire (R),

3. time (t) for which current is passed.

Thus, the heat produced in the wire by current I in time ‘t’ is

= 0.45 + 0.27

= 0.72 A

**Question**. What is the formula of (a) Resistance (R) of an electric appliance, (b) Safe current (I) in terms of power rating (P) and voltage rating (V).**Answer : **(a) Resistance of appliance

**Question**. Identify the components used in circuit diagrams represented by the following symbols :

**Answer :** (a) An electric cell (b) A fixed resistance

(c) A variable resistance (d) Plug key (open)

(e) Plug key (closed)

(f) A battery or combination of cells

(g) Ammeter (h) Voltmeter

**Question**. Write two points of difference between electric energy and electric power.**Answer :**

**Question**. Distinguish between kilowatt and kilowatt hour.**Answer :**

**Question**. A torch bulb when cold has a resistance of 2·5 Ω. It draws a current 450 mA, when connected to a 6 V battery and glows brightly. Calculate the resistance of the bulb when glowing and explain the reason for the difference in resistance.**Answer : **While glowing, I = 450 mA = 0.45 A, V = 6 volt

Using Ohm’s law,

The reason for the difference in resistance of bulb when cold (R = 2·5 Ω) and while glowing (R = 13·33 Ω), is that the resistance of filament of bulb increases with the increase in temperature.

**Question**. Two 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 ?**Answer : **Given : Power of one lamp, P1 = 100 W

Power of second lamp, P2 = 60 W

Since, both the lamps are connected in parallel, thus,

potential difference will be equal.

Thus, potential difference = 220 V

We know that,

Power (P) = VI

Thus, the total current through the circuit

**Question**. What is (a) the highest, (b) the lowest total resistance that can be secured by combination of four resistors of 1 Ω, 10 Ω, 100 Ω and 1000 Ω ?**Answer :**

(a) To obtain the highest resistance, the resistors must be connected in series.

∴ Highest resistance, Rs = (1 + 10 + 100 + 1000) Ω

= 1111 Ω

(b) To obtain the lowest resistance, the resistors must be connected in parallel.

The lowest resistance is given by

**Question**. Why is a series arrangement not used for domestic circuits ?**Answer : **Series arrangement is not used for domestic circuits for the following reasons :

(a) The voltage of the source gets divided in all the appliances connected in series, in the ratio of their resistances, so each appliance does not operate at its rated voltage.

(b) The resistance of the circuit increases and it reduces the current in the circuit, so each appliance gets less power.

(c) If any one appliance in series arrangement is switched off (or gets spoilt), no other appliance connected with it in series will then operate.

**Question**. State the energy conversion taking place in the following electric appliances :**(a) Electric heater, (b) Electric motor, (c) Loud-speaker,****(d) Electrolysis.****Answer : **(a) Electrical energy gets converted into heat energy in an electric heater.

(b) Electrical energy changes into mechanical energy in an electric motor.

(c) Electrical energy gets converted into sound energy in a loudspeaker.

(d) Electrical energy changes into chemical energy during electrolysis.

**Question**. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ?**Answer : **Advantages of connecting electrical appliances in

parallel instead of connecting in series :

(a) Voltage remains same in all the appliances.

(b) The total effective resistance is less.

(c) Switching ON/OFF of one device doesn’t affect others.

**Question**. Consider the following circuit :

**What would be the readings of the ammeter and the voltmeter when key is closed ? Give reason to justify your answers.****Answer :** R = R_{1} + R_{2} + R_{3}

R = 5 W + 8 W + 12 W = 25 W

V = 6 V

V = IR

Since the ammeter is connected in series hence the ammeter reading will be 0.24A.

**Question. What is electrical resistivity ? Derive its SI unit. In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 100 mA. If the length of the wire is doubled, how will the current in the circuit change ? Justify your answer.****Answer : **Electrical resistivity of the material of conductor is the resistance offered by the conductor of length 1 m and area of cross-section 1 m^{2}

**Question: Consider the following food chain which occurs in a forest : Grass ?? Deer ?? Lion If 10000 J of solar energy is available to the grass, how much energy would be available to the deer to transfer it to the lion?****Answer:** The energy available to the deer is 1000 J to transfer it to the lion. This can be depicted as follows:

Grass →10% Deer→10% Lion

10,000 J 1000J 100 J

**Question. Name a device that helps to maintain a potential difference across a conductor.****Answer. **A source of electricity such as cell, battery, power supply, etc. helps to maintain a potential difference across a conductor.

**Question. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?****Answer.** 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.

**Question. Define the unit of current.****Answer. T**he 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.

**Short Answer: I**

**Question. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series. ****Answer.** Three cells of potential 2 V, each connected in series, is equivalent to a battery of potential 2 V + 2 V + 2 V = 6V. The following circuit diagram shows three resistors of resistances 5 Ω , 8 Ω and 12 Ω respectively connected in series and a battery of potential 6 V.

**Question: Give an example to illustrate that indiscriminate use of pesticides may result in the degradation of the environment.****Answer:** Pesticides are the chemicals used to kill plant and animal pests. They are non-biodegradable and toxicants. For example : Excessive use of DDT resulted in reduced population of fish eating birds. DDT accumulated in such birds through the food chain. It interfered with the egg shell formation.

The shell being thin broke due to weight of the bird during incubation. Hence, their population declined.

**Question. Calculate the number of electrons constituting one coulomb of charge. ****Answer. **One electron possesses a charge of 1.6 × 10^{−19} C, i.e., 1.6 × 10^{−19} C of charge is contained in 1 electron.

**Question. What 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 ?****Answer.** There are four coils of resistances 4 Ω , 8 Ω , 12 Ω , and 24 Ω respectively.

(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 Ω

(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by

**Question. Why does the cord of an electric heater not glow while the heating element does?****Answer.** The heating element of an electric heater is a resistor. The amount of heat produced by it is proportional to its resistance. The resistance of the element of an electric heater is very high. As current flows through the heating element, it becomes too hot and glows red. On the other hand, the resistance of the cord is low. It does not become red when current flows through it.

**Question. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.****Answer. **The amount of heat (H) produced is given by the Joule’s law of heating as

H = VIt

Where,

Voltage, V = 50 V

Time, t = 1 h = 1 × 60 × 60 s

**Question. An electric iron of resistance 20 G takes a current of 5 A. Calculate the heat developed in 30 s .** **Answer. **The amount of heat (H) produced is given by the joule’s law of heating as

H = VIt

Where,

Current, I = 5 A

Time, t = 30 s

Voltage, V = Current × Resistance = 5 × 20 = 100 V

H = 100 X 5 X 30 = 1.5 X 10^{4} J

Therefore, the amount of heat developed in the electric iron is 1.5 X 10^{4} J

**Question. What determines the rate at which energy is delivered by a current?****Answer. **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.

**Question. How is a voltmeter connected in the circuit to measure the potential difference between two points?****Answer. **To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.

**Question. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10 ^{−8} Ω m. What will be the length of this wire to make its resistance 10 Ω ? How much does the resistance change if the diameter is doubled?**

**Answer.**Resistance (R) of a copper wire of length l and cross-section A is given by the expression,

**Question: (a) What is the shape of the graph obtained by plotting potential difference applied across a conductor against the current** **flowing through it?****(b) What does the slope of this V-I graph at any point represent?Answer: (a) The shape of the graph obtained by plotting potential difference applied across conductor against the current**

**flowing through it will be a straight line.**

**(b) According to ohm’s law,V = IR or I= V/R So, the shape of V-I graph at any point represents the resistance of the given conductor.**

**Question: A piece of wire of resistance 20 Ω is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.Answer:** As we know that if R, P, l and A are respectively

the resistance, resistivity, length and crosssectional

area of a given wire, then

R=P,l/A

Given : R = 20 Ω

If l is increased to 2l, then l = 2l

R’=P,/A’

So,R’=P,2l/A=2(Pl/A)

R’=2R

∴R’=2X20 Ω=40 Ω

So, the resistance of wire in the new situation is 40 Ω.

**Short Answer: II **

**Question: In the circuit diagram given below. **

**Calculate :****(a) the current through each resistor****(b) the total current in the circuit****(c) the total effective resistance of the circuit.Answer: **

**Question: Two lamps, one rated 60 W at 220 V and the other 40 W at 220 V, are connected in parallel to the electric supply at 220 V.****(a) Draw a circuit diagram to show the connections.****(b) Calculate the current drawn from the electric supply.****(c) Calculate the total energy consumed by the two lamps together when they operate for one hour.Answer:** (a)The circuit diagram for the given connection can be drawn as follows:

**Question. Use the data in Table 12.2 to answer the following Table Electrical resistivity of some substances at 20°C **

**(a) Which among iron and mercury is a better conductor?****(b) Which material is the best conductor?****Answer.**

Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.

(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.

**Question. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?****Answer. **

Where,

ρ = Resistivity of the material of the wire

l = Length of the wire

A = Area of cross-section of the wire

Resistance is inversely proportional to the area of cross-section of the wire. Thicker the wire, lower is the resistance of the wire and vice-versa. Therefore, current can flow more easily through a thick wire than a thin wire.

**Question. On what factors does the resistance of a conductor depend?****Answer. **The resistance of a conductor depends upon the following factors:

(a) Length of the conductor

(b) Cross-sectional area of the conductor

(c) Material of the conductor

(d) Temperature of the conductor

**Question. What is meant by saying that the potential difference between two points is 1 V?****Answer.** 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.

**Question. 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?****Answer. **The change in the current flowing through the component is given by Ohm’s law as,

V = IR

I = V / R

Where,

Resistance of the electrical component = R

Potential difference = V

Current = I

The potential difference is reduced to half, keeping resistance constant.

Let the new resistance be R’ and the new amount of current be I ‘.

Therefore, from Ohm’s law, we obtain the amount of new current.

Therefore, the amount of current flowing through the electrical component is reduced by half.

**Question. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below − ****I (amperes ) 0.5 1.0 2.0 3.0 4.0****V (volts) 1.6 3.4 6.7 10.2 13.2****Plot a graph between V and I and calculate the resistance of that resistor.****Answer. **The plot between voltage and current is called IV characteristic. The voltage is plotted on x-axis and current is plotted on y-axis. The values of the current for different values of the voltage are shown in the given table.

The IV characteristic of the given resistor is plotted in the following figure.

**Question. When 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.****Answer.** Resistance (R) of a resistor is given by Ohm’s law as,

V = IR

R = V / I

Where,

Potential difference, V = 12 V

Current in the circuit, I = 2.5 mA = 2.5 X 10^{-3} A

Therefore, the resistance of the resistor is 4.8 kΩ

**Question. A battery of 9 V is connected in series with resistors of 0.2 Ω , 0.3 Ω , 0.4 Ω , 0.5 Ω and 12 G, respectively. How much current would flow through the 12 Ω resistor?****Answer. **There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as

V = IR

I = V / R

Where,

R is the equivalent resistance of resistances . These 0.2 Ω , 0.3 Ω , 0.4 Ω , 0.5 Ω and 12 Ω

are connected in series. Hence, the sum of the resistances will give the value of R.

R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

Potential difference, V = 9 V

I = 9 / 13.4 = 0.671 A

Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.

**Long Answer:**

**Question**. (a) For the combination of resistors shown in the following figure, find the equivalent resistance between M & N.

**(b) State Joule’s law of heating.****(c) Why we need a 5 A fuse for an electric iron which consumes 1 kW power at 220 V ?****(d) Why is it impracticable to connect an electric bulb and an electric heater in series ?****Answer : **(a) (R_{3} × R_{4}/R_{3} + R_{4}) + R_{1} + R_{2}

(b) Joule’s law of heating implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance for a given current and

(iii) directly proportional to the time for which the current flows through the resistor. H = I_{2}Rt

where

I = Current

R = Resistance

t = Time taken

(c) For an electric iron which consumes 1 kW electric power when operated at 220 V, a current of (1000/220) A, i.e., 4.54 A will flow in the circuit. In this case, a 5 A fuse must be used.

(d) In a series circuit the current is constant throughout the electric circuit. Thus, it is obviously impracticable to connect an electric bulb and an electric heater in series, because they need currents of widely different values to operate properly.

**Question**. (a) How will you infer with the help of an experiment that the same current flows through every part of a circuit containing three resistors in series connected to a battery ?**(b) Consider the given circuit and find the current flowing in the circuit and potential difference across the 15 Ω resistor when the circuit is closed.**

**Answer :**

(a) Let three resistors R_{1}, R_{2} and R_{3} are connected in series which are also connected with a battery, an ammeter and a key as shown in figure.

When key is closed, the current starts flowing through the circuit. Take the reading of ammeter. Now change the position of ammeter to anywhere in between the resistors and take its reading. We will observe that in both the cases reading of ammeter will be same showing same current flows through every part of the circuit above.

(b) Given,

R1 = 5 Ω, R2 = 10 Ω, R3 = 15 Ω, V = 30 V

Total resistance, R = R1 + R2 + R3

[…5 Ω, 10 Ω and 15 Ω are connected in series]

= 5 + 10 + 15

= 30 Ω

Potential difference, V = 30 V

Current in the circuit, I = ?

From Ohm’s law,

Current flowing in the circuit = 1 A

Potential difference across 15 Ω resistors = IR_{3}

= 1 A × 15 Ω

= 15 V

**Question**. (a) Three resistors R_{1}, R_{2} and R_{3} are connected in parallel and the combination is connected to a battery, ammeter, voltmeter and key. Draw suitable circuit diagram and obtain an expression for the equivalent resistance of the combination of the resistors.**(b) Calculate the equivalent resistance of the following network :**

**Answer :**

(a) Let R_{1}, R_{2} and R_{3} are three resistance connected in parallel to one another and R is the equivalent resistance of the circuit. A battery of V volts has been applied across the ends of this combination.

When the switch of the key is closed, current I flows in the circuit such that,

**Question**. (a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.**(b) In an electric circuit two resistors of 12 Ω each are joined in parallel to a 6 V battery. Find the current drawn from the battery. ****Answer : **(a) It is observed that total current I is equal to the sum of separate currents.

I = I_{1 }+ I_{2} + I_{3} …(i)

Let Rp be the equivalent resistance of the parallel combination of resistors.

**Question**. Calculate the value of x if the equivalent resistance between the points P and Q as shown in figure is 5 Ω.

**Answer :**

Equivalent resistance of 6 Ω and 4 Ω in series,

R_{1} = (4 + 6) Ω = 10 Ω

Equivalent resistance of 8 Ω and x Ω in series,

R_{2} = (8 + x) Ω.

Now, R_{1} and R_{2} are in parallel.

Therefore, the equivalent resistance of R_{1} and R_{2} connected in parallel can be calculated as

**Question**. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor and a plug key, all connected in series. Now, connect the ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistors. What would be the readings in the ammeter and the voltmeter ?**Answer :**

Since, resistances are connected in series, thus electric current remains the same through all resistors.

Here we have,

Electric current, I = 0.24 A

Resistance, R = 12 Ω

Thus, potential difference (V) through the resistor of 12

Ω is given by

V = I × R

= 0.24 × 12 = 2.88 V

∴ Reading of ammeter = 0.24 A

Reading of voltmeter through resistor of 12 Ω = 2.88 V.

**Question**. (a) List the factors on which the resistance of a conductor in the shape of wire depends.**(b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ?****Give reason.****(c) Why are alloys commonly used in electrical heating devices ? Give reason.****Answer : **(a) Resistance of a conductor depends directly on its length and is inversely proportional to the area of cross-section.

(b) Metals have free electrons and they move and conduct electricity, whereas glass does not allow electrons and charges to flow freely as it is an insulator.

(c) The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at higher temperatures. Therefore, conductor of electric heating devices, such as toasters and electron irons are made of an alloy rather than pure metal.

**Question**. (a) Applied Ohm‘s law to obtain the relation for combined resistance when three resistors R_{1}, R_{2} and R_{3} are connected in series.**(b) Write any three difference between the series and parallel combination of resistance.****(c) A set of ‘n‘ identical resistors each resistance R are connected in series and the effective resistance is found to be ‘x‘. When these are connected in parallel, the effective resistance is found to be ‘y‘. Find the ratio of X and Y.****Answer :** (a) Series combination :

Consider a system in which three resistances R_{1},

R_{2} and R_{3} respectively are connected is series with the cell as shown in figure.

Let V be the potential difference maintained by the cell across the combination and I be the current following through each resistors.

using Ohm‘s law, potential difference across each resistor is

**(b)**

**Question: List the factors on which the resistance of a conductor depends. Answer:** Resistance of a conductor depends upon the following factors:

(i) Length of the conductor : Greater the length

(l) of the conductor more will be the resistance (R).

R ∞ l

(ii) Area of cross-section of the conductor:

Greater the cross-sectional area of the conductor,

less will be the resistance.

R∞ 1/A

(iii) Nature of conductor.

**Question: Explain with the help of a labelled circuit diagram how you will find the resistance of a combination of three resistors of resistances R _{1}, R_{2} and R_{3}, joined in parallel. Also mention how you will connect the ammeter and the voltmeter in the circuit when measuring the current in the circuit and the potential difference across one of the three resistors of the combination.** The following circuit diagram depicts the parallel connection of three resistors R

Answer:

_{1}, R

_{2}and R

_{3}. 17

**Question: (a) Two resistors R _{1} and R_{2} may form (i) a series combination or (ii) a parallel combination, and the combination may**

**be connected to a battery of six volts. In which combination will the potential difference across R**

_{1}and across R_{2}be the same and in which combination will the current through R_{1}and through R_{2}be the same?**(b) For the circuit, shown in this diagram,**

**Calculate(i) the resultant resistance (ii) the total current(iii) the voltage across 7 Ω resistorAnswer**:

In series combination, current remains same in all resistors and in parallel combination, voltage potential difference remains constant. So, in circuit (ii), the potential difference will be the same across R_{1} and R_{2}. In circuit (i) the current will be the same across R_{1} and R_{2}.

**Question: Derive the expression for the heat produced due to a current ‘I’ owing for a time interval ‘t’ through a resistor ‘R’ having a potential difference ‘V’ across its ends. With which name is the relation known? How much heat will an instrument of 12 W produce in one minute if it is connected to a battery of 12 V?Answer: **

**Question: Why does the connecting cord of an electric heater not glow red hot while the heating element does?Answer** Connecting cord has very low resistance than heating element. Since heat produced in a material due to electric current is directly proportional to its resistance. Hence, heating element will glow red hot and connecting cord does not.

**Question: A 4 kW heater is connected to a 220 V source of power. Calculate****(i) the electric current passing through the heater.****(ii) the resistance of the heater.****(iii) the electric energy consumed in a 2 hours** **use of the heater.Answer:**

**Long Answer Questions**

**Question: A hot plate of an electric oven, connected to a 220 V line. It has two resistance coils A and B each of the 30 W which may be used separately, in series or in parallel. Find the value of the current required in each of the three cases.****Answer:** V = 220 Volt

R_{A} = R_{B} = 30 Ω

a. When both are used separately.

Then current drawn by each

I=V/ R=220/30=7.3A

b. When two coils are connected in series then total resistance of coils

= 30 + 30 = 60 Ω

I=220/60=3.66A 60

c. When two coils are connected in parallel then total resistance

= 30/2 = 15 A

Now current, I=220/15=14.67A

**Question: A hot plate of an electric oven, connected to a 220 V line. It has two resistance coils A and B each of the 30 W which may be used separately, in series or in parallel. Find the value of the current required in each of the three cases.****Answer:** V = 220 Volt

R_{A} = R_{B} = 30 Ω

a. When both are used separately.

Then current drawn by each

I=V/ R=220/30=7.3A

b. When two coils are connected in series then total resistance of coils

= 30 + 30 = 60 Ω

I=220/60=3.66A 60

c. When two coils are connected in parallel then total resistance

= 30/2 = 15 A

Now current, I=220/15=14.67A

**Question: An electric lamp is marked 100 W, 220 V. It is used for 5 hours daily. Calculate: (i) its resistance while glowing, (ii) energy consumed in kWh per day.****Answer:**

Electric Iron 100W –– 220 V

Used for time, t = 5 hr. daily

(i) R=V^{2}/P=220 X 220/100=484Ω

(ii) Energy consumed by iron per day

= PXt = 100X5 Wh

= 500Wh = 0.5 kWh

**Question: A piece of wire having resistance R is cut into four equal parts,****a. How will the resistance of each part compare with the original resistance?****b. If the four parts are placed in parallel, how will the joint resistance compare with the resistance of the original wire?****Answer:** a. When wire is cut in equal pieces then resistance on one piece =R/4

=

b. Effective resistance in parallel combination of these four pieces of resistance R/4

each.

1/R_{P}=4/R + 4/R +4/R +4/R =16/R

RP= R/16

**Question: Three resistors of 5 W, 10 W, and 15 W are connected in series and the combination is connected to battery of 30 V. Ammeter and voltmeter are connected in** **the circuit. Draw a circuit diagram to connect all the devices in proper correct order. What is the current flowing and potential difference across 10 W resistance?****Answer: **Effective resistance of the circuit

R = 5 + 10 + 15 = 30 Ω

Current in the circuit, I=V/R=30/30A

I = 1A

Current in 10 Ω resistor = 1A

Potential difference across 10 Ω resistor

V = IR = 1X10

V = 10 Volt

**Question: Two identical resistors, each of resistance 2 W are connected in turn: (i) in series, and (ii) in parallel to a battery of 12 V. Calculate the ratio of power consumed in the two cases.****Answer:** Given: R_{1} = R_{2} = 2 Ω

(i) In series Rs. = 2 + 2 = 4 Ω

Power consumed in series:

P_{S} =V^{2}/R_{S}=12X12/4W

P_{S}=36W

(ii) In parallel combination 1/RP=R_{2}=1/2+1/2

R_{P}=1Ω

R_{P} = V^{2}/RP=12X12/1W

P_{P} = 144 W

**Question: A torch bulb is rated 5 V and 500 mA. Calculate its (i) power (ii) resistance and (iii) energy consumed when it is lighted for 4 hours.****Answer: **Given: 5V–500 mA

(i) Power = VXI = 5X 500X10-^{3}

= 2.5W

(ii) Resistance =V/I =5/500X10^{3}=10Ω

(iii) Energy consumed in four hrs.

= 2.5X4X60X60

= 10X3600 W.s.

= 3.6X104 J

**Question:An electric iron has a rating of 750 W; 200 V. Calculate: a. the current required.****b. the resistance of its heating element,****c. energy consumed by the iron in 2 hours.****Answer:** Rating of iron is 750 W–200V.

P = 750W, V = 200 Volt.

a. P = VI or I =P/V

I=750/ 200 = 3.75 A

b. Resistance =V/I=200/3.75=53.3Volt

c. Energy consumed in 2 hr = PXt

E = 750X2Wh = 1500Wh

**Question:The potential difference across the terminals of a cell is 1.5 volt. If it is connected with a resistance of 30 ohms, calculate the current flowing through the circuit.**** Answer:** Given: V = 1.5 volt.

R = 30 Ω

I = V/R=1.5/30=1/20=0.05A

**Question: A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.****Answer:** Given: I = 0.5 A

t = 10X60 s

t=10X60_{S} I=Q/t or Q=It

Q = 0.5X10X60 C

Q = 300 C

**Question: (a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 ohm?****(b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 ohm?****Answer: **

(a) Given:V = 220 V

R = 1200 ohm

As V = IR

220 = IX1200 or I = 220/1200A

I = 0.18 A

(b) V = 220 V

R = 100 ohm

I=V/ R=220 A/100 A

I = 2.2 A

**Question: The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw, if** **the potential difference is increased to 120 V?****Answer:**

V = 60 V, I = 4A

R=V/ I=60/4=15 Ω

when V = 120 V, R = 15Ω

then, I’=V/ R=120/15

I’ = 8A

**Question: Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature?****Answer:**

We have l = 1m

R = 26 ohm

t = 20°C

r = 0.15 mm

P = ?

As R=P l/A

P=RXA/ =RXΠr^{2}/l

P=26X22/7X(0.15X10-^{3})2/1

P = 1.84#10-^{6} Wm

**Question. 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 Ω resistor. What would be the readings in the ammeter and the voltmeter?****Answer. **To measure the current flowing through the resistors, an ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the 12 Ω resistor, a voltmeter should be connected parallel to this resistor, as shown in the following figure.

The resistances are connected in series.

Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm’s law,

V = IR,

Where,

Potential difference, V = 6 V

Current flowing through the circuit/resistors = I

Potential difference across 12 Ω resistor = V_{1}

Current flowing through the 12 Ω resistor, I = 0.24 A

Therefore, using Ohm’s law, we obtain

V_{1 }= IR = 0.24 X 12 = 2.88 V

Therefore, the reading of the ammeter will be 0.24 A.

The reading of the voltmeter will be 2.88 V.

**Question. How 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 ?****Answer.** There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.

(a) The following circuit diagram shows the connection of the three resistors.

Here, 6 Ω and 3 Ω resistors are connected in parallel.

Therefore, their equivalent resistance will be given by

This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.

Therefore, equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω

Hence, the total resistance of the circuit is . 4Ω

2. The following circuit diagram shows the connection of the three resistors.

All the resistors are connected in series. Therefore, their equivalent resistance will be given as

Therefore, the total resistance of the circuit is . I Ω

**Question. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.****Answer.** Power (P) is given by the expression,

P = VI

Where,

Voltage, V = 220 V

Current, I = 5 A

P = 220 X 5 = 1100 W

Energy consumed by the motor = Pt

Where,

Time, t = 2 h = 2 × 60 × 60 = 7200 s

P = 1100 × 7200 = 7.92 × 10^{6} J

Therefore, power of the motor = 1100 W

Energy consumed by the motor = 7.92 × 10^{6} J

**Question. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?****Answer. **For x number of resistors of resistance 176 Ω , the equivalent resistance of the resistors connected in parallel is given by Ohm’s law as

V = IR

R = V / I

Where,

Supply voltage, V = 220 V

Current, I = 5 A

Equivalent resistance of the combination = R,given as

From Ohm’s law,

Therefore, four resistors of 176 Ω are required to draw the given amount of current.

**Question. Show how you would connect three resistors, each of resistance 6 Ω , so that the combination has a resistance of (i) 9 Ω , (ii) 4 Ω .****Answer. **If we connect the resistors in series, then the equivalent resistance will be the sum of

the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω , which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be

6 / 2 = 3 Ω which is also not desired

Hence, we should either connect the two resistors in series or parallel.

(i) Two resistors in parallel

Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be

The third 6 Ω resistor is in series with 3 Ω . Hence, the equivalent resistance of the circuit

is 6 Ω + 3 Ω = 9 Ω .

(ii) Two resistors in series

Two 6 Ω resistors are in series. Their equivalent resistance will be the sum The third 6 Ω resistor is in parallel with 12 Ω . Hence, equivalent resistance will be

Therefore, the total resistance is . 4 Ω

**Question. 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?****Answer.** (a) The melting point and resistivity of tungsten are very high. It does not burn readily at a high temperature. The electric lamps glow at very high temperatures. Hence, tungsten is mainly used as heating element of electric bulbs.

(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. It produces large amount of heat.

(c) There is voltage division in series circuits. Each component of a series circuit receives a small voltage for a large supply voltage. As a result, the amount of current decreases and the device becomes hot. Hence, series arrangement is not used in domestic circuits.

(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A), i.e., R α 1/A

(e) Copper and aluminium wires have low resistivity. They are good conductors of electricity. Hence, they are usually employed for electricity transmission.

**Question. Judge the equivalent resistance when the following are connected in parallel −****(a) 1 Ω and 10 ^{6}Ω,**

**(b) 1 Ω and 10**

^{3}Ω and 10^{6}Ω.**Answer.**(a) When 1 Ω and 10

^{6}Ω are connected in parallel:

Let R be the equivalent resistance.

Therefore, equivalent resistance 1 Ω

(b) When 1 Ω, 10^{3}Ω, and are connected in parallel:

Let R be the equivalent resistance.

Therefore, equivalent resistance = 0.999 Ω

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