# Electricity Class 10 Science Important Questions

Please refer to **Electricity Class 10 Science Important Questions with answers** below. These solved questions for Chapter 12 Electricity in NCERT Book for Class 10 Science have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these solved problems properly as these will help them to get better marks in your class tests and examinations. You will also be able to understand how to write answers properly. Revise these questions and answers regularly. We have provided Notes for Class 10 Science for all chapters in your textbooks.

## Important Questions Class 10 Science Chapter 12 Electricity

All Electricity Class 10 Science Important Questions provided below have been prepared by expert teachers of Standard 10 Science. Please learn them and let us know if you have any questions.

**ONE MARK QUESTIONS**

**Question: Which kind of charge moves: positive charges (protons) or negative charges (electrons)?** **Answer:** Negative charge moves while protons can’t move because they are stuck in the nucleus so they can’t move due to “strong nuclear force”.

**Question: What is electric potential?** **Answer:** The electric potential, or voltage, is the difference in potential energy per unit charge between two locations in an electric field.

**TWO MARKS QUESTIONS**

**Question: How does the resistance of an ohmic conductor depend on the applied voltage?** **Answer:** Resistance do not depend on the voltage it depends on the material of the resistor whether the intermolecular force of attraction is strong or loose which decides the resistance.

**Question: Electrons are responsible for conduction in a conductor. The speed of electron is not very high. Then, how the bulb is on immediately even through the switch and bulb are far away from each other in a household circuit?****Answer:** This is because the number of the free electrons constituting current in a typical conductor is very large of order 10^{18} or more. This large electron number density keeps the good amount of current flowing through.

**Question:** **State the laws of refraction of light. Explain the term ‘absolute refractive index of a medium’ and write an expression to relate it with the speed of light in vacuum**.**Answer:** Laws of refraction:

a. The incident ray, refracted ray and normal to the point of incidence all lie in the same plane.

b. The ratio of sin of incident angle to sin of angle of refraction for a given pair of medium is constant.

sin i/sinr = constant

Absolute refractive index of a medium is the ratio of speed of light in air or vacuum to speed of light in the medium.

Absolute refractive index

= Speed of light in air/ vacuum/ Spee **Question: Five resistors are connected in a circuit as shown. Find the ammeter reading when circuit is closed.**

**Answer:** Labelling each of the resistance as shown in the figure below.** **

Resistance R_{1} and R_{2} are in series

Hence the current reading in ammeter is 1 A.

**Question: Three V-I graphs are drawn individually for two resistors and their series combination. Out of A, B, C which one represents the graph for series combination of the other two? Give reason for your answer. **

**Answer:**

Therefore, the slope of the V-I graph represents the effective resistance.

Now in series combination of the resistance all the resistances get summed up

Therefore, the series combination will have the maximum resistance than the other resistors.

Hence the line with maximum slope (line C) will give the series combination of the other two resistors.

**THREE MARKS QUESTIONS**

**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 = 22 Ω

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

**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. An electric circuit consisting of a 0.5 m long nichrome wire XY, and an ammeter, a voltmeter, four cells of 1.5 V each and a plug key was set up.**

(i) Draw a diagram of this electric circuit to study the relation between the potential difference maintained between the points ‘X’ and ‘Y’ and the electric current flowing through XY,

(ii) Following graph was plotted between V and I values :

What would be the values of 1/V ratios when the potential difference is 0.8 V, 1.2 V and 1.6 V respectively? What conclusion do you draw from these values?**Answer:** (i) Electric circuit to study the relation between potential diFFerence and the current.

**Question. The electrical resistivity of silver is 1.60 × 10–6 Ω. What will be the resistance of a silver wire of length 10 m and crosssectional area 2 × 10–3 m2? ****Answer:** Given, electrical resistivity of silver,

Ρ = 1.6 × 10–6 Ω m

Length of silver wire, l = 10 m

Area of cross-section, A = 2 × 10–3 m2

As we know,

resistance of wire is given by

R = P,l/A

Putting the values of l, Ρ and A, we get

R= 1.6 x 10-6 x 10/2 x 10-3 = 8.0 x 10-3 Ω

**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:** Power of the heater, P = 4 kW = 4000 W

Voltage of source of power, V = 220 V

(i) Since, P = VI (I is the current)

or I = P/V = 4000/220 = 18.18 A

(ii) Power, P is also given by P = I2R

R = P/I2 = 4000/(18.18) = 12 Ω

(iii) Since, electric energy consumed is given by

E = P × t (where t is time)

E = 4 kW × 2h = 8 kWh

E = 8 unit

**Question. (a) Two resistors R1 and R2 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 R1 and across R2 be the same and in which combination will the current through R1 and through R2 be the same?**

(b) For the circuit, shown in this diagram, 19 Calculate

(i) the resultant resistance

(ii) the total current

(iii) the voltage across 7 Ω resistor**Answer:** (a)

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 R1 and R2. In circuit (i) the current will be the same across R1 and R2.

**Question. What is meant by saying that the potential difference between two points is 1 volt? Name a device that helps to measure the potential difference across a conductor?****Answer: **When we say that the potential diFFerence between two points is 1 V, it means that 1 J of work is being done to move a unit charge between that two points.

Voltmeter is a device that helps to measure the potential diFFerence.

**Question. What is an electric circuit? Distinguish between an open and a closed circuit.****Answer:** Electric circuit : Closed and continuous path of an electric current is called an electric circuit. 1

**Question. Name an instrument that measures electric current in a circuit, define the unit of electric current.****Answer:** Ammeter is an instrument that measures electric current in the circuit.

One ampere is defined as the flow of one coulomb of charge per second. i.e., 1 A = 1 C s–1.

**Question. Electrical resistivities of some substances at 20°C are given below:**

Silver 1.60 × 10–8 Ω m

Copper 1.62 × 10–8 Ω m

Tungsten 5.20 × 10–8 Ω m

Iron 10.0 × 10–8 Ω m

Mercury 94.0 × 10–8 Ω m

Nichrome 100 × 10–8 Ω m

Answer the following questions in relation to them :

(i) Among silver and copper, which one is a better conductor? Why?

(ii) Which material would you advise to be used in electrical heating devices? Why?**Answer:** (i) Silver is the best conductor of electricity because of low resistivity.

(ii) Nichrome should be used in electrical heating devices, due to very high resistivity. It has a high resistance and produces a lot of heat on passing current.

**Question. Out of 60 W and 40 W lamps, which one has a higher electrical resistance when in use?****Answer:** We know that : p = V2/R

Power is inversely proportional to the resistance.

∴ 40 W has a higher electrical resistance.

**Question. Why is a series arrangement not used for connecting domestic electrical appliances in a circuit? ****Answer:** (1) In series arrangement, same current will flow through all the appliances, which is not required for domestic electric circuit.

(2) Total resistance of domestic circuit will be sum of the resistance of all appliances and hence current drawn by the circuit will be less.

(3) We cannot use independent on/of switches with individual appliances.

(4) All appliances are to be used simultaneously even if we do not need them.

**Question. Two resistors, with resistances 5 Ω and 10 Ω respectively are to be connected to a battery of emf 6 V so as to obtain :**(i) minimum current flowing

(ii) maximum current fkowing

(a) How will you connect the resistances in each case?

(b) Calculate the strength of the total current in the circuit in the two cases.

**Answer:**(a) In order to make the flow of minimum current in the circuit, we can connect 5 Ω and 10 Ω resistors in series.

And to get maximum flow of current in the circuit, we can connect 5 Ω and 10 Ω resistors in parallel.

**Question. Draw a schematic diagram of an electric circuit consisting of a battery of two cells each of 1.5 V, 5 Ω, 10 Ω and 15 Ω resistors and a plug key, all connected in series.****Answer: **The required circuit diagram is as below :

**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:**

Let R1, R2 and R3 be the three resistances with resistance value 2 Ω, 5 Ω and 10 Ω respectively.

(a) Let V1, V2 and V3 be the potential differences

across the resistors R1, R2 and R3.

Let currents I1, I2, I3 flow through the resistors R1,

R2, R3 respectively.

Since R1, R2 and R3 are connected in parallel,

V1 = V2 = V3 = 6 V

Now, current through R1 is

I1 = V1/R1 = 6/2 = 3A

Current through R2 is I2 = V2/R2 = 6/5 = 1.2A

Similarly, current through R3, I3 = V3/R3 = 6/10 = 0.6A

(b) Total current in circuit is given by

I = I1 + I2 + I3 = 3 + 1.2 + 0.6 = 4.8 A

(c) Total eFFective resistance (Rp) of the circuit is given as

**Question. (a) Define the term “volt”.**

(b) State the relation between work, charge and potential difference for an electric circuit.

Calculate the potential difference between the two terminals of a battery if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to the other.**Answer:** (a) ‘Volt’ is the SI unit of electric potential difference, it is defined as the amount of work done in moving a unit positive charge from one

point to another.

(b) The relationship between work (W), charge

(Q) and potential difference (V) for an electric circuit is given by

V = W/Q

Given : W = 100 J, Q = 20 C, V = ?

Using the relation, V = W/Q = 100/20 = 5V

So, 5 V of potential diFFerence between the two terminals to transfer 20 C of charge when work done is 100 J.

**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 ∞ l/A

(iii) Nature of conductor.

**Question. What do the following symbols mean in circuit diagrams? **

**Answer:** (i)

rheostat/variable resistance

**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, Ρ, 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 , l’/A’

So, R’ = P,2l/A = 2(P,l/A)

R’ = 2R

∴ R’ = 2 × 20 Ω = 40 Ω

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

**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. Derive an expression for the heat produced in a conductor of resistance R when a current I flows through it for time t.**

Two identical resistors of resistance R are connected in series with a battery of potential difference V for time t. The resistors are then connected in parallel with the same

battery for the same time t. Compare the heat produced in the two cases. **Answer: **If resistors are connected in series

Total resistance of the combination would be,

RS = R + R = 2R

Let Hs be the heat produced in series combination.

So, HP = 4HS

So, heat produced in parallel combination will be four times the heat produced in series.

**Question. Explain with the help of a labelled circuit diagram how you will find the resistance of a combination of three resistors of resistances R1, R2 and R3, 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.****Answer:** The following circuit diagram depicts the parallel connection of three resistors R1, R2 and R3.

The value of resistances are R1, R2 and R3.

Let currents I1, I2 and I3 flow through the resistors R1, R2 and R3 respectively.

It is observed that total current I is equal to the sum of the separate currents through each branch.

So,

I = I1 + I2 + I3 …(i)

Let RP be the equivalent resistance. Then,

**Question. Deduce the expression for the equivalent resistance of the parallel combination of three resistors R1, R2 and R3.Consider the following electric circuit. **

(i) Which two resistors are connected in series?

(ii) Which two resistors are connected in parallel?

(iii) If every resistor of the circuit is 2 Ω what current will flow in the circuit?**Answer: **Given circuit can be redrawn as follows :

**Question. Derive the expression for the heat produced due to a current ‘I’ flowing 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:** (a) As we know, from the definition of potential difference (V),

V = W/Q ,

here, ‘W’ is work done in moving charge from one point to another, ‘Q’ is the amount of charge.

W = V × Q

W = V X Q/t x t (On multiplying and dividing by time ‘t’)

∴ W = VIt

Since this work done is converted into heat energy, so, we can write

H = VIt …(ii)

Where ‘H’ is heat energy produced by electrons.

From Ohm’s Law,

V = IR, here, R is the resistance of the resistor.

Putting this in equation (ii), we get

H = I2Rt

Above relation is also known as Joule’s law of heating.

(b) Since, heat developed = power × time

∴ H = P × t

Given, P = 12 W; t = 1 min = 60 s

So, heat developed in 1 min = 12 × 60 = 720 J

**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: The resistivity of copper is 1.7 × 10-8 Ωm. What length of copper wire of diameter 0.1 mm will have a resistance of 34Ω?****Answer:** Given

Resistivity (ρ) = 1.7 × 10-^{8} Ωm

Diameter (d) = 0.1 mm = 0.1×10-^{3} m

(∵ 1m = 1000 mm)

Resistance(R) = 34 Ω

Hence the length of the wire is 15.7 m. E

**Question: Calculate the work done in moving a charge of 5 coulombs from a point at 20 to another at 30 V.****Answer:** Given;

Charge (Q) = 5C;

Potential difference = 30 – 20 = 10V;

Work done = Charge × Potential difference

⇒ Work done = 5 × 10 = 50 J.

Hence the work done is 50 Joule. E

**Question: A wire of resistance 20 Ω is bent to form a closed square. What is the resistance across a diagonal of the square?****Answer:** Since the wire is bend in form of the square and we know that sides of square are equal so each side will have same resistance i.e 5 ohm. Now if we make diagonal two faces in each side of the diagonal will have two sides and resistance will be 10 ohm. As shown below.

So total resistance will be

⇒ R_{eq} = 5 ohm

Hence equivalent resistance is 5 ohm.

**Question: The electric power consumed by a device may be calculated by the using either of the two expressions P = I _{2}R or P = V_{2}/R. The first expression indicates that it is directly proportional to R whereas the second expression indicates inverse proportionality. How can the seemingly different dependence of P on R in these expressions be explained.**

**Answer:**Both the expressions are correct. In the first case where P = I

^{2}R current (I) remains constant. Whereas in the second expression where P = V

^{2}/R the Voltage (V) remains constant.

Therefore P = I

^{2}R is used where the resistors are connected in series because in the series connection the current in each resistance is constant.

P = V

^{2}/R is used where the resistors are connected in parallel because in the parallel connection the voltage in each resistance is constant. So power becomes inversely proportional to resistance.

**Question: State difference between the wire used in the element of an electric heater and in a fuse wire.****Answer:**

**Question: An electrical appliance draws a current of 0.5 A when applied potential difference across it is 220V. Calculate the amount of charge flowing through it in 30 minutes.****Answer:** Given that;

Current = 0.5 A

Voltage = 220 V

Time = 30 min = 30 × 60 sec = 1800 sec

∵ 1 min = 60 sec)

We know that current=Charge/Time

⇒ Charge = Current × Time

⇒ Charge = 0.5 × 1800 C

⇒ Charge = 180 × 5 C

⇒ Charge = 900 C

Hence the charge flowing through appliance in 30 minutes is 900 Coulomb.

**Question: A 9 Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.****Answer:** When the wire is cut into three equal parts the length of each part will be l/3.

And resistance is directly proportional to the length.

∴ The resistance of each part will be 9/3 = 3.

Now we have three resistors each of 3Ω connected parallel as shown in figure below.

Hence the equivalent resistance of the combination is 1 Ω

**Question: Draw a schematic diagram of an electric circuit consisting of battery of five 2V cells, a 20 Ω resistor, a 30 Ω resistor, a plug key, all connected in series. Calculate the value of current flowing through the 20 Ω resistor and the power consumed by the 30 Ω resistor.****Answer:** The schematic diagram is shown below.

Current Flowing Through 20Ω resistor

Hence the current flowing is 0.5 Ampere.

Power consumed by the 30 Ω resistor/Power = Voltage × current.

Given Voltage = 10V; current = 0.5A ⇒ Power = 10 × 0.5 = 5W Hence power consumed by 30 Ω resistor is 5W.

Also, the resistance will become half because resistance is directly proportional to the length and the length is halved.

Resistance of each part connected in parallel = 5Ω. As shown below.

Resistance of parallel combination

Hence the resistance between ends of the diameter is 2.5Ω.

**FIVE MARKS QUESTIONS**

**Question: A. What do the following circuit symbols represent? **

**Answer:** (i) The above symbol represents that the two wires are not joined. As shown in the figure below.

Such symbol is used when there are two or more wires in the circuit.

(ii) The above symbol represents the Rheostat or variable resistance as Shown in figure below.

The Rheostat is used when there is need to vary the current or resistance in the circuit. The original rheostat used in the circuit is shown below in the figure below.

**Question: B.** The potential difference between the terminals of an electric heater is 60V when it draws a current of 4A from the source. Find the resistance of heater when in use.

Answer: According to the question

Potential Difference (V) = 60V

Current drawn (I) = 4A By Ohm’s Law

V = IR

⇒ 60 = 4 × R

⇒ Hence the resistance is 15Ω.

**Question: A. Find the value of current I in the circuit given below:**

**Answer:**

Resistors across AC and DE are in parallel i.e. RAC and RDE are in parallel.

So

**Question:** **B.You have four resistors of 8 Ω each. Show how would you connect these resistors to have effective resistance of 8 Ω. ****Answer:** Two parallel combinations must be connected in series with each other to get the effective resistance of 8Ω. As shown in the figure below.

The effective resistance of each of the parallel combination is 4Ω. And this two 4Ω resistors are added together to get 8 Ω effective resistance.

**Question: A wire of 3Ω resistance and 15cm in length is stretched to 45 cm length. Calculate (i) New resistance.Assuming the wire has uniform cross section area.****Answer:** Given:

Resistance R = 3Ω

Original length L1 = 15cm = 0.15m

New length, L2 = 45 cm = 0.45 m

Formula Used:

Where, ρ is resistivity of the wire L is the length of the wire A is the area of the wire.

Hence, the new resistance is 9 Ω**Question: A. Shushant was doing an experiment by using an ammeter. Unfortunately, it fell from his hand and broke. Fearing the scolding of teacher, his group mates advised him not to tell the teacher, but he told her. On listening to him patiently, the teacher did not scold him as it was just and accident and used the opportunity to show the whole class the internal structure of ammeter.****What are the values displayed by Shushant?****Answer:** Shushant told his teacher that ammeter was broke by him he did not hide from his teacher this shows that Shushant is very honest and respectful towards his teacher. Also, in spite of his group mates telling him not to tell to the teacher he developed enough courage to tell everything to his teacher this show that he is courageous and confident student who accept his mistake.

**Question: B. Shushant was doing an experiment by using an ammeter. Unfortunately, it fell from his hand and broke. Fearing the scolding of teacher, his group mates advised him not to tell the teacher, but he told her. On listening to him patiently, the teacher did not scold him as it was just and accident and used the opportunity to show the whole class the internal structure of ammeter.****What is the use of ammeter? How is it connected in the circuit? ****Answer:** An ammeter is used to measure the electric current in the unit of Ampere (A).It can also be used to verify the Ohm’s law.

It is always connected in the series. The figure below shows the ammeter and its connection in the circuit.

Ammeter

Ammeter connected in series.

The ammeter is connected in the series because it has very low resistance If it is connected in parallel across any load then all current in circuit will choose lower resistive path (i.e. ammeter) which will cause the circuit to be damaged. Hence it is used in series. Also the current in the series is always same.

**Question: 29.C. Shushant was doing an experiment by using an ammeter. Unfortunately, it fell from his hand and broke. Fearing the scolding of teacher, his group mates advised him not to tell the teacher, but he told her. On listening to him patiently, the teacher did not scold him as it was just and accident and used the opportunity to show the whole class the internal structure of ammeter.****State the aim of any one experiment where Shushant could have used the ammeter.****Answer:** Shushant could have used the ammeter to verify the ohm’s law which states that “current flowing through a conductor is directly proportional to the potential difference across the conductor”

l∝V.

After setting up the circuit (as shown in figure below) and drawing the graph between V and I will get the straight line passing through origin.

The Ammeter will measure the current and the Voltmeter will measure the potential difference across the resistor and the ratio of voltage and current will give the resistance which is constant.

**Question: Bulb is rated at 200 V, 100 W. Calculate its resistance. Five such bulbs are lighted for 4 hours daily. Calculate the units of electrical energy consumed per day. What would be the cost of using these bulbs per day at the rate of Rs. 4.00 per unit?****Answer:** We know that for an electric appliance

**Question: A. In the given circuit, connect a nichrome wire of length ‘L’ between points X and Y and note the ammeter reading. **

**(i) When this experiment is repeated by inserting another nichrome wire of the same****thickness but twice the length (2L), what changes are observed in the ammeter reading?****(ii) State the changes that are observed in the ammeter reading if we double the area of cross-section without changing the length in the above experiment. Justify your answer in both the cases.****Answer**: (i) The ammeter reading will decrease. This is because Resistance is directly proportional to the length; hence with the increase in the length the resistance of the wire will increase. So the current will decrease due to increase in the resistance (current is inversely proportional to the resistance).

(ii) The ammeter reading will increase because. R=P l/A

Thus the resistance is inversely proportional to the area of the cross section, so with the increase in the area the resistance of the wire will decrease.

Therefore with the decrease in the resistance the current will increase (current is inversely proportional to the resistance).hence the ammeter reading will increase

**Question: B. “Potential difference between points A and B in an electric field is 1 V”. Explain the above statement.****Answer:** Given that the potential difference is 1 V. The potential difference is the Difference in the potential between two points that represents the work involved or energy released in transfer of a unit quantity of electricity from one point to the other. The figure below illustrates it ** **

VA = potential at A

VB = potential at B

From the above definition of the potential difference the above statement means that 1Joule of work is being done to move a charge of 1 coulomb from point A to point B.

**Question: A. When a high resistance voltmeter is connected directly across a resister its reading is 2 V. An electric cell is sending the current of 0.4A, (measured by an ammeter) in the electric circuit in which a rheostat is also connected to vary the current.****Draw an equivalent labelled circuit for the given data.****Answer:** An equivalent labelled circuit for the given data is given below.

**Question: B. When a high resistance voltmeter is connected directly across a resister its reading is 2 V. An electric cell is sending the current of 0.4A, (measured by an ammeter) in the electric circuit in which a rheostat is also connected to vary the current. ****Find the resistance of the resister.****Answer:** Since the voltmeter reading shows 2V. Thus V = 2.

Also, Current send by electric cell = 0.4A

Thus I = 0.4A

By Ohm’s law V = IR

⇒ 2 = 0.4 × R

⇒ R=2/0.4=20/4=5Ω

The resistance of the resistor is 5Ω.

**Question: C. When a high resistance voltmeter is connected directly across a resister its reading is 2 V. An electric cell is sending the current of 0.4A, (measured by an ammeter) in the electric circuit in which a rheostat is also connected to vary the current.****Name and state the law applicable in the given case. A graph is drawn between a set of values of potential difference (V) across the resister and current (I) flowing through it. Show nature of graph thus obtained.****Answer:** The law applicable here is Ohm’s Law.

It states that the current flowing through a conductor is directly proportional to the potential difference across the conductor

l∝v.

The proportionality constant is the resistance of the conductor; it is constant for the known range of the temperature.

V = IR.

The Graph obtained is the straight line with positive slope passing through the origin as shown in the figure below.

**Question: A wire of given material having length l and area of cross-section A, has a resistance of 2 Ω. Find the resistance of another wire of same material having length 2l and are of cross-section A/2.****Answer:** Given

Hence the new resistance will be four times the previous resistance i.e. 8Ω.

**Question: B. Calculate the ratio of the equivalent resistance of the above two wires in parallel combination to that in series combination.****Answer:** The resistance of the two wires are

R_{1} = 2Ω and R_{2} = 8Ω

When they are connected in series.

Equivalent resistance = R_{1} + R_{2}

⇒ Req = 2 + 8 = 10 Ω.

Req in series = 10 Ω.

When they are connected in parallel.

Hence the ratio of the Equivalent resistances in parallel to series is 4/25

**Question: A. An electric lamp is marked 26 W, 220V. It is used for 10 hours daily. Calculate its resistance while glowing.****Answer:** Given

Power (P) = 26W

Voltage (V) = 220V

We know that

**Question: B. An electric lamp is marked 26 W, 220V. It is used for 10 hours daily. Calculate energy consumed in kWh per day.****Answer:** Energy = Power × time

⇒ Energy = 26 × 10 = 260 W

We divide by 1000 for converting in Kw

Energy Consumed =260/1000=0.26 KW

Hence the energy consumed is 0.26 KW

**Question: A. Identify the type of combination in which R1 and R2 are connected in the given circuit diagram. **

**Answer:** The resistors R_{1} and R_{2 }are connected in the parallel combination because the potential difference across both the resistors is same.

**Question: B. Find the effective resistance of the combination.**

**Answer:** Effective resistance

Effective resistance of the combination is 2Ω.