Students should refer to the below Electricity Class 10 notes prepared as per the latest curriculum issued by CBSE and NCERT. These notes and questions are really useful as they have been developed based on the most scoring topics and expected questions in upcoming examinations for Class 10. Electricity is an important topic in Science Class 10 which if understood properly can help students to get very good marks in class tests and exams.

## Electricity Class 10 Notes and Questions PDF Download

Read the notes below which will help you to understand all important and difficult topics in this chapter. There are some topics in Electricity chapter which you should understand carefully as many questions can come from those parts. Our team of teachers have designed the revision notes so that its helpful for students to revise entire course prior to the class tests.

**Question: In the given figure what is the ratio of current in A _{1} ,and A_{2}**

Ans-V=IR V=const.

I ∞ 1/R I_{1}/I_{2} = R/2R I_{1}/I_{2} = 1/ 2

**Question: A charge of 6 C is moved between two points P and Q having , potential 10V and 5V respectively. Find the amount of work done.**Ans-W=q(V

_{2}-V

_{1})=6(10-5)=30 joule

**Question: A wire of resistance R is bent in form of a closed circle, what is the resistance** **across a diameter of the circle?**Ans-1/R’ =1/(R/2)+1/(R/2) R’=R/4.

**Question: Name the physical quantity whose SI unit is JC ^{-1}.**Ans-Potential

**Question: In the given figure what is ratio of ammeter reading when J is connected to A and then to B**

Ans. when J is connected to A

I=V/R=3/5A=O.6A

When J is connected to B

V=1+2+3+4=10V

I=10/5=2A

**Question: Two cubes A and B are of the same material. The side of B is thrice as that of A.** **Find the ratio RA/RB.**

Ans-R_{A} = ρL/A R_{B=}ρ3L/9A

R_{A} : R_{B} =3:1

**Question: 3 X 1011 electrons are flowing through the filament of bulb for two minutes. Find** **the current flowing through the circuit. Charge on one electron=1.6X10-19 C.**

Ans-q=ne=3×10^{11}x1.6×10^{-19}=4.8×10^{8}C

I=q/t=4.8×10^{8}/(2×60)=4×10^{7}A.

**Question: Two wires of equal cross sectional area , one of copper and other of manganin have same resistance. Which one will be longer?**Ans-R=ρL/A (R,A=const .L=1/ρ)

ρ

_{manganin}> ρ copper

L

_{copper}>L manganin

**Question: Three equal resistances are connected in series then in parallel. What will be the ratio of their Resistances?**Ans-R

_{series}=3R.

R

_{parallel}=R/3

R

_{series}/ R

_{parallel}=3R/(R/3)=9

**Question: A Rectangular block of iron has dimensions L X L X b. What is the resistance of the block measured between the two square ends ? Given ρ= resistivity.**Ans-R=ρ b/L

^{2}

**Question: Justify for any pair of resistance the equivalent resistance in series is greater equivalent resistance in parallel .**

Ans- Since, R=V/I

R_{A}>R_{B}

A=Series, B=Parallel

**Question: How many bulbs of 8Ώ should be joined in parallel to draw a current of 2A from a battery of 4 V?**Ans-R=V/I=4/2=2 Ώ, let ‘n’ be the no of bulbs.

1/R=1/R

_{1}+ 1/R

_{2}+……..1/Rn =n/8

½=n/8, n=4

**Question: A nichrome wire of resistivity 100X10-6ohm- m and copper wire of resistivity** **1.62X10-8 ohm-m of same length and same area of cross section are connected** **in series , current is passed through them, why does the nichrome wire gets** **heated first?**

Ans. Q=I^{2} Rt

Q= I^{2} { ρ L/A}t

Nichrome wire has higher resistivity than copper wire . Therefore, it is heated first

### Important Questions Electricity Class 10 Science

**Very Short Answer Type Questions :**

**Question. Why does the cord of an electric heater not glow while the heating element does? ****Answer:** The cord of the electric heater is made up of copper or aluminium which has lower resistance and higher conductance, therefore very less heat energy is produced.

Heating element is made up of alloys which have high resistance and lot of heat is produced, and only a little amount of electric energy makes it to glow.

**Question. Out of the two wires ‘X’ and ‘Y’ of the same material as shown below, which one has 12. The material should have (i) high resistivity, (ii) high melting point.reater resistance? **

**Answer: **‘Y’ has more resistance because resistance is directly proportional to the length of the wire.

**Question. Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line, if the maximum allowable current is 5 A? ****Answer:** Total power = N × P where ‘N’ is the number of bulbs, ‘P’ power of each bulb.

∴ I = N X P /V ⇒ 5 = N × 10/220 ⇒ N = 5 X 220 /10 = 110

**Question. Why heat is produced when current is passed through a conductor?****Answer:** The electrons collide with each other while moving and loses some kinetic energy which is converted into heat energy.

**Question. An electric heater of resistance 8 W draws 15 A current from the supply mains for 2 hours. Calculate the rate at which heat is developed in the heater.****Answer:** R = 8 W, I = 15 A, t = 2 hours

Rate of heat developed = H/t = I2 X R X t/t = I2 X R

= 152 × 8 = 1800 J s–1 = 1800 Watt = 1.8 kW

**Question. A potential difference of 4 V derive a current of 3 A through a resistor. How much electrical energy will be converted into heat during 10 seconds?****Answer:** R = V/I = 3/4 = 1.33 Ω

H = I2 × R × t = 3 × 3 × 4/3 × 10 = 120 J

**Question. Explain the role of fuse wire connected in series with any electrical appliance in an electric circuit.****Why should a fuse with a defined rating for an electric circuit not be replaced by the one with a larger rating?****Answer:** Fuse wire is a safety device connected in series with the live wire of the circuit, since it has high resistivity and low melting point. It melts when a sudden urge of large current passes through it and disconnects the entire circuit from the electrical supply. But in case if we use a larger rating fuse wire instead of a defined rating fuse wire then it will not protect the circuit as high current will easily pass through it and it will not melt.

**Question. Why are heating elements made of alloys rather than metals? ****OR****Why are alloys commonly used in electrical heating devices like toasters and electric iron? Give reason.****Answer:** Alloys have high resistivity/high melting point/alloys do not oxidise (or burn readily at high temperatures).

**Question. Find the resistance of bulb rated as 100 W at 250 V. ****Answer:** P = V2/R ⇒ R = V2/P = 250 X 250/100 = 625 Ω

**Question. What is the cost of running an AC with average power of 1000 W for 8 hours for 30 days. The cost of electric energy is `4.70 per kW h.****Answer:** E = P × t = 1000 W × 8 × 30 = 240000 W h = 240 kW h

Cost of electric energy = 240 × 4.70 = ` 1128.

**Question. Why do we get electric shock in damp conditions?****Answer:** Water provides conducting path for a current to flow through the human body in damp conditions like bathroom. We get electric shock if we are bare footed. Wet body has low resistance, high current can easily pass through, leading to electric shock.

**Question. V-I graph for two wires ‘A’ and ‘B’ are shown in the figure. If both the wires are of same length and are of same thickness, which of these two is made of****the material of higher resistivity? Give justification for your answer.****Answer:** ‘A’ has greater resistivity than ‘B’. It depends upon the nature of material. High resistivity means lower value of electric current flowing through the wire.

**Question. An electric iron has a rating of 750 W, 220 V. Calculate the**

(i) Current flowing through it and (ii) Its resistance when it is in use.**Answer:** (i) P = 750 W, V = 220 V

P = V × I ⇒ 750 = 220 × I ⇒ I = 750/220 = 3.409 A

(ii) P = V2/R ⇒ R = V2/P = 220 X 220/750 ⇒ R = 64.53 Ω

**Question. Name the commercial unit of energy. Convert it into Joules. What is other name of commercial unit of energy?****Answer:** The commercial unit of electrical energy is kW h. (Kilowatt hour)

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

**Question. Find the minimum rating of fuse that can be safely used on a line on which two 1.1 kW rating electric geysers are to be run simultaneously. The supply voltage is 220 V.****Answer:** I = P/V = 1 1 X 1000/220 = 1100/220 = 5A

The rating of fuse that can be used is more than 5 A.

It is also called Board of Trade Unit (BOTU)

**Short Answer Type Question :**

**Question. Resistance of a metal wire of length 1 m is 26 W 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:** R = 26 Ω, l = 1 m, d = 0.3 mm = 3 × 10–4 m 1

**Question. The figure below shows three cylindrical copper conductors along their face areas and lengths. Discuss in which geometrical shape the resistance will be the highest.**

**Answer:**

Thus in figure (b) the resistance is maximum.

So, from equations (i), (ii) and (iii), we get R2 > R1 > R3.

Hence, the cylinder having length (2L) and area of cross-section (A/2) has the highest resistance followed by the cylinder having length (L) and area of cross section (a), while cylinder with length (L/2) and area of cross-section (2A) has the least resistance. This is because resistance of a substance is directly proportional to the length and is inversely proportional to the area of cross-section.

**Question. B1, B2, B3 are three identical bulbs connected as shown in the figure.**

When all the three bulbs glow, a current of 3 A is recorded by the ammeter A.

(i) What happens to the glow of the other two bulbs when bulb B1 gets fused?

(ii) What happens to the reading of A1, A2, A3 and A when bulb B2 gets fused. **Answer:** (i) The glow of bulbs B2 and B3 will remain same because they are in parallel connection with bulb B1.

(ii) A1 shows 1 ampere reading, A2 shows zero and A3 show 1 ampere reading. ‘A’ will show 2 A reading.

**Question. Compare the power used in the 2 W resistor in each of the following circuits: (i) a 6 V battery in series with the 1 W and 2 W resistors, and (ii) a 4 V battery in parallel with 12 W and 2 W resistors.****Answer:** (i) Rs = R1 + R2 = 1 W + 2 W = 3 Ω, V = 6 V

I = V/R = 6/3 = 2A

P = I2 × R (Since same current flows through each resistor) = 2 × 2 × 2 = 8 Ω (P R = 2 Ω)

(ii) 1/RP = 1/R1 + 1/R2 ⇒ 1/RP = 1/12 + 1/2 = 1+6/12 = 7/12 ⇒ RP = 12/7 Ω

Resistance will change, voltage remains the same.

Power across 2 W resistor, P = V2/R = 4 X 4/2 = 8W

Power used are in the ratio 1 : 1.

**Question. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electrical mains supply. What current is drawn from the line if the supply voltage is 220 V?****Answer:** R100 Ω = V2/P = 220 X 220/100 ; R60 Ω = V2/P = 220 X 220/60

I = V/R100 = 220 X 100/220 X 220 = 100/220 = 0.45 A ; I = V/R60 = 220 X 60/220 220 = 6/22 = 0.27 A

∴ Total current = 0.45 A + 0.27 A = 0.72 A

**Question. (i) Explain what is the difference between direct current and alternating current. Write one important advantage of using alternating current.**

(ii) An air conditioner of 2 kW is used in an electric circuit having a fuse rating of 10 A. If the potential difference of the supply is 220 V, will the fuse be able to withstand when the air conditioner is switched on? Justify your answer. **Answer:** (i) The current whose direction gets reversed after every half cycle is called alternating current or A.C.

There is no change in the direction of D.C. D.C. is a uni directional current.

The most important advantage of using A.C. over D.C. is that in the A.C. mode electric power can be transmitted over long distances at a high voltage and low current with very little loss of power.

(ii) Here P = 2 kW = 2000 W, V = 220 Volt

P = VI, so the current, I = P/V = 2000/220 = 9.09 A

As the current is 9.09 A below the rating of fuse, therefore the fuse will withstand, i.e. it will not blow off when A.C. is on.

**Question. Calculate the resistance of 1 km long wire of copper of radius 1 mm. (Resistivity of copper is 1.72 × 10–8 W m.) ****Answer:**

**Question. The resistance of a wire of 0.01 cm radius is 10 W. If resistivity of the material of wire is 50 × 10–8 ohm metre, find the length of the wire. ****Answer:**

**Question. What is meant by ‘electrical resistance’ of a conductor? State how resistance of a conductor is effected when**

(i) Low current passes through it for a short duration.

(ii) A heavy current passes through it for 30 seconds. **Answer: **It is in opposite direction offered to the flow of electric current.

(i) No effect on resistance, low current, hence no appreciable rise in temperature so there no change in resistance.

(ii) Heavy current for 30 seconds may increase the temperature so resistance will increase.

**Question. An electric kettle of 2 kW works for 2 hours daily. Calculate the**

(a) Energy consumed in SI unit and commercial unit.

(b) Cost of running it in the month of June at a rate of ` 3.00 per unit. **Answer:** (a) P = 2 kW, t = 2 h

E = P × t = 2 × 2 h = 4 kW h

(b) Total energy consumed per month = 4 kW h × 30 = 120 kW h

Total cost = 120 × 3 = ₹ 360.

**Question. A piece of wire of resistance 20 W is drawn out of a material so that its length is increased to twice of its original length. Calculate the resistance in the new situation. **

**Answer:**

**Question. An electric fan has a rating of 460 W on the 230 V mains line. What fuse should be fitted in the plug?****Answer:** P = V × I ⇒ 460 = 230 × I ⇒ I = 2 Ampere

A 3A fuse should be fitted (It should be slightly more than 2A).

**Question. Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 4 W in series with a combination of two resistors (8 W each) in parallel and a voltmeter across parallel combination. Each of them dissipate maximum energy and can withstand a maximum power of 16 W without melting. Find the maximum current that can flow through the three resistors.****Answer:**

**Question. Two lamps, one rated 100 W at 220 V and the other 200 W at 220V are connected (i) in series and**

(ii) in parallel to electric main supply of 220V. Find the current drawn in each case.**Answer:**

**Question. A student has two resistors- 2 W and 3 W. She has to put one of them in place of R2 as shown in the circuit. The current that she needs in the entire circuit is exactly 9A. Show by calculation which of the two resistors she should choose.****Answer:** The overall current needed = 9A. The voltage is 12V

Hence by Ohm’s law V=IR,

The resistance for the entire circuit = 12/9 = 4/3 Ω. = R

Hence, R = (R1R2)/(R1+R2) = 4R2/(4 + R2) = 4/3

**Long Answer Type Questions :**

**Question. Explain the following:**

(i) Why is tungsten used almost exclusively for filament of electric lamps?

(ii) Why are the conductors of electric heating devices, such as bread-toaster and electric iron, made of an alloy rather than a pure metal?

(iii) Why is series arrangement not used for domestic circuits?

(iv) How does the resistance of a wire vary with its area of cross-section?

(v) Why are copper and aluminium wires usually used for the transmission of electric current?**Answer:** (i) It has high resistance and high melting point. So it does not melt when current is passed through it.

(ii) Alloys have more resistivity and higher melting point.

(iii) It increases the resistance and current decreases. Also if one component fails to work, others will also not work

(iv) Resistance is inversely proportional to the area of cross section.

(v) It is because Cu and Al have low resistivity and it allows current to flow.

**Question. (a) Define electric power. Express it in terms of potential difference, V and resistance, R.**

(b) An electrical fuse is rated at 2 A. What is meant by this statement?

(c) An electric iron of 1 kW is operated at 220 V. Find which of the following fuse that are respectively rated at 1 A, 3 A and 5 A can be used for it.**Answer:** (a) Electric power is the rate of doing work by an energy source or the rate at which the electrical energy dissipated or consumed per unit time in the electric circuit.

So, Power (P) = Work done (W)/Time (t)

= Electrical energy dissipated/Time (t)

⇒ VI or P = V2/R

(b) It means the maximum current will flow through it is only 2 A. Fuse wire will melt if the current flowing through it exceeds 2 A value.

(c) Given: P = 1 kW = 1000 W, V = 220 V

Current drawn, I = P/V = 1000/220 = 50/11 = 4.54 A

To run electric iron of 1 kW efficiently, rated fuse of 5 A should be used.

**Question. (a) Write two points of differences between electric energy and electric power.**

(b) Out of 60 W and 40 W lamps, which one has higher electrical resistance when in use.

(c) What is the commercial unit of electric energy? Convert it into the units of Joule.**Answer:** (a) Difference between electric energy and electric power:

(b) For the same applied voltage, P ∝ 1/R , i.e. smaller the power of electrical device, higher is its electrical resistance.

So a 40 W lamp has higher electrical resistance than a 60 W lamp.

(c) Kilowatt hour – Commercial unit of electrical energy.

1 kW h = 1000 W h = 1000 J/s × 3600 s = 3600000 J = 3.6 × 106 J

**Question. Two identical wires, one of nichrome and other of copper are connected in series and a current (i) is passed through them. State the change observed in the temperature of the two wires. Justify your answer. State the law which explains the above observations.****Answer:** The resistivity of nichrome is more than that of copper, so its resistance is also high. Therefore large amount of heat is produced in the nichrome wire for the same current passed as compared to that of the copper wire. Accordingly greater change in temperature is observed in case of nichrome wire. This can be explained by Joule’s law of heating.

Joule’s law of heating states that the amount of heat produced in a conductor is:

(i) Directly proportional to the square of current flowing through it, i.e.

H ∝ I2

(ii) Directly proportional to the resistance offered by the conductor to the flow current, i.e.

H ∝ R

(iii) Directly proportional to the time for which current is flowing through the conductor, i.e.

H ∝ t

Combining these, we get, H ∝ I2Rt, or H = KI2Rt

where K is a proportionality constant and in SI units, it is equal to unity.

**Question. (i) Write an expression for the resistivity of a substance.**

(ii) State the SI unit of resistivity.

(iii) Distinguish between resistance and resistivity.

(iv) Name two factors on which the resistivity of a substance depends and two factors on which it does not depend. **Answer:** (i) Resistivity (r) = RA/l (ii) Its SI unit is Ω m.

(iii) Resistivity is a characteristic property of a material that does not depend upon the dimensions of the material whereas resistance depends upon the dimensions of the material.

(iv) Resistivity does not depend on the:

(a) length of conductor, (b) area of cross section of conductor Resistivity depends on the:

(a) nature of material of conductor (b) temperature of conductor

**Question. An electric circuit two resistors of 20 W and a conductor of resistance 4W are connected to a 6 V vattery as shown in the figure. Calculate: **

(a) The total resistance of the circuit,

(b) the current through the circuit,

(c) the potential difference across the (i) electric lamp and (ii) conductor, and

(d) power of the lamp.**Answer: **(a) Rs = R1 + R2 = 20 Ω + 4 Ω = 24 Ω

(b) I = V/R = 6V/24 Ω = 0.25A

(c) (i) V = I × R = 0.25 A × 20 W = 5 V

(ii) V = I × R = 0.25 A × 4 = 1V

(d) (i) P = I2 × R = (1/4)2 X 20 = 1/16 X 20 = 1.25 W

**Question. (a) State Ohm’s law. Derive the relation and give graphical representation for it.**

(b) An electric oven rated at 500 W is connected to a 220 V line and used for 2 hours daily. Calculate the cost of electric energy per month at the rate of ` 5 per kW h.**Answer:** (a) Linear equation, hence a straight line passing through the origin is obtained as follows:

Ohm’s law states that potential difference across the given metallic wire in an electric circuit is directly proportional to the electric

current flowing through it, if the temperature remains constant.

Mathematically, V ∝ I.

or V/I = Constant

or V = IR (where R is a constant known as the

constant of proportionality).

(b) Energy consumed per day = 1 kW h.

∴ P = V × I, E = P × t ⇒ E = V × I × t = 220 × 2.27 × 2 = 1 kWh

So, I = V/P = 500/220 = 2.27 A (∴ P = V X I ∴ I = 500/220)

**Question. (a) What is the function of fuse wire in an electric circuit?**

(b) What would be the rating of the fuse for an electric kettle which is operated at 220 V and consumes 500 W power?

(c) How is the SI unit of electric energy related to its commercial unit?**Answer:** (a) Fuse wire protects the electrical circuits from over voltage and high current.

(b) 2.2 A flows through the circuit, fuse should be rated 3 A.

(c) 1 kW h = 3.6 × 106 J

**Question. (i) Consider a conductor of resistance ‘R’, length ‘L’, thickness ‘d’ and resistivity ‘r’.**

Now this conductor is cut into four equal parts. What will be the new resistivity of each of these parts? Why?

(ii) Find the resistance if all of these parts are connected in: (a) Parallel (b) Series

(iii) Out of the combinations of resistors mentioned above in the previous part, for a given voltage which combination will consume more power and why?**Answer:** (i) Resistivity will not change as it depends on the nature of the material of the conductor.

(ii) The length of each part becomes L/4. While r, A remain constant. ∴ R = rL/A.

Resistance of each part = Rpart = (rL/4)/A = R/4.

(a) In parallel connection, 1/Reqv + 1/Rpart + 1/Rpart + 1/Rpart + 1/Rpart = 4/Rpart = 16 /R ⇒ Reqv = R/16 Ω

(b) In series connection, Reqv = R/4 + R/4 + R/4 + R/4 = = R Ω (ohms)

(iii) P = V2/R.

If Reqv is less, power consumed will be more.

In the given case, Reqv is lesser in the parallel connection and the power consumed will be more.

**Question. List two distinguishing features between the resistance and resistivity of a conductor. A wire is stretched so that its length becomes 6/5 times of its original length. If its original resistance is 25 Ω, find its new resistance and resistivity. Give justification for your answer in each case. ****Answer:**

Since resistance is directly proportional to the length and is inversely proportional to the area of cross section.

∴ New Resistance: R = 6/5 X 25/(5/6)2 = 150/5 X 25 X 36 = 216/5 = 43.20 ohms

When the wire is stretched, its length becomes 6/5 times of its original length, and area of cross-section will become 5/6. Resistivity will remain the same because it does not depend upon the length and area of cross section. It depends on the nature of material of the substance and the temperature.

**Question. (i) What is meant by series combination and parallel combination of resistances? 1**

(ii) In the circuit diagram given below five resistances of 5Ω, 20Ω, 15Ω, 20Ω and 10 W are connected as given in the figure to a 6 V battery.

Calculate: (a) Total resistance in the circuit.

(b) Total current flowing in the circuit.**Answer:** (i) Resistors in Series: When resistors are joined end to end, it is called series combination of resistances.

In series combination of resistors, the current remains the same through each resistor. Therefore the value of current in the ammeter remains the same, independent of its position in the electric circuit.

In the above circuit, V is equal to the sum of V1, V2, V3, that is the total potential difference (V) across a combination of resistors in series is equal to the sum of potential differences across the individual resistors.

V = V1 + V2 + V3

The current through each resistor is I.

V = IR

V1 = IR1 V2 = IR2 V3 = IR3

As V = V1 + V2 + V3 ⇒ \ IR = IR1 + IR2 + IR3 ⇒

RS = R1 + R2 + R3

∴ We can conclude that when several resistors are joined in series, then equivalent resistance of the combination is equal to the sum of their individual resistances and is thus greater than any of the individual resistance.

∴ Resistors in Parallel:

When the resistances, says, R1, R2, R3 are connected in in parallel combination as shown in the diagram, then the potential difference remains the same across each resistor in parallel combination of resistors. In parallel combination the current is distributed in different resistances in different branches.

**Question. You have been assigned a duty to create awareness in your school about saving electricity.**

(i) Write any two ways by which you will create awareness among your schoolmates about saving electricity.

(ii) Explain how saving electricity is important at individual level as well as at national level. **Answer:** (i) Speaking over it in the morning assembly in school.

• Putting posters over it on the school notice board.

(ii) If we save electricity, it can be used by those villages which do not have electricity.

• It can be used in industries, agriculture and for other useful purposes.

• It improves the national economy because high speed trains, industries, development in villages depends upon electricity.

**Question. (i) Draw a closed circuit diagram consisting of a 0.5 m long nichrome wire XY, and ammeter, a voltmeter, four cells of 1.5 V each and a plug key. **

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

What would be the values of V/I 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:**

**Question. Study the following electric circuit and find (i) the current flowing in the circuit and (ii) the potential difference across the 10 W resistor. **

**Answer:** 10 W and 20 W resistances are connected in series, therefore their equivalent resistance is given by: R1

= R1 + R2 = 10 + 20 = 30 W

(i) Current flowing in the circuit, I = V/RS = 3/30 = 1/10 = 0.1 A

(iii) Potential difference across the 10 W resistor, V = IR = 1/10 X 10 = 1 volt

**Question. In the given circuit, A, B, C and D are four lamps connected with a battery of 60V. 7**

Analyse the circuit to answer the following questions.

(i) What kind of combination are the lamps arranged in (series or parallel)?

(ii) Explain with reference to your above answer, what are the advantages (any two) of this combination of lamps?

(iii) Explain with proper calculations which lamp glows the brightest?

(iv) Find out the total resistance of the circuit.**Answer:** (i) The lamps are in parallel.

(ii) Advantages:

If one lamp is faulty, it will not affect the working of the other lamps.

They will also be using the full potential of the battery as they are connected in parallel.

(iii) The lamp with the highest power will glow the brightest.

P = VI

In this case, all the bulbs have the same voltage. But lamp C has the highest current.

Hence, for lamp C, P = 5 × 60 Watt = 300 W. (the maximum).

(iv) The total current in the circuit = (3 + 4 + 5 + 3) A = 15A

The Voltage = 60V

V = IR and hence R = V/I

R = 60/15 W = 4 W

**Question. Find the net current flowing through the following electric circuit: **

**Answer:** Series combination of 1 W and 3 W resistance is in parallel combination with 6 W resistor. Their equivalent resistance is

1/RP = 1/6 + 1/3+1 = 1/6 + 1/4 = 2 + 3/12 ⇒ RP = 12/5 2.4 Ω

Now, 3.6 W, 2.4 W and 3 W resistors are connected in series, therefore their equivalent resistance be given by: RS = R1 + R2 + R3 = 3.6 + 2.4 + 3 = 9 Ω

Hence, the current flowing through the circuit is, I = V/R = 4.5/9 = 45/90 = 1/2 = 0.5 A

**Question. (a) With the help of 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 reciprocal of the individual resistances.**

(b) In an electric circuit two resistors of 12 W each are joined in parallel to a 6V battery. Find the current drawn from the battery.

**Answer:** (a) Refer to Answer 1(i) — Long Answer Type Questions.

(b) 1/RP = 1/R1 – 1/R2 = 1/12 + 1/12 = 1+1/12 = 1/12 ⇒ Rp = 6 Ohms

I = V/R = 6V/6ohm = 1 Ampere

**Question. Figure shows a 3 W resistor and a 6 W resistor connected in parallel across a 1.5 V cell.**

Calculate the current across: (i) 3 Ω resistor, (ii) 6 Ω resistor, (iii) the cell, (iv) calculate the resistance of the parallel combination of 3 Ω and 6 Ω resistors.**Answer:** (i) I1 = V/R ⇒ I1 = 1.5 /3 = 0.5 A , (ii) I2 V/R ⇒ I2 = 1.5/6 = 0.25 A

(iii) ITotal = I1 + I2 = 0.50 + 0.25 = 0.75 A (iv) 1/RP = 1/R1 + 1/R2 = 1/3 + 1/6 = 3/6 = 1/2 ⇒ Rp = 2Ω

**Question. State Ohm’s law. Draw a labelled circuit diagram to verify this law in the laboratory. If you draw a graph between the potential difference and current flowing through a metallic conductor, what kind of curve will you get? Explain how would you use this graph to determine the resistance of the conductor.****Answer:** Ohm’s Law: It states that ratio of potential difference and current is constant and is equal to the resistance of the conductor at a particular temperature. In other words, the current flowing through a conductor is directly proportional to the potential difference at a constant temperature.

The graph between V and I will be a straight line.

Slope = tan θ = ‘R’

Slope = BC/AC = R