Please refer to Electricity Class 10 Science Important Questions given below. These solved questions for Electricity have been prepared based on the latest CBSE, NCERT and KVS syllabus and books issued for the current academic year. We have provided important examination questions for Class 10 Science all chapters.

**Class 10 Science** **Electricity** **Important Questions**

**Very Short Answer Type Questions :**

**Question. When is potential difference between two points said to be 1 volt?****Answer : ** Potential difference betweeen two point is said to be 1 Volt if the amount of work done in bringing 1 C charge between them is 1 J.

**Question. An electric iron draws a current of 0.5 A when the voltage is 200 volt. Calculate the amount of electric charge flowing through it in one hour.****Answer : ** We have I = 0.5 A

V = 200 volt

t = 1 hr = 3600 s

Q = I x t = 0.5#3600 A – s = 1800 C

**Question. An electric appliance draws a current of 0.4 A when the voltage is 200 volt. Calculate the amount of charge flowing through it in one hour.****Answer : ** Q = It

Given, I = 0.4 A

V = 200 Volt

t = 1 hr = 3600 s

` Q = 0.4 x 3600 C

Q = 1440 C

**Question. State Ohm’s law.****Answer : ** If the physical conditions of a conductor remain same then current through a conductor is directly proportional to the potential difference b/w the two ends of the conductor.

I ? V & V = IR

**Question. Draw a schematic diagram of a circuit consisting of a cell of 1.5 V, 10 ohm resistor and 15 ohm resistor and a plug key all connected in series.****Answer : ** Schematic diagram is shown below.

**Question. Why do we use copper and aluminium wire for transmission of electric current?****Answer : ** Copper and aluminium wires are used for electric transmission due to their low resistivity.

**Question. If the charge on an electron be 1.6#10-19C, find the approximate number of electrons in 1 C.****Answer : ** 1.6×10-19C charge is of = 1 electron and 1 C charge is of = 1/1.6 x 10^{-19} electron No. of electrons = 6.25×10^{18}

**Question. List any two factors on which resistance of a conductor depends.****Answer : ** Resistance of a conductor:

a. is directly proportional to its length R ? r …(1)

b. is inversely proportional to its area of cross section.

R ∝ 1/A

Combining (1) and (2), we get R ∝ P/A

**Question. What is the SI unit of electric potential?****Answer : ** Volt is the SI unit of electric potential.

**Question. Give an example of a metal which is the best conductor of heat.****Answer : ** Gold, Silver, Copper etc. metals are good conductor of heat.

**Question. Define electric circuit. Distinguish between open and closed circuit.****Answer : ** Electric circuit is the arrangement in which electric current can flow when circuit is switched on. In open circuit there is no flow of current as the switch is off.

In closed circuit a current flows in the circuit when switch is on.

**Question. What is the lowest resistance that can be obtained by combining four coils of resistances 4 W , 8 W , 12 W and 24 W ?****Answer : ** If the resistances are combined in parallel then we shall get the lowest resistance. In parallel combination:

**Question. Nichrome is used to make the element of electric heater. Why?****Answer : ** Nichrome is used to make element of electric heater because nichrome is an alloy which has high melting point and high resistances.

**Question. happens to the resistance of a conductor when the length of the conductor is reduced to half?****Answer : ** Resistance is directly proportional to the length of the conductor. If length becomes half the resistance also become half of its initial value.

**Question. What happens to resistance of a conductor when its area of cross-section is increased?****Answer:**

**Question. Two resistors of 10 Ω and 15 Ω are connected in series to a battery of 6 V. How can the values of current passing through them be compared?****Answer:** In series, same current flows through each resistor. So, ratio of current is 1 : 1.

**Question. Name a device that helps to maintain a potential difference across a conductor.****Answer:** Cell or battery

**Question. Write relation between heat energy produced in a conductor when a potential difference V is applied across its terminals and a current I flows through for ‘t’****Answer:** Heat produced, H = VIt

**Question. A given length of a wire is doubled on itself and this process is repeated once again. By what factor does the resistance of the wire change?****Answer:** Am. Length becomes one-fourth of the original length and area of cross-section becomes four times that of original.

**Question. State difference between the wire used in the element of an electric heater and in a fuse wire.****Answer: **The wire used in the element of electric heater has a high resistivity and have a high melting point, i.e. even at a high temperature element do not burn while fuse wire have a low melting point and high resistivity.

**Question. How is an ammeter connected in a circuit to measure current flowing through it?****Answer:** In series

**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:**

**Question. Write S.I. unit of resistivity.****Answer:** Ohm-metre (Ωm).

**Question. What happens to the resistance of a conductor when temperature is increased?****Answer : ** The resistance of a conductor increases with rise in temperature.

**Short Answer Type Questions :**

**Question.The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if it operates on a 6 V battery (Neglect the change in resistance due to unequal heating of the filament in the two cases).****Answer:**

**Question. Consider the following circuit diagram. If R _{1 }= R_{2} =R_{3} = R_{4}= R_{5} = 3 Ω, find the equivalent resistance(Rs) of the circuit.**

**Answer:**

**Question.The charge possessed by an electron is 1.6 X 10-19 coulombs. Find the number of electrons that will flow per second to constitute a current of 1 ampere.****Answer:**

**Question. Explain the role of fuse in series with any electrical appliance in an electric circuit. Why should a fuse with defined rating for an electric circuit not be replaced by one with a larger rating?****Answer:** Fuse wire is a safety device connected in series with the live wire of circuit. 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 instead of a defined rating, then it will not protect the circuit as high current will easily pass through it and it will not melt.

**Question. Draw a schematic diagram of an electric circuit consisting of a battery of five 2 V 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 Ωresistor.****Answer:** Req = 20 + 30 = 50 Ω img

Current through both 20 Ωand 30 Ω= I =10/50

= 0.2 A

Power consumed by 30 Ω = I2R = (0.2)2 x 30=1.2W

**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:**

**Question. How much current will an electric bulb draw from 220 V source if the resistance of the bulb is 1200Ω? If in place of bulb, a heater of resistance 100 Ω is connected to the sources, calculate the current drawn by it.****Answer:**

**Question. What is an electric circuit? Distinguish between an open and a closed circuit.****Answer: ** An arrangement for maintaining the continuous flow of electric current by the electrical energy source through the various electrical components connected with each other by conducting wires is termed as electric circuit.

An open circuit does not carry any current, while a closed circuit carries current.

**Question. (i) Draw a closed circuit diagram consisting of a 0.5 m long nichrome wire XY, an 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 10 Ω resistor.**

**Answer:**

**Question. Find the current drawn from the battery by the network of four resistors Shown in the figure.**

**Answer:**

**Question. Define 1 volt. Express it in terms of SI unit of work and charge calculate the amount of energy consumed in carrying a charge of 1 coulomb through a battery of 3 V.****Answer:** When 1 joule of work is done in carrying 1 coulomb of charge, from infinity to a point in the electric field, then potential at that point is called 1 volt. Potential difference between two points is

**Question. V-I graph for two wires A and B are shown in the figure. If both wires are of same length and same thickness, which of the two is made of a material of high resistivity? Give justification for your answer.**

**Answer: **Greater than slope of V-I graph, greater will be the resistance of given metallic wire. In the given graph, wire A has greater slope then B. Hence, wire A has greater resistance.

For the wires of same length and same thickness, resistance depends on the nature of material of the wire,

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

**Answer:**

**Question. V-I graphs for two wires A and B are shown in the figure. If both the wires are made of the same material and are of equal thickness, which of the two is of more length? Give justification for your answer.****Answer : ** We know for identical wire more length more resistance and vice versa slope of wire A is more than B. Hence resistance of A is more and its length also.

**Question. In an experiment to study the relation between the potential difference across a resistor and the current through it, a student recorded the following observations: **

**On examine the above observations, the teacher asked the student to reject one set of readings as the values were out of agreement with the rest. Which one of the above sets of readings can be rejected? Calculate the mean value of resistance of the resistor based on the remaining four sets of readings.****Answer: **The third reading for V = 3.0 volt and I — 0.6 A will be rejected as it has larger deviation from the rest of the readings.

The value of resistance in the other four observations will be I (using R = V/I) 10Ω, 11 Ω, 10 Ω and 10.67 Ω.

So, the mean value of resistance = 41.67/4 = 10.417 = 10.42 Ω

**Question. What is meant by “electrical resistance” of a conductor? State how resistance of a conductor is affected when****a. a low current passes through it for a short duration;****b. a heavy current passes through it for about 30 seconds.****Answer : ** Electrical resistance is the property of a conductor by virtue of which it opposes the flow of current through it. It is equal to the ratio of the potential difference applied across its ends to the current flowing through it.

R = V/I

a. When a low current is passed for a short duration, through a conductor, heat produced is almost negligible and hence no appreciable change in its resistance.

b. When heavy current is passed through the conductor for 30 s. Conductor may be get heated and its resistance and resistivity change.

**Question. Name and define the SI unit of current. Calculate the number of electrons that flow through a conductor in 1 second to constitute a current of 1 ampere. (Charge on an electron = 1.6 X 10 ^{-19} coulomb)**

**Answer :**SI unit of current is Ampere (A)

I = q/t

If q = 1 C, t = 1 s

then I = 1A

If 1 C charge flows in 1 s in a conductor then magnitude of current is said to be 1A.

q = ne

n = q/e = I x t/e

= 1A x 1e/1.6 x 10

^{-19}= 100/16 x 10

^{18}

= 6.25 x 10

^{18}

**Question. Electric current flows through three lamps when arranged in (a) a series (b) a parallel. If the filament of one lamp breaks. Explain what happens to the other two lamps in both the cases.****Answer : **

a. In series combination if the filament of one lamp breaks then the circuit will be broken and hence other lamps stops glowing.

b. In parallel combination of lamps if the element of one lamp breaks then other two will continue to glow.

**Question. Study the V-I graph for a resistor as shown in the figure and prepare a table showing the values of I (in amperes) corresponding to four different values of V (in volts). Find the value of current for V = 10 volts.****How can we determine the resistance of the resistor from this graph? **

**Answer : ** When V = 10 volt from the graph

**Question. Find the number of electrons transferred between two points kept at a potential difference of 20 V if 40 J of work is done.****Answer : ** Given: V = 20 Volt

W = 40 J

W = P x t

= Vx I x t

= V X Q/t x t

or W = VxQ

40 = 20xQ

or Q = 2C

ne = Q

n = Q/e = 2/1.6 x 10^{-19}

= 1.25 x 10^{19}

**Question. Name and define SI unit of resistance. Calculate the resistance of a resistor if the current flowing through it is 200 mA, when the applied potential difference is 0.8 V.****Answer : **

a. SI unit of resistance is ohm Ω

1 Ω = 1V/1A

b. The resistance of a conductor is said to be 1 ohm if a current of one amp flows through it when a potential differences of 1 volt is applied across it.

c. R = V/I = 0.8/200 X 10-5 = 4 Ω

**Question. (a) Explain why a conductor offers resistance to the flow of current.****(b) Differentiate between conductor, resistor and resistance.****Answer : **

a. When a current is passed through a conductor, the atoms or molecule of the conductor produce an hindrance in the path of flow of electron. This hindrance in the path of flow of charge is called resistance of the conductor.

b. A substance which allow to pass the charges through them easily is called a conductor.

Resistor : A conductor having some value of resistance is called a resistor.

Resistance : It is the property of any conductor by virtue of which it opposes the flow of charge through it.

**Question. A piece of wire of resistance 6 W is connected to battery of 12 V. Find the amount of current flowing through it. Now, the same wire is redrawn by stretching it to double its length. Find the resistance of the new (redrawn) wire.****Answer : ** Given: R = 6 Ω

V = 12 Volt

I = ?

**Question. A wire of length l and resistance R is stretched so that the length is doubled and area of cross-section is halved. How will its:****a. Resistance change?****b. Resistivity change?****Answer : **

a. Let initial length, area of cross-section and resistance of wire are l , A and R respectively.

When length is stretched two times, let its area of cross-section becomes Al. Its initial volume of wire = final volume of the wire.

Al = Al2l

or A’/A = 1/2 or A’ = A/2

R’ = P,(21)/A’ = P,21 X 2/A

= 4P,l/A

= 4 R

b. Resistivity does not change because it is property of the material of a conductor.

**Question. Show how would you join three resistors, each of resistance 9 W so that the equivalent resistance of the combination is (i) 13.5 W (ii) 6 W ?****Answer : ** or

(a) Write Joule’s law of heating.

(b) Two bulbs, 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. Joules Law of Heating : When a current I is passed through a resistor R for time t then heat produced

H ∝ I2

∝ R

∝ t

H ∝ I2Rt

or H = I2Rt

b. Given two bulbs of 100 W–200 V and 60W–220 V.

Current in 1st bulb:

I1 = P1/V = 100/220 A = 5/11 = 0.45 A

Current in second bulb.

I2 = P2/V = 60/220 A = 3/11 A = 0.27 A

**Question. (a) List the factors on which the resistance of a conductor in the shape of a 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. Factors on which resistance of a wire depends:

i. Resistance is directly proportional to length.

ii. Resistance is inversely proportional to area of cross-section.

R ∝ l , R ∝ 1/A

R ∝ 1/A or R = P,1/A

b. Metal are good conductor due to having large number of free electrons and their low resistivity.

Glass is a bad conductor because it has no free electrons and its resistivity is higher.

c. Alloys are commonly used in electrical heating devices due to their high resistivity and high melting point which produces more heat.

**Question. A hot plate of an electric oven connected to a 220 V line has two resistors A and B each of 22 W resistance.****These resistors may be used separately, in series or in parallel. Find the current flowing in all the three cases.****Answer : **Given, V = 220 Volt

RA = 22 W

RB = 22 W

a. Current in resistances separately?

= 220/22 A = 10 A

b. Current in series combination

RS = 22 + 22 = 44 Ω

IS = 220/44 = 5 A

c. Current in parallel combination

1/RP = 1/22+ 1/22 = 2/22 Ω

RP = 11Ω

IP = 220/11 = 20 A

**Question. An electric kettle of 2 kW works for 2 h daily. Calculate the (a) energy consumed in SI and commercial units (b) cost of running it in the month of June at the rate of `3.00 per unit.****Answer : ** (a) Given: P = 2 kW = 2000W

t = 2 h

Electric energy, E = P#t = 2×2 = 4 kWh

(b) Total energy consumed in month of June (having 30 days)

Electric kettle = Ω 4X30h kWh = 120 kWh

= 120 units.

Cost of running electric kettle:

= `120 X 3 = `360

**Question. Two lamps, one is rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to a 220 V supply. Find the current drawn from the supply line.****Answer : ** Given: Lamp one 100 W, 220 V, Lamp 2 60 W, 220 V.

Let their resistances are R1 and R2

**Long Answer Type Questions :**

**Question. a. What is meant by saying that the potential difference between two points is 1 volt?****b. Why does the connecting cord of an electrical heater not glow while the heating element does?****c. Electrical resistivity of some substance at 20°C are given below: **

**Answer the following questions in relation to them:****i. Among silver and copper which one is better conductor? Why?****ii. Which material would you advise to use in electrical heating device? Why?****Answer : **

a. If the amount of work done in bringing a unit positive charge from one point to another is 1 J in an electric field then potential difference b/w the two points is said to be 1 volt.

1V = 1J/1C

b. Connected cord of an electrical heater does not glow because this is of copper metal, which has low resistivity and good conductivity. Heating element glows due to its high resistivity or poor conductivity.

c. (i) The resistivity of silver is lesser than copper so it is a good conductor.

(ii) The resistivity of nichrome is maximum in the given table so nichrome will be advised to use in electrical heating device.

**Question. (a) Name an instrument that measures electric current in a circuit. Define unit of electric current.****(b) What are the following symbols mean in an electric circuit? **

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

**Answer : **

a. Ammeter.

If IC charge flows in an electric circuit is 1 s then the current is said to be 1 A.

b. (i) Rheostat (ii) Closed key 99

**Question. What is meant by resistance of a conductor? Name and define its SI unit. List the factors on which the resistance of a conductor depends. How is the resistance of a wire affected if:****a. its length is doubled,****b. its radius is doubled?****Answer : ** Property of any conductor by virtue of which it opposes the flow of current in the conductor is called its resistance.

SI unit of resistance is Ohm. If by applying a potential difference of 1 volt the current in the conductor is 1 A.

Then the resistance of the conductor is said to be 1 ohm.

Factor affecting resistances:

a. If length is double then resistance also becomes doubled.

b. If radius is doubled then area A = p π2rh2 becomes

4 times, then the resistance becomes 1/4.

**Question. What do you mean by heating effect of electric current?****Explain the production of heat in a resistor by flow of electric current through it. Name two devices based on heating effect of current.****Answer : ** When current is passed through a conductor, heat is produced. Production of heat in a conductor on passing the current is called heating effect of current.

When a current is passed in a conductor by applying a potential difference, electrons get accelerated and collide with the atoms of the conductor. During the collision there is a loss of kinetic energy. This loss in K.E. appears in the form of heat energy in the conductor 1. Electric heater, 2. Bulb.

**Question. (a) Name an instrument that measures potential difference between two points in a circuit. Define the unit of potential difference in terms of SI unit****of charge and work. Draw the circuit symbols for (i) variable resistor, (ii) a plug key which is closed one.****(b) Two electric circuits I and II are shown below “**

**(i) Which of the two circuits has more resistance?****(ii) Through which circuit more current passes?****(iii) In which circuit, the potential difference across each resistor is equal?****(iv) If R1 > R2 > R3 in which circuit more heat will be produced in R1 as compared to other two resistors?****Answer : **

a. Voltmeter

The amount of work done in bringing a unit positive charge from one point to another in an electric field is said to be potential difference

b. (i) In series combination resistance is more than parallel combination.

(ii) Lesser the resistance more the current in circuit i.e., in parallel (II) current is max.

(iii) In parallel combination (II)

(iv) More heat in (I) across R1

**Question. When a high resistance voltmeter is connected directly across an electric bulb, its reading is 2 V. An electric cell is sending the current of 0.4 ampere (measured by an ammeter) in the electric circuit.****a. Draw the circuit.****b. Find the resistance of the electric bulb,****c. State the law that is applied for making these calculation. If a graph is plotted between V and I, show the nature of the graph obtained.****Answer : **

c. Ohm’s law: If the physical conditions of a conductor is kept constant then current in the circuit is directly proportional to the potential difference applied across the ends of the conductor.

Since V ? I

Hence graph between V and I is a straight line for a conductor. Which passes through the origin O of the graph.

**Question. The value of current (I) flowing through a given resistor of resistance (R), for the corresponding values of potential difference (V) across the resistor are as given below :**

**Plot a graph between current (I) and potential difference (V) and determine the resistance (R) of the resistor.****Answer : **

**Question. (a) Explain how does a cell maintain current in a circuit.****(b) In the circuit given below the resistance of the path xTy = 2 W and that of xZy = 6 W****(i) Find the equivalent resistance between x and y.****(ii) Find the current in the main circuit.****(iii) Calculate the current that flows through the path xTy and xZy. **

**Answer : **

a. Potential difference b/w the two terminals of a cell is produced due to chemical reaction in the cell. This potential difference maintains a current in the circuit.

b. (i) The resistance b/w XZY and XTY are in parallel combination. Their equivalent resistance.

**Question. (a) State the commercial unit of electric energy and find its relation with its SI unit.****(b) The current through a resistor is made three times its initial value. Calculate how it will affect the heat produced in the resistor.****(c) Find the increase in the amount of heat generated in a conductor if another conductor of double resistance is connected in the circuit keeping all other factors unchanged.****Answer : **

a. Commercial unit of electric energy = kWh

1 kWh = 3.6X106 J

b. Initial heat generated in the resistor = I2Rt

H1 = I2Rt …(1)

when current is made three times i.e. 3I now heat generated

H2 = ^3I h2Rt …(2)

H2 = 9I2Rt

In later case, heat generated is 9 times the initial heat

generated.

c. If another conductor of 2R is connected in series

then total resistance = R + 2R = 3R.

Now heat generated H = I2 ^3Rht

H = 3I2Rt

In this case, heat generated is three times.

**Question. (a) Derive an expression to find the equivalent resistance of three resistors connected in series.****Also, draw the schematic diagram of the circuit.****(b) Find the equivalent resistance of the following circuit: **

**Answer : **

**Question. (a) Name and state the law that gives relationship between the current through a conductor and the potential difference across its two terminals. Also, express this law mathematically.****(b) Draw the V-I graph for this law. Justify your answer.****(c) Write the name and use of the circuit components whose symbols are given below. **

**Answer : **

a. The law is Ohm’s law.

If the physical conditions of a conductor is kept constant then current through it is directly proportional to the potential difference applied across it.

V ∝ I or V = RI

b. Since V ∝ I so a graph b/w V and I is a straight line.

c. (i) Symbol is of variable resistor and it is used to regulate the current.

(ii) Plug key is closed. When plug key is closed current flows through the circuit.

**Question. Draw a schematic diagram of an electric circuit (in the “on” position) consisting of a battery of five cells of 2 V each, a 5 W resistor, a 8 W resistor, a 12 W resistor and a plug key, all connected in series. An ammeter is put in the circuit to measure the electric current through the resistors and a voltmeter is connected so as to measure the potential difference across the 12 W resistor.****Calculate the reading shown by the: (a) ammeter (b) voltmeter in the below electric circuit.****Answer : ** Resistors of 5 W , 8 W , 12 W all the connected in series.

Hence,

RS = R1+ R2+ R3

= ^5 + 8 + 12hW

RS = 25 W

As V = IR

**Question. Define resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit.****Calculate the resistance of 50 m length of wire of cross-sectional area 0.01 square mm and of resistivity 5X10 ^{-8} Wm .**

**Answer :**

Resistance is the opposition offered in the path of flow of current by the atoms or molecules of a conductor.

Factors affecting resistances:

a. length R ∝ l

b. area of cross-section R ∝ 1/A

c. nature of material.

Rheostat is used to change the current in the circuit without changing the voltage source.

Given: l = 50 m

A = 0.01mm2 = 0.01 X 10-6 m2

r = 5X10-8 Wm

As R = P,lA = 5 X 10-5 X 50/0.01 X 10-5

R = 250 Ω

**Question. (a) Calculate the resistance of the wire using graph. **

**(b) How many 176 W resistors in parallel are required to carry 5 A on a 220 V line?****(c) Define electric power, Derive relation between power, potential difference and resistance.****Answer : **

b. Resistance of the circuit to carry a current of 5A on 220V.

**Question. (a) Three resistors R1, R2 and R3 are connected in parallel and the combination is connected to battery, ammeter, voltmeter and key. Draw suitable circuit diagram. Obtain an expression for the effective resistance of the combination of resistors in parallel.****(b) Why are electric bulbs filled with chemically inactive nitrogen or argon?****(c) What is meant by the statement that the rating of a fuse in a circuit is 5 A?****Answer : **

a. Let equivalent (effective) resistance is R then I = V/R

1/R = 1/R1 + 1/R2 + 1/R3

b. To prevent oxidising the filament due to high temperature.

c. The current in the fuse wire must not exceed 5 A otherwise it will melt.

**Question. A current of 10A exists in a conductor.Assuming that this current is entirely due to the flow of electrons (a) find the number of electrons crossing the area of cross section per second, (b) if such a current is maintained for one hour, find the net flow of charge.****Answer : ** Current, I = 10 A

Charge flowing through the circuit

in one second,

(a) We know, Charge on an electron

= 1.6 × 10^{–19}C

So, No. of electrons crossing per second

=10C/( 1.6× 10^{-19}) = 6.25 × 10^{19}

(b) Net flow of charge in one hour

= Current × Time

= 10 A × 1 h

10 A × (1 × 60 × 60 s) = 36000 C

**Question. A current of 5.0 A flows through a circuit for 15 min. Calculate the amount of electric charge that flows through the circuit during this time.****Answer : **Given : Current, I = 5.0 A

Time, t = 15 min. = 15 × 60 s = 900 s

Then, Charge that flows through the circuit,

Q = Current × Time

= 5.0A × 900 s

= 4500 A.s = 4500 C

**Question. If four resistances each of values 1 ohm are connected in series. Calculate equivalent resistance.****Answer : **In series,

R_{1} = R_{2} = R_{3} = R_{4} = 1ohm

putting values, we get,

R_{s} = 1 + 1 + 1 + 1 = 4

**Question. A piece of wire is redrawn by pulling it until its length is doubled. Compare the new resistance with the original value.****Answer : **Volume of the material of wire remains same.

So, when length is doubled, its area of crosssection will get halved. So, if l and a are the original length and area of cross-section of wire,

Original value of the resistance, R = Ρ×λ/a

and,

New value of the resistance,

**Question. A TV set shoots out a beam of electrons. The beam current is 10μA.****(a) How many electrons strike the TV screen in each second ?****(b) How much charge strikes the screen in a minute?****Answer : **Beam current, I = 10 μA = 10 ×10^{–6}A

Time, t = 1 s

So,

(a) Charge flowing per second,

Q = I × t = 10 × 10^{–6}A × 1s = 10 × 10^{–6}C

We known,

Charge on an electron = 1.6 × 10^{–19}C

So, No. of electrons striking the TV screen

(b) Charge striking the screen per min

= (6.25 × 10^{14} × 60) × 1.6 × 10^{–19}C

= 6.0 × 10^{–3}C

**Question. Suppose a 6-volt battery is connected across a lamp whose resistance is 20 ohm the current in the circuit is 0.25 A, calculate the value of the resistance from the resistor which must be used.****Answer : **Lamp resistance, R = 20 ohm

Extra resistance from resistor, R = ?

(to be calculated)

For R and R’ in series,

Total circuit resistance, Rs = R + R’

= 24 ohm

But Rs = R + R’

Hence R’ = R_{s} – R

= 24 – 20 = 4 ohm

Extra resistance from resistor,

R’ = 4 ohm.

**Question. A resistance of 6 ohms is connected in series with another resistance of 4 ohms. A potential difference of 20 volts is applied across the combination. Calculate the current through the circuit and potential difference across the 6 ohm resistance.****Answer : **For better understanding we must drawn a proper circuit diagram. It is shown in fig.

We use proper symbols for electrical components.Resistances are shown connected in series, with 20 V battery across its positive and negative terminals. Direction of current

flow is also shows from positive terminal of the battery towards its negative terminal.

Potential difference, V = 20 V

Potential difference across 6Ω

V_{1} = ? (to be calculated)

Total circuit resistance = 10Ω

From Ohm’s law, R_{s} =V/R_{S}

Circuit current, I = 2 ampere or (2A)

Putting values, we get, V_{1} = 2 × 6 = 12 volts

Potential difference across 6Ω resistance = 12 V

**Question. Resistors R _{1}, R_{2} and R_{3} having values 5Ω,10Ω, and 30Ωrespectively are connected in parallel across a battery of 12 volt. Calculate**

**(a) the current through each resistor (b) the total current in the circuit and (c) the total circuit resistance.**

**Answer :**Here,

R

_{1}= 5Ω R

_{2}= 10Ω, R

_{3}= 30Ω, V = 12 V

(a) I

_{1}= ? I

_{2}= ? I

_{3}= ?

(b) I = I

_{1}+ I

_{2}+ I

_{3}= ?

(c) R

_{p}= ?

**Question. Calculate the resistance of 100 m long copper wire. The diameter of the wire is 1 mm.****Answer : **Using the relationship,

**Question. For the circuit shown in the following diagram what is the value of**

** (i) current through 6 Ω resistor****(ii) potential difference (p.d.) across 12 Ω****Answer : ** (i) For current through 6Ω Current from 4 V battery flows through first parallel branch having 6 Ω and 3 Ω in series.

Current in this branch

**(ii) For p.d. across 12Ω**

Current through second parallel branch

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

**(i) total resistance of the circuit****(ii) total current flowing in the circuit****(iii) potential difference across R1****Answer : (i) For total resistance**

8Ω and 12Ω are connected in parallel.

Their equivalent resistance comes in series with 7.2Ω resistance as shown in fig.

With 7.2Ω and 4.8Ω in series

R_{s} = 7.2 + 4.8 = 12Ω

Total circuit resistance = 12 ohms.

**(ii) For total current**

Total circuit resistance, R = 12 ohm

Potential difference applied, V = 6 V

I = ?

From Ohm’s law R =V/I

I =V/R

I =6/12

= 0.5

Total circuit current = 0.5 A Ans.

**(iii) For potential difference across R _{1}**

R =V/I

V = IR

V

_{1}= IR

_{1}

V

_{1}= 0.5 × 7.2

= 3.6 V

Potential difference across, V

_{1}= 3.6 V. Ans

**Question. Two resistances are connected in series as shown in the diagram.**

**(i) What is the current through the 5 ohm resistance ?****(ii) What is the current through R ?****(iii) What is the value of R ?****(iv) What is the value of V ?****Answer : **First resistance, R_{1} = 5 Ω

(i) Current through 5 ohm resistance, I = ?

(ii) Current through R, I = ?

(iii) Value of second resistance, R = ?

(iv) Potential difference applied by the battery,

V = ?

Current through 5Ω resistance = 2 ampere (2A).

(ii) Since R is in series with 5Ω same

current will flow through it,

Current through R = 2 A.

(iv) From relation, V = V_{1} + V_{2}

V = 10 + 6 = 16 volts

V = 16 volts

**Question. Three resistances are connected as shown in diagram through the resistance 5 ohms, a current of 1 ampere is flowing :**

**(i) What is the current through the other two resistors?****(ii) What is the potential difference (p.d.) across AB and across AC ?****(iii) What is the total resistance.****Answer : (i) For current in parallel resistors**

For same potential difference across two parallel resistors,

Current is 0.6 A through 10 Ω

**(ii) For p.d. across AB**

From Ohm’s law, R =V/I, V = IR

V = 1 × 5 = 5V

P.D. across AB = 5 V. Ans

For parallel combination of 10Ω and 15Ω P.D. across BC, V = I1R1 = 0.6 × 10 = 6 V

P.D. across AC = P.D. across AB + P.D. across BC.

= 5 + 6 = 11 V

**(iii) For total circuit resistance**

For 10Ω and 15Ω in parallel

**Question. In the circuit diagram.**

** Find (i) total resistance****(ii) current shown by the ammeter A.**

** Answer : ** 3 Ω and 2 Ω in series become 5Ω.Equivalent circuit is shown in fig.

Circuit resistance = 2.5 ohm

**(ii) For circuit current**

Potential difference, V = 4 V

Circuit resistance R_{p} = 2.5 Ω

Circuit current, I = ? (to be calculated)

**Question. Resistors R _{1} = 10 ohms, R_{2} = 40 ohms,R_{3} = 30 ohms, R_{4} = 20 ohms, R_{5} = 60 ohms and a 12 volt battery is connected as shown.**

**Calculate :**

**(a) the total resistance and (b) the total current flowing in the circuit.**

**Answer :**The situation is shown in (figure).