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**SHORT ANSWER TYPE QUESTIONS**

**Q.1. On what factors does the resistance of a conductor depend ?OrList the factors on which the resistance of a conductor in the shape of a wire depends. [CBSE 2019]Ans:** Resistance of a conductor (i) is directly proportional to its length, (ii) inversely proportional to its cross-section area and depends on the material of the conductor. Resistance also depends on the temperature.

(b) If the given blub is replaced by a blub of rating 25 W, 220 V, will there be any change in the value of current and resistance ? Justify your answer and determine the change. [CBSE 2019]

∴ Current drawn by the bulb

and resistance of the bulb R =

(b) On taking another bulb of power P' = 25 W and voltage V = 220 V, there is a change in the value of current and resistance because their values depend on the power of the bulb.

New current

and new resistance

A device of resistance R is connected across a source of V voltage and draws a current I . Derive an expression for power in terms of voltage (or current) and resistance. [CBSE 2019]

Ans:

W = QV.

But Q = It ∴ W = VIt

As power is defined as the rate of doing work, hence

If R be the value of resistance of the conductor, then V - RI and hence

P = VI = (RI)I = I

Again

Thus, in general we can say that electric power is given by

∴ Current in the circuit I

∴ Power used in the 2Ω resistor P

(ii) When 2 Ω resistor is joined to a 4 V battery in parallel with 12 Ω and 2 Ω resistors, current flowing in 2 Ω resistor is independent of the other resistors.

∴ Current flowing through 2Ω resistor

∴ Power used in the 2Ω resistor

∴

In series combination, the same current flows in all the resistances but the potential difference across each of the resistance is different.

According to Ohm ’s law, we have

If the total potential difference between A and B is V, then

Let the equivalent resistance be R, then

V = IR

and hence

⇒

Ans:

Ans.

Ans

V ∝ I or V = IR

Here, constant R is known as the resistance of given conductor.

(b) Resistance of a metal wire of length 5 m is 100 Ω. If the area of crosssection of the wire is 3 x 10

where ρ is a constant, which is known as the electrical resistivity of the material of conductor.

Thus, resistivity ρ = RA/l

∴ SI unit of resistivity ρ shall be = Ω-m

(b) Here l = 5 m, R = 100 Ω and A = 3 x 10

∴ Resistivity of the material of wire ρ = RA/l =

(b) In an electric circuit two resistors of 12Ω each are joined in parallel to a 6 V battery. Find the current drawn from the battery. [CBSE 2019]

In parallel combination of three resistances R

I = I

The potential difference across each of these resistances is the same.

Thus, from Ohm’s law

If R is the equivalent resistance then,

I = V/R

∴

and

(b) Here R

Net resistance R o f the parallel grouping is :

∴ Current drawn by the circuit from the battery I =

The effective resistance of parallel combination of R

∴ Net resistance of the circuit R = R

∴ Current through 12 Ω resistor = Circuit current

(ii) Both ammeters A

(ii) Calculate the electric current passing through the above circuit w hen the key is closed. (iii) Potential difference across 15 Ω resistor. [CBSE 2019]

(i) The schematic diagram is given in Fig. 12.25.

(ii) Here total voltage V = 5 x 2 = 10 V

and total resistance R = R

= 5 + 10 + 15 = 30 Ω

∴ Current passing through the circuit when the key is closed

(iii) Potential difference across resistor R3 of 15 Ω

(i) To obtain an equivalent resistance R

(ii) To obtain equivalent resistance R

Ans:

Heat H = I

Ans:

Ans:

In parallel combination of three resistances R

I = I

The potential difference across each of these resistances is the same.

Thus, from Ohm’s law

If R is the equivalent resistance then,

I = V/R

∴

and

Ans:

Ans:

Ans. (a) The resistance of a conductor is a property of the conductor, which affects the flow of current through it on maintaining a potential difference across its ends.

Unit of resistance is ohm. Resistance of a conductor is said to be 1 ohm, if a potential difference of 1 V is to be applied across its ends for m aintaining flow of 1 A current.

(b) If given resistance wire is replaced by another thicker wire o f same material and same length then cross-section area of wire is increased and consequently its resistance decreases So the current flowing in the circuit increases.

(b) Here radius of wire r = 0.01 cm = 0.01 x 10

⇒

Electric power P = VI = I

(b) We know that slope of V-I graph for a given wire gives its resistance and in given figure slope of graph is more for wire A. It means that R

In series arrangement same current I flows through both the resistance.

As heat produced per unit time is given by I

Ans. (a) A rheostat, (b) A closed plug key.

Ans:

Ans:

Ans:

Ans:

**Ans:** 3.6 x 10^{6} J = 1 kW h.**Q.32. Define an electric circuit. Draw a labelled, schematic diagram of an electric circuit comprising of a cell, a resistor, an ammeter, a voltmeter and a closed switch. Distinguish between an open and a closed circuit. [CBSE 2015]****Ans:** A continuous and closed path of an electric current is called an electric circuit.

A labelled, schematic diagram of an electric circuit showing a cell E, a resistor R, an ammeter A, a voltmeter V and a closed switch S is shown here.

An electric circuit is said to be an open circuit when the switch is in ‘off’ mode (or key is unplugged) and no current flows in the circuit.

The circuit is said to be a closed circuit when the switch is in 'on' mode (or key is plugged) and a current flows in the circuit.**Q.33. (a) n electrons, each carrying a charge -e, are flowing across a unit crosssection of a metallic wire in unit time from east to west. Write an expression for electric current and also give its direction of flow. Give reason for your answer.****(b) The charge possessed by an electron is 1.6 x 10 ^{-19} coulomb. Find the number of electrons that will flow per second to constitute a current of 1 ampere. [CBSE 2015]**

electric current I =

As electrons are flowing from east to west, the direction of electric current is from west to east.

(b) Here current I = 1 A, time t = 1 s and charge on each electron e = 1.6 x 10

Hence, number of electrons flowing n =

(b) How will the resistivity of a conductor change when its length is tripled by stretching it ? [CBSE 2015]

Ans. (a) The resistance of a cylindrical conductor i.e., a wire (R) is (i) directly proportional to its length L, (ii) inversely proportional to its cross-section area A and (iii) depends on the nature of material of wire. Mathematically,

Here, ρ is known as the resistivity of given material. It is defined as the resistance offered by a unit cube of given material when current flows perpendicular to the opposite faces. (b) The resistivity of the conductor remains unchanged.

(b) A wire of length L and resistance R is stretched so that the length is doubled and area of cross-section halved.

(b) (i) The new resistance of wire Thus, resistance increases to four times its original value.

(ii) The resistivity remains unchanged because it does not depend on the dimensions of a conductor of given metarial.

R = R

∴ Current flowing in the circuit

(ii) The potential difference across R 1 = 10 Ω resistor is V

⇒ R = 2 Ω

∴ Current drawn from the battery I=

⇒ R

Similarly R

Now R

R

Similarly R

Finally R

(i) the current flowing through the resistor.

(ii) the energy that dissipates as heat in 10 s. [CBSE 2015]

(i) The current flowing through the resistor I =

(ii) Energy dissipated as heat in time t = 10 s is

∴ H = I

Ans: Here charge transferred Q = 90000 C, potential difference b etw een the terminals o f battery V = 40 V and time t = 1 h = 3600 s.

Current =

Amount of heat generated H = Vlt = 40 x 25 x 3600 = 3600000 J = 3.6 x 10

and power expended

As the electric line is o f rating 220 V, 5 A, hence we can connect n lamps in parallel where

Thus, we can safely connect 27 lamps of 40 W, 220 V rating to a 220 V, 5 A line.

**LONG ANSWER TYPE QUESTIONS**

**Q.1. An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery as shown in the circuit. Calculate(а) 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. **Here voltage of battery V = 6 V, resistance of electric lamp = R

(a) Since R

R = R

(b) The current through the circuit I = V/R = 6/24 = 0.25 A

(c) (i) Potential difference across the electric lamp V1 = IR

(ii) Potential difference across the conductor V

(d) Power of the lamp P = I

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

(b) Calculate the equivalent resistance of the network shown in Fig. 12.38 [CBSE 2019]

(a) The arrangement is shown in circuit diagram of Fig. 12.39.

In parallel combination of three resistances R

I = I

The potential difference across each of these resistances is the same.

Thus, from Ohm’s law

If R is the equivalent resistance then,

I = V/R

∴

and

(b) In the network resistors of 20 Ω and 20 Ω are joined in parallel and make a resistance R

This combined resistance R

(b) Two identical resistances of 12 Ω each are connected to a battery of 3 V.

(ii) In parallel arrangement, equivalent resistance R

(b) Here = R

For minimum resistance, two resistors must be connected in parallel so that

Hence power

For maximum resistance, two resistors must be connected in series so that

R

So, the power

⇒

(а) current flowing through each resistance is same, and

(b) total potential difference is equal to the stun of potential differences across individual resistors. [CBSE 2019]

(a) Plug the key and note the ammeter reading.

Then change the position of ammeter to anywhere in between the resistors and again note the ammeter reading. We find that ammeter reading remains unchanged. It shows that in series arrangement same current flows through each resistor.

(b) Insert a voltmeter across the ends X and Y of the series combination of resistors. Plug the key so as to complete the circuit and note the voltmeter reading V across the series combination of resistors.

Take out plug from key K and disconnect the voltmeter. Now insert the voltmeter across the ends of first resistor R

V = V

It shows that in series arrangement of resistors total potential difference is equal to the sum of potential differences across individual resistors.

(a) The arrangement is shown in circuit diagram of Fig. 12.39.

In parallel combination of three resistances R

I = I

The potential difference across each of these resistances is the same.

Thus, from Ohm’s law

If R is the equivalent resistance then,

I = V/R

∴

and

(b) Here series combination of R

⇒ r = 4 Ω

(b) Calculate the equivalent resistance R of a combination of three resistors of 2 Ω, 3 Ω and 6 Ω joined in parallel. [CBSE 2016]

In series combination, the same current flows in all the resistances but the potential difference across each of the resistance is different.

According to Ohm ’s law, we have

If the total potential difference between A and B is V, then

Let the equivalent resistance be R, then

V = IR

and hence

⇒

(b) Here = R

The equivalent resistance R for parallel combination of resistors will be

SI unit of electric current is an ampere (A). Current is said to be one ampere, if rate of flow of charge through a cross-section of conductor be 1 coulomb per second.

Direction of conventional current is taken as the direction of flow of positive charge or opposite to the direction of flow of negative charge. If negatively charged in a conductor flow from B to A then the direction of conventional current will be from A to B.

Here current I = 1 A, time t = 1 s and charge on electron e = 1.6 x 10

∴

If V be the potential difference maintained across the ends of a wire then, by definition, the amount of work done for flow of 1 C charge through the wire is V.

∴ Work done for flow of Q charge

W = VQ = VIt [∵ Q = It]

where I is the current flowing in time t.

As V = IR, hence

W= VIt = (IR)It = I

This work done (i.e., electrical energy dissipated) is converted into heat. Hence, the amount of heat produced,

Q = I

This is known as Joule’s law of heating.

Incandescent lamps, electric iron, electric stove, toaster, geyser, electric room heater etc., are the appliances based on heating effect of electric current.

(ii) Calculate the resistance of that resistor.

(iii) What does the graph represent ? [CBSE 2015]

Ans:(i) The plotted V-I graph is shown in Fig. 12.37.

(ii) We take two points A and B on the graph for which currents are 1.5 A and 2.5 A respectively and the corresponding values of V are 4.6 V and 7.8 V.

∴ Resistance of resistor

(iii) The V-I graph is linear one. It shows that the given resistor strictly obeys Ohm’s law.

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