By | April 23, 2017



Ques.1 What is current ? Write its S.I unit?

Ans.1 Electric Current is expressed by the amount of charge flowing through a particular area. S.I unit – Ampere (A)


Ques.2  What the positive is the direction of current conventionality?

Ans.2  From the positive terminal to negative terminal.


Ques.3 Give the expression of current & explain.

Ans.3  I – Current

Q- Charge

T- Time

            I = Q/T

The current is said to be 1 A when a charge of 1C flow in 1 second.

Ques.4 What is S.I unit of electric charge? Define.

Ans. 4 S.I unit = coulomb

Q= I *T

Charge is 1C contains 6*108 electrons.


Ques.5 Define Ampere?

Ans.5   The current is said to be 1A when a charge if 1C flows per second.


Ques.6 Name the instrument that measures electric current. How is it connected in circuit?

Ans.6 Ammeter measures current. It is always connect in series.


Ques.7 What is electric circuit?

Ans.7 Its a path through which electric current can flow. Its consist of a cell, key, bulb & an ammeter or a voltmeter.


Ques.8 What is no. of electrons in 1C of Charge.

Ans.8   6*108 electrons.


Ques.9 What is potential difference?

Ans.9 Potential difference is given by work done to move a charge from one point to another an electric field.

 V =  W/Q

It is signified by ‘V’

So, potential difference = work done / charge

SI unit of potential difference is volt.


Ques.10 Define 1 volt?

Ans. When 1 joule of work is done in moving 1 coulomb of charge from one point to another in an electric field, the potential difference is said to be one volt.


Ques.11 How is the potential difference measure?

Ans.11 It is measured by an instrument ‘voltmeter’.


Ques.12 How is voltmeter attached in a circuit?

Ans.12 A voltmeter is attached in parallel with the circuit across two points.


Ques.13 Name the device that helps to maintain a potential difference across a conductor?

Ans.13 Cell, Battery etc.


Ques.14 How much energy is given to each coulomb of charge passing through a 6V battery?

Ans.14       Charge= 1 Coulomb

V= 6V



W=6 Joules

Work Done= Energy= 6J


Ques15. Make an electric circuit with the help of the following instruments?

A battery, conducting wires, Ammeter, Voltmeter, Key Bulb


Ques.16 State ohm’s law? Give its mathematical representation?

Ans.16 Ohms law states that potential difference ‘V’ across the ends of a given metallic wire in an electric circuit is directly proportional to current flowing through it, provided its temperature remains same. So, V α I

V= I*R

R is the resistance and its SI unit is ohms (Ω)


Ques.17 What is resistance?

Ans.17 Its a property of the conductor to resist the flow of charges through it. SI unit is ohm and represented by Ω




Ques.18 Define 1 ohm?

Ans.18 R=V/I


When a current of 1 ampere flows through a conductor a potential difference of 1 volt, the resistance of conductor is said to be 1 ohm.


Ques.19 What happens to current when the resistance is doubled?

Ans.19 It becomes half.


Ques.20 Name an electrical component that regulate current without changing the voltage?

Ans.20 A component used to regulate the current without changing the voltage source is called variable resistance or rheostat.


Ques.21 What are the factors on which resistance of conductor depends?


  • Length of Conductor
  • Area of cross section
  • Nature of material


Ques.22 How does resistance vary with length and area of cross section of conductors?

Ans.22 Resistance is directly proportional to length & inversely proportional to its area of cross section.

  R α L      (1)

  R α 1/A   (2)

Combining 1 & 2 we get

R  α  l/A

R =  Ρ l/A


This constant of proportionality ‘Rho’ is called resistivity of the material of the conduction. It is the characteristics property of material. Its SI unit is Ω m.


Ques.23 What is the range of resistivity of conductors and insulators?

Ans. The range of resistivity of conductor is from 10-8   to 10-6  ohm meter and the range of resistivity in insulators is from 1012 to 1017   ohm meter. Both resistance and resistivity vary with temperature.


Ques.24 Why are alloys used as electrical heating elements rather than individual metals?

Ans.24 Alloys are used as electrical heating element because their resistivity is generally higher than that of its constituent metals , they do not oxidise or burn readily at high temperatures. That’s why they are used in electrical heating devices like in electric iron, toasters, geezers etc.


Ques.25 Will current flow more easily through a thick wire or thin of the same material , when connected to same source, why?

Ans.25 More current will flow easily in a thick  wire as resistance is inversely proportional to the cross section area of wire , a thick wire will have less resistance than thin wire.


Ques.26 Let the resistance of an electrical component remains the constant while potential difference is decreasing. what will happen to the current?

Ans.26 Current will decrease as it is directly proportional to the potential difference across two ends.


Ques.27 Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Ans.27  The coil of electric toaster and electric irons are made up of alloy because their resistivity is generally higher than that of pure metals. They do not oxidise or burn readily at  high temperatures.


Ques.28 Which among iron and mercury is better conductor?

Ans.  Iron


Ques.29 Which material is best conductor?

Ans, silver.


Ques.31 What happens to the current when length of wire is double?

Ans.31 When length of wire is double, its resistance also gets double and so, its current becomes half.


Ques.32 How much current will an electric bulb draw of resistance 1000Ω from a 290 V source.

Ans.32    V  =   IR

       220 v = I × 1000Ω

            I    =   220 v/1000Ω

            I     =  0.22 amperes


Ques. 33 How much current an electric heater of 100Ω resistance will draw from  230 volt.

Ans.  V=IR 

   230 = I × 100Ω

  I  =  2.3 A


Ques.34 When a 12 V battery is connected across an unknown resistor , there is a current of 2.5 mA in the circuit. Find  the value of the resistance of the resistor.

Ans. 34  current  = 2.5 mA

                Voltage = 12V

                R = I/V

               R = 12/2.5 × 10-3

               R =12 × 400

               R = 4800Ω


Ques. 35 A copper wire has diameter 0.5 mm and resistivity at 1.6 × 10-8 Ω m. what will be the length of wire to make its resistance 10Ω. How much does the resistance  change if the diameter is doubled.

Ans.35   Diameter = 0.5mm =.5 ×10-3 m

Resistivity = 1.6 × 10-8  Ωm

Resistance = 10Ω

R = pl1/A         (p= resistivity)

R = 1.6 × 10-8 × p/22(×.25 × 10-6)

P = 10×22×2.5×10-5/1.6 × 10-18

                                           P =122.7m


Ques.36 Derive an expression for total resistance in the circuit when resistance are connected in series and parallel.

Ans. In series     



 V= V1 + V 2 + V3——- (equation 1)

V = IR

On applying ohm’s law to the 3 resistors

   V1 = 1R1 , 1R2  = 1R3

  From equation (1)

  1R = 1R1 +1R2+ 1R3

 R = R1 +R2 + R3


  In parallel-

 I = I1 +I2 + I3 ————   (equation 1)

I =V/R

I1 = V/R1,   I2= V/R2, I = V/R3

 From equation 1

V/R  = V/R1 + V/R2 +V/R3

        = 1/R= 1/R1 +1/R2 + 1/R3


Ques.37 Draw  a schematic diagram of a  circuit consisting of a battery of three cells of 12v each,  a 5Ω resistor, and  a 12 Ω resistor, and a plug key, all connected  in series .

Ans. Diagram.




Ques. Redraw the circuit in above question putting in an ammeter to measure the current through the resistors  and a voltmeter to measure potential difference  across  12 Ω resistors . what would be the reading in the ammeter and voltmeter.

Ans. Total resistance:

                Rs = R1 + R2 + R3

                     = 5+ 12+ 8 Ω

                     = 25 Ω

V = IR

6v = I ×25 Ω

   I = 6/25

I  = 6 × 4/25 × 4

  = 0.24 A


Ques. 39 Judge the equation resistance when the following are connected in parallel (a) 1 Ω and 10 Ω (b) 1 Ω and 103 Ω, and 106 Ω .

Ans. 39

                            1/Rp = 1/1  +1/106

                                    = 106 + 1/ 106

                                                    = 1000001/106

                                                    =106 /1000001

                             Rp  = 0.9


           1/Rp = 1/1 + 1/106 + 1/103

                             = 106 + 1 + 103/106


               Rp =  106/1001001



Ques.40 An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of 500 Ω are connected in parallel to a 220v source. What is the resistance of electric iron connected to same source that takes as much current as all three appliances, and what is current through it?

Ans.40   I/Rp = 1/R1 +  1/R2 +1/R3

                 I/Rp =1/100 +1/50+1/500

                          = 5+10+1/500+     

                          = 16/500

Rp = 500 Ω/16

     = 31.25 Ω

I = V/R

I = 220 ×16/ 500

 = 7.04A


Ques.41 What are advantages  of connecting electrical device in parallel ?

Ans. 41

  • They all work on the same potential difference to which ever they are connected.
  • If one device shot circuits, the other will continue
  • Each one of them will draw the current according to their own resistance & there will be no wastage of current.
  • Each device can be controlled by a separated switches.


Ques. 42 How can three resistors  of resistance 2 Ω,3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω (b) 1 Ω.

Ans. 1/Rp =1/3 +1/6

                  = ½

           Rp = 2

          Rs  = 2

      Total =  4 Ω

   (b)   1/Rp =  1/2 + 1/3 +1/6

                     = 6/6

                     = 1

              Rp  =  1 Ω

They should be connected in parallel.


Ques.43 A piece  of wire  of resistance R is cut in 5 equal parts and connected in a parallel. If the equivalent resistance of this  combination is R , then ratio R’/R is .

Ans. 43  1/R’ = 1/R/5  + 1/R/5 + 1/R/5  + 1/R/5 + 1/R/5 + 1/R/5

               1/R’ = 5/R + 5/R + 5/R + 5/R + 5/R

R/R’ = 25

R =25

1/R’= 25/R




Ques.44 A battery of 9v is connected in series with resistors of 0.2Ω,0.3 Ω,0.4 Ω,0.5 Ω and 12 Ω. How much current would flow through 12 Ω resistance.

Ans.44 Total resistance :

                                 RS = 13.4 Ω

                 VOLTAGE   =  9V

                                 I    = V/R

                                  I  = 9/13.4

                                    = 90/134



Ques.45 How many 176 Ω resistors ( in parallel) are required  to carry 5A on a 2220v line?

Ans.45  5A  =  220/R     (V= IR)

                        R = 220/5

                        R = 44 Ω

 No. of resistors =176/44 =4

4 resistors are required.


Ques.46 Show how you would connect 3 resistors each of resistance 6 Ω, so that the combination has resistance of 4 Ω.

Ans.46       1/R = 1/12  +  1/6

                          = 3/12


                   Rp =4 Ω




Ques.47 What is  electrical power ? give its expression and its SI unit?

Ans.47 The rate at which electrical energy is consumed in an electric circuit is called power. Its SI unit is watt(w).

     Watt  = Volt × Ampere


Ques.48 Define 1 watt.

Ans. 48    Watt  = Volt × Ampere

When 1 ampere of current flows through a circuit of  1 volt potential difference, the power is used to be 1 watt.

P= I2R

P=  V× V/R= V2/R



Ques.49 A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24Ω resistance, which may be used separately, in series, or in parallel. What are the currents in 3 cases?

Ans.49 Voltage=220 V

RA=24Ω    RB=24 Ω

In series, I=V/R

I=220/24 = 55/6 A = 9.1 A

RS= 24Ω + 24Ω

    =48 Ω

I=V/R = 220/48 A= 4.5A

RP= 1/24 + 1/24

    = 2/24 = 1/12 Ω

R=12 Ω

I= 220/12= 55/3A = 18.3 A


Ques.50 Compare the power used in the 2 Ω resistor in each of the following circuits

  1. A 6V battery in series with 1 Ω and 2 Ω resistors
  2. A 4V battery in parallel with 12 Ω and 2 Ω resistor.

Ans.50 (i) P.d=6 V

Total R = 1+2 = 3 Ω

P=V2/R = 6*6/3 = 12W


I=P/V = 12/6 = 2A



=2*2= 4V


=4*2= 8W


(ii) P.D= 4V

TOTAL 1/R = 1/12 + 1/2 = 7/12


I= V/R = 4/12/7 = 28/12 = 7/3 = 2.3A

I= V/R = 4/2 = 2A

P= V2/R = 16/2 = 8 W


Ques.51 Several electric bulbs designed to be used on a 220V electric supply line, are rated 10W. How many lamps can be connected in parallel with each other across the two wires of 220V line if maximum allowable current is 5A?

Ans.51 Power= V*I


Power= 1100W

No. of bulbs connected(each of 10W)=1100/10 = 110 bulbs

110 bulbs can be used


Ques.52 Two lamps one rated 100W at 220V and other 60W at 220V are connected in parallel mains supply. What current is draw from the line if the supply voltage is 220V.

Ans.52 (Case I)



P=V*I = 220*I = 5/11= I

(Case II)


P=V*I = 60=220*I

I=60/220 = 3/11

Total current = I1 + I2

=3/11 + 5/11

=3+5/11 = 8/11 = 0.72A


Ques.53 Which uses more energy, a 250W TV in 1 hour or a 1200W toaster in 10 minutes?

Ans.53 P=W/T


250*60= Energy

=15000 Joules

P=W/T             1200=W/10

=12000 Joules= Energy

250W TV in 1 hour will use more energy


Ques.54 An electric heater of resistance of 8 Ω draws 15A from a service mains 2 hours. Calculate the ratio at which heat is developed in the heater.

Ans.54  Current=15A,  Resistance=8 Ω,  Time=2 hours


= (15)2*8*2

=3600 Joules

Hence, heat produced in electric heater is 3600 Joules


Ques.55 What is the heating effect of electric current?

Ans.55 If the electric current is purely resistive i.e. the resistors are only connected to the battery, the source energy continuously gets dissipated entirely in the form of heat. This is known as heating effect of electric current. This effect utilized in devices such as heaters, toasters, geysers, electric iron etc.


Ques.56 State joule’s law of heating?

Ans.56 The SI unit of heat is joules and Joules law of heating states that heat produces in a resistor is

  1. Directly proportional to square of the current flowing through the circuit.
  2. Directly proportional to resistance for a given current.
  • Directly proportional to the time in which current flows.



Ques.57 Why does the cord of electric heater does not glow while element glows?

Ans.57 The cord of the electric heater does not glow while the heating element does because the resistivity of copper wire is very less and the resistivity of heating element is quite high as compared to cord.


Ques.58 Compute the heat generated while transferring 96000C of charge in 1 hour through a potential difference of 50 volts.

Ans.58 Heat= V*I*T

Heat= 50* 96000/1 *1 = 48*106

H= 4.8*106 Joules

So heat generated is 4.8*106 Joules


Ques.59 An electric iron of resistance 20Ω takes a current 5A. Calculate the heat developed in 30 sec.

Ans.59 H=I2RT

= 25*20*30

=15000 Joules


Ques.60 (i) Why is tungsten used almost exclusively for electric lamps?

Ans.(i) Tungsten is exclusively used for filament in electric lamps because it has high melting point, it can be converted into very thin wires it has a comfortably low resistivity, it does not oxidise easily in an inert environment and it produces enough light on getting heated.


Ques.61 (ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than pure metal?

Ans.(ii) This is because the resistivity of alloys is much more than pure metal so it can produce large amount of heat.


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

Ans.(iii) This is because

  • All the devices will work of different potential difference.
  • All the devices will use the same amount of current.
  • We will not be able to control each device with separate switch.
  • If one device gets short circuited then all the rest will also not work.


(iv) How does the resistance of wire varies with its area of cross section?

Ans.(iv) The resistance of wire is inversely proportional to its area of cross-section so, the resistance of thin wire will be more and the resistance of thick wire will be less of the same material allowing more current to flow through a thick wire.


(v) Why are copper and aluminium wires usually employed for electricity transmission?

Ans.(v) This is because

  • Copper and aluminium are readily available and cheap.
  • They are highly ductile.
  • They are very good conductors of electricity.
  • There is no loss of electricity when transmitted through copper and aluminium wires.


Ques.61 Why are bulbs filled with inert gases like argon, nitrogen etc.?

Ans.61 The bulbs are usually filled with inert gases like nitrogen and argon to prolong the life of filament which provides light. In the inert environment, the tungsten filament will not oxidise easily.


Ques.62 What is an electric fuse? State its uses?

Ans.62 An electric fuse is a piece of wire made of a metal or an alloy of appropriate melting point like aluminium, copper, iron, lead etc. It is placed in series with the device. The purpose of fuse wire is to protect the circuits and appliances by stopping the flow of any unduly high electric current. If a current larger than the specified value flows through circuit, the temperature of fuse wire increases and breaks the circuit. The fuses used for domestic purposes are 1A, 2A, 3A, 5A and 10A.


Ques.63 What determines the rate at which energy is delivered by a circuit?

Ans.63 Power determines the rate at which more energy is delivered by a circuit.


Ques.64 An electric motor takes 5A from a 220V line. Determine the power of motor and the energy consumed in 2 hours.

Ans.64 P=V*I

P= 220*5 WATTS

P= 1100 Watts

Energy consumed in 2 hours = 1100*2 hours

= 2200 Wh

=2.2 kWh

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