Q 1 :

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. The ratio of power produced in series and parallel combinations would be:

  • 1 : 2

     

  • 2 : 1

     

  • 1 : 4

     

  • 4 : 1

     

(3)

As the two conducting wires are of the same material, equal lengths and equal diameters. They will have the same resistance.

Let the resistance of each wire be R and applied potential difference, V

Rseries=R1+R2            = R+R=2RPseries=V2Rseries            = V2(2R)                  (i)

1Rparallel=1R1+1R2                = 1R+1RRparallel  = R2Pparallel = R2Pparallel = V2Rparallel              =  2V2RDividing equation (i) by equation (ii) PseriesPparallel=14

 



Q 2 :

A student plots V–I graphs for three samples of nichrome wire with resistances R, R, and R.


Choose the correct statement that holds true for this graph:

  • R = R = R

     

  • R > R > R

     

  • R > R > R

     

  • R > R > R

     

(4)

As it is clear From the graph, the current for conductor A is less than that for A, and A is less than A.
we can say , I < I < I.

We know, R = V/I or R  1/I.

If I is less, then R will be more

I < I < IR > R > R.

 



Q 3 :

The instrument used for measuring electric current is:

  • galvanometer

     

  • ammeter

     

  • voltmeter

     

  • potentiometer

     

(2)

Ammeter is a device used for measuring electric current in amperes.

 



Q 4 :

Which of the following does not apply to silver?

  • The resistance provided is directly proportional to its length.

     

  • The resistance provided is inversely proportional to the area of cross section.

     

  • Their resistivity is in the range 10-8Ω-m to 10-6Ω-m

     

  • The movement of electrons on their outermost orbital is tightly held together.

     

(4)

For a given continuous piece of uniform wire, the resistance is directly proportional to its length. Thus, silver which is a good conductor of electricity and the resistance in it is directly proportional to its length.

 



Q 5 :

R1 and R2 are two resistors and r1 and r2 are equivalent resistances in series and parallel respectively, then R1R2:

  • r1r2r1+r2

     

  • r1+r2r1r2

     

  • r1+r1-4r1r2r1+r1+4r1r2

     

  • r1+r12-4r1r2r1-r12-4r1r2

     

(4)

Hence r1 = R1 + R2                   iand r2 = R1 R2R1 + R2                  ii R1 R2 = r2 r1Now,     R1 - R22 = R1 + R22 - 4 R1 R2                                          = r12 - 4r2 r1R1 - R2 = r12 - 4r2 r1                  iiiand R1 + R2 = r1                               ivsolving equation (iii) and (iv) for R1 and R2.R1 = r1 + r12 - 4r1 r22and R2 = r1 - r12 - 4r1 r22R1R2=r1 + r12 - 4r1 r2r1 - r12 - 4r1 r2

 



Q 6 :

There are three resistors connected in parallel, the resistance of each resistor is 3 ohm. What is the total resistance of all the three resistors?

 

  • 1 Ω

     

  • 6 Ω

     

  • 15 Ω

     

  • 3 Ω

     

(1)

It is given that the three resistors are connected in parallel and the resistance of each resistor is 3 Ω.

R1=R2=R3=3Ω

From the formula given below, we can calculate the total resistance of all the three resistors:

1Rp=1R1+1R2+1R3=13+13+13=33 Rp=1Ω

Hence, the total resistance is  1Ω



Q 7 :

If a person has five resistors each of value 15Ω then the maximum resistance he can obtain by connecting them is:

  • 1 Ω

     

  • 5 Ω

     

  • 10 Ω

     

  • 25 Ω

     

(1)

Resistance of one resistor = 15Ω

Number of resistors = 5

Maximum resistance can be obtained by combining the resistors in series: 

Rs=R1+R2+R3+R4+R5=15+15+15+15+15=1+1+1+1+15=55=1Ω

Hence, a person on combining five resistors in series gets resistance  1 Ω

 



Q 8 :

Match the Symbols in column A with the Components in column B.

  • A-3; B-4; C-3; D-2

     

  • A-1; B-2; C-1; D-4

     

  • A-4; B-1; C-2; D-3

     

  • A-2; B-3; C-4; D-1

     

(4)

A-2; B-3; C-4; D-1

 



Q 9 :

The resistance of the wire when the length of the wire increases two times:

  • becomes 2 times

     

  • becomes 3 times

     

  • becomes 6 times

     

  • becomes 4 times

     

 (1)

The electrical resistance of a wire can be expressed as:R=ρLA

Where, A = Area of cross-section of the conductor, L = Length of the conductor, ρ = Resistivity

From this relation, it is clear that the resistance is directly proportional to the length and inversely proportional to area of cross-section.

If length becomes 2L, then

R'=ρ2LA=2ρLASo, R' = 2R

Thus, the resistance becomes 2 times if the length of the wire is doubled.



Q 10 :

Which among the following is the correct way of connect ammeter and voltmeter in the circuit to determine the equivalent resistance of two resistors in series?

  • Both ammeter and voltmeter in series

     

  • Both ammeter and voltmeter in parallel

     

  • Ammeter in parallel and voltmeter in series

     

  • Ammeter in series and voltmeter in parallel

     

(4)

The correct way of connecting ammeter and voltmeter in the circuit to determine the equivalent resistance of two resistors is connecting ammeter in series and voltmeter in parallel. Ammeter is connected in series, so that the whole current passes through it and voltmeter is connected in parallel so that it could measure the complete voltage of the circuit.

 



Q 11 :

The proper representation of series combination of cells for obtaining maximum potential is:

  •  

  •  

  •  

  •  

(1)

The maximum potential is obtained when cells are connected in series such that the negative terminal of the first cell is connected to the positive terminal of the second cell and so on.

 



Q 12 :

The equivalent resistance of a series combination of two resistances is X ohm. If the resistances are of 10 Ω and 40 Ω respectively, the value of X will be:

  • 10 Ω

     

  • 20 Ω

     

  • 50 Ω

     

  • 40 Ω

     

(3)

R=R1+R2R=10+40=50Ω

Hence, the value of X is 50 Ω.

 



Q 13 :

A cylindrical conductor of length l and uniform area of cross-section A has resistance R. The area of cross-section of another conductor of same material and same resistance but of length 2l is:

  • 0.5 A

     

  • 1.5 A

     

  • 2 A

     

  • 3 A

     

(3)

For a conductor, R=ρlA

R1R2=L1L2×A2A1A2=R1R2×L2L1A1A2=RR×2llAA2=2A

 



Q 14 :

The maximum resistance which can be made using four resistors each of resistance 12Ω is

  • 2 Ω

     

  • 1 Ω

     

  • 2.5 Ω

     

  • 8 Ω

     

(1)

The maximum resistance which can be made using four resistors each of resistance 12Ω is 2Ω. This can be explained as:  In series combination, the current in each resistor remains constant and the voltage gets added up. As a result, the individual resistances also get added up.

So, Equivalent Resistance =R1+R2+R3+R4

                                           =12+12+12+12=42=2Ω

 



Q 15 :

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R' then the ratio RR' is 

  • 1/25

     

  • 1/5

     

  • 5

     

  • 25

     

 (4)

Given, a piece of wire with resistance R is cut into 5 equal parts

1R'=5R+5R+5R+5R+5R1R'=25R

now, these pieces of wire are connected in parallel then the resistance is (R')

1R'=5R+5R+5R+5R+5RSo, 1R'=25RR       R'=R25Now,  RR'=RR25RR'=25RRR'=25

 



Q 16 :

On which of the given factors, resistance does not depend:

  • length of conductor

     

  • area of cross-section

     

  • temperature

     

  • density

     

 (4)

The resistance of a wire can be expressed as: 

R=ρLA

Where,
A = Area of cross-section of the conductor
L = Length of the conductor
ρ = Resistivity

From the above relation, we can see that resistance of a wire is directly proportional to its length and inversely proportional to the area of cross-section. Hence, resistance does not depend on the density.



Q 17 :

Which of the following obeys Ohm’s law?

  • Filament of a bulb

     

  • LED

     

  • Nichrome

     

  • Transistor

     

(3)

In conductors, resistance remains constant when the current passing through them is increased; they are known as Ohmic conductors. Nichrome, which is an alloy, is made in such a way that its resistance remains constant for a wide range of temperatures. Hence, nichrome obeys Ohm’s Law. Whereas, transistor, LED, bulb filament do not obey Ohm’s law because with the varied change in temperature, their resistance changes.

 



Q 18 :

Electrical resistivity of an alloy of copper and nickel is _______ when compared with the electrical resistivity of an alloy of copper, manganese and nickel.

  • same

     

  • double

     

  • more

     

  • less

     

(3)

Electrical resistivity of an alloy of copper and nickel is more when compared with the electrical resistivity of an alloy of copper, manganese, and nickel. The electrical resistivity of Cu-Ni alloys with increasing temperature rises steeply. At 200°C, the electrical resistivity of Cu-Ni alloy is 49×10-8Ω.m whereas that of copper, manganese, and nickel is49×10-8Ω.m

 



Q 19 :

There is a dual of 8 ohm resistance on the aerial. Determine the aerial’s new resistance.

  • 2 Ω

     

  • 4 Ω

     

  • 7 Ω

     

  • 10 Ω

     

(1)

Let ( l ) be the length and ( A ) be the area of cross-section.

R=ρlA=8Ωl'=l2A'=2AR'=ρl'A'=ρl22A=14ρlA=14×8=2Ω



Q 20 :

Electrical resistivity of a given metallic wire depends upon:

  • its length

     

  • its thickness

     

  • its shape

     

  • nature of the material

     

(4)

The resistivity of a material is constant at a constant temperature. Resistivity of material does not depend on length, thickness, and shape of the material. It only depends on the temperature.



Q 21 :

An electric heater is rated at 2 kW. Electrical energy costs Rs 4 per kWh. What is the cost of using the heater for 3 hours?

  • 12

     

  • 24

     

  • 36

     

  • 48

     

(2)

Consumption of electrical energy in 3 hours can be calculated by using the formula:

E=P×t  = 2  kW × 3  hours = 6  kWh 

Unit cost of electrical energy

=4per kWh

Therefore, the cost of energy used for three hours will be:
4×6=24



Q 22 :

The values of mA and µA are:

  • 10-6 and  10-9 A respectively

     

  • 10-3 and  10-6 A respectively

     

  • 10-3 and  10-9 A respectively

     

  • 10-6 and  10-3 A respectively

     

(2)

An ampere is the SI unit of electric current.

1 A = 1000 mA or 1 mA =11000 A=10-3 AThus, 1μA=10-3×10-3 A=10-6 A

 



Q 23 :

The resistance of a resistor is reduced to half of its initial value. In doing so, if other parameters of the circuit remain unchanged, the heating effects on the resistor will become:

  • two times

     

  • half

     

  • one-fourth

     

  • four times

     

(1)

Resistance of a resistor RΩ

New resistance of a resistor R2Ω

All other parameters of the circuit remain unchanged.

By applying Joule’s law of heating: H=I2Rt

As per Ohms law  𝑉 = 𝐼R

or  𝐼 = 𝑉/𝑅  H=VR×VR×R×t=V2R×tCase I: H=V2R×tCase II: H'=V2R2×t=V2×2R×tH'=H×2

Hence, the heating effect in the resistor will become two times if all other parameters of the circuit remain the same.



Q 24 :

An electric fuse is connected with:

  • live wire

     

  • earthing

     

  • neutral wire

     

  • parallel to the line wire

     

(1)

An electric fuse is connected with the live wire because it gets blown up when an excess current tries to pass through it in order to save the electrical appliances by restricting the flow of that current.



Q 25 :

Which of the following terms does not represent electrical power in a circuit?

  • I2R

     

  • IR2

     

  • VI

     

  • V2/R

     

(2)

We know that, P=VI…(i)

Where,
P =  Power
V = Potential difference
I = Current
V = IR …(ii)

On substituting equation (ii) in equation (i), we get:

P=I2R         (iii)

Again, from Ohm's law:

I=V/R  (iv)

On substituting equation (iv) in equation (i), we get:

P=V2R

Hence, option (b) IR2 does not represent power



Q 26 :

Let us consider the flow of the current through a metallic wire. If the temperature of the entire system increases, what will happen from the following options?

  • Potential difference (V) increases

     

  • Resistance (R) decreases

     

  • Potential difference (V) decreases

     

  • V and R remain the same

     

(3)

If the temperature of the entire system increases when the current is flowing through a metallic wire, the power (VI) is dissipated in the form of heat. Hence, the potential difference (V) decreases.

 



Q 27 :

Which of the following terms does not represent electrical energy in a circuit?

  • I2Rt

     

  • IR2t

     

  • VIt

     

  • V2tR

     

(2)

Electric power, P=VI=I2R=V2R

Electrical Energy =Pt=VIt=I2Rt=V2tR