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MCQs On Direct Current (DC) Electrical Circuit

28
MARCH, 2018

1) A DC circuit has ___________ as a load.

1. Resistance
2. Inductance
3. Capacitance
4. All of the above
EXPLANATION
DC Circuit has no field changing effect. So, there will be no inductance and capacitance effect in the circuit.

2) The purpose of load in an electric circuit is to

1. Increase the circuit current
2. Utilize electrical energy
3. Decrease the circuit current
4. None of the above
EXPLANATION
An electrical load is an electrical component or portion of a circuit that consumes electric power.

3) Electrical appliances are not connected in series because

1. Series circuit is complicated
2. Power loss is more
3. Appliances have different current ratings
4. None of the above
EXPLANATION
Electrical appliances are not connected in series because current will be same through all the appliances connected whereas appliances have different current ratings and also that voltage decreases as more no. Of appliances are connected.

4) Electrical appliances are connected in parallel because it

1. Is a simple circuit
2. Draws less current
3. Results in reduced power loss
4. Makes the operation of appliances independent of each other
EXPLANATION
When appliances are connected in a parallel arrangement, each of them can be put on and off independently.

5) Three resistance 14.5 Ω, 25.5 Ω and 60 Ω are connected in series across 200 V. What will be the voltage drop across 14.5 Ω

1. 29 V
2. 13.5 V
3. 14 V
4. 18 V
EXPLANATION
In this circuit, total current is calculated as
R = 14.5+25.5+60
R = 100 Ω
I = V/R
I = 200/100
I = 2 A
Now, voltage drop across 14.5 Ω resistance is
V = IR
V = 2×14.5
V = 29 V

6) Two bulbs B1 100 W, 200 V and B2 40 W, 200 V are connected in series across 200 V battery, the total circuit resistance will be

1. 1000 Ω
2. 400 Ω
3. 1400 Ω
4. 135 Ω
EXPLANATION
In this circuit,
For B1
P1 = V2/R
R1 = V2/P
R1 = (200)2 / 100
R1 = 400 Ω
For B2
P2 = V2/R
R2 = (200)2 / 40
R2 = 1000 Ω
Total resistance in circuit R = R1 + R2
R = 1400 Ω

7) When the number of resistance are connected in parallel, the total resistance is

1. Greater than the smallest resistance
2. Between the smallest and greatest resistance
3. Less than the smallest resistance
4. None of the above
EXPLANATION
If two resistances in parallel are equal and of the same value, then the total or equivalent resistance is equal to half the value of one resistor.

8)  When cells are arranged in parallel

1. The current capacity increases
2. The current capacity decreases
3. The e.m.f decreases
4. The e.m.f increases
EXPLANATION
In a parallel circuit there is more than one path for the current to flow along. When cells are connected in parallel the total value of the cells is the same as each individual cell. For two 1.5 V cells in parallel, the voltage stays at 1.5 V, but the life (current capacity) of the battery is doubled.

9) A cell of e.m.f E is connected across a resistance R. The potential difference between the terminals of the cell is found to be V. The instant resistance of the cell is

1. 2(E – V)V/r
2. 2(E – V)r/E
3. (E – V)r
4. ((E – V)/V)r
EXPLANATION
e=I(R+r)
V=IR
∴ eV=R+VR
=1+rR
=rR
=(ev−1)
∴ r=R(eV−1)

10) An external resistance R is connected to a cell of internal resistance r. The maximum current flows in the external resistance when

1. R < r
2. R > r
3. R = r
4. None of the above
EXPLANATION
The external resistance (R) is connected with the internal resistance (r) of the cell in series connection. So, the total resistance is ( R + r ). When the total resistance is low, then there is a maximum current flow in that circuit. If we apply the option (A) and (B), then the overall resistance increases than the option (C). Hence, if the external resistance (R) and the internal resistance (r) are equal, then the maximum current flows through the circuit.

11) A wire has a resistance of 12 Ω. It is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is

1. 6 Ω
2. 3 Ω
3. 12 Ω
4. 24 Ω
EXPLANATION
Here, wire is connected by bending it in circle. After then, Taking two points at any diameter means circle is divided in two equal half that makes it parallel connected. So, 12 Ω is divided into two equal i.e. 6 Ω resistance each. By solving these two parallel connected resistance we get 3 Ω as equivalent resistance.

12) The total resistance of circuit by connecting 50 resistance of 1/4 Ω in parallel is

1. 50/4 Ω
2. 4/50 Ω
3. 200 Ω
4. 1/200 Ω

EXPLANATION
1/4=0.25 ohm
1/R=1/0.25+1/0.25+••••••••••50 resisters
1/R=1/0.25*50=200 ohm
R= 1/200 ohm

13) The internal resistance of a cell of e.m.f 2 V is 0.1 Ω. It is connected to a resistance of 3.9 Ω. The voltage across the cell is

1. 0.5 V
2. 1.95 V
3. 1.9 V
4. 2 V
EXPLANATION
Here Current I = (E / (r+R) )
I = 2 / (3.9+0.1)
I = 2 / 4
I = 0.5

14) Three 2 Ω resistors are connected to form a triangle. The resistance between any two corners is

1. 6 Ω
2. 2 Ω
3. (3/4) Ω
4. (4/3) Ω
EXPLANATION
The total resistance of the triangle and between any two corners will be 1.33 ohm.
2 || (2+2)
2 || 4
(2×4)/(2+4)
8/6 = 1.33333

15) A cell with zero internal resistance and e.m.f of 2 V is connected across 2, 3 and 5 Ω. The potential difference across 3 Ω resistor will be

1. (2/3) V
2. 0.6 V
3. 3 V
4. 6 V
EXPLANATION
First find the total current within it.
Rt = R1 + R2 + R3
Rt = 10 Ω
V = IRt
I = V / R
I = 2/10
I = 0.2 A
Now, voltage drop across 3 ohm resistor.
V = IR2
V = 0.2×3
V = 0.6 V

16) Two bulbs 200 W and 100 W are connected across 220 V supply in series. Find the power consumed by both bulbs

1. 300 W
2. 200 W
3. 66 W
4. 50 W
EXPLANATION
Here first calculate the resistance of the circuit. i.e.
Rt = R1 + R2
R1 = (V)2 / P1
R1 = (220)2 / 200
R1 = 242
R2 = (V)2 / P2
R2 = (220)2 / 100
R2 = 484
Rt = 242 + 484
Rt = 726

Now, Calculate the total Power
Pt = (V)2 / Rt
Pt = (220)2 / 726
Pt =66 W

17) An electrical fan and heater are marked as  100 W and 1000 W respectively at 220 V supply. What will be the resistance of heater

1. Zero
2. Greater than that of the fan
3. Less than that of the fan
4. Equal to that of the fan
EXPLANATION
For constant voltage, we know that
P ∝ 1/R
So higher the power, lower will be the resistance.

18) If the internal resistance of the cell connected in circuit is greater than the external circuit, higher current will be obtained by arranging cell in ______

1. Series
2. Parallel
3. Series-Parallel
4. Any connection
EXPLANATION
As per the above statement, the internal resistance of the cell is larger than the external resistance. So, if we connect the cells in parallel, then the overall internal resistance decreases. We know that, in low resistance path the current flow is more. Hence, we need to group the cells in parallel to make the current flow as high.
19) Conductivity is the reciprocal of

1. Current density
2. Reactance
3. Resistivity
4. Inductance
EXPLANATION
Electrical conductivity or specific conductance is the reciprocal of electrical resistivity. It represents a material’s ability to conduct electric current.
20) The potential difference of an energy source that provides 5 mJ of energy for every micro coulomb of charge that flows is

1. 5 V
2. 50 V
3. 500 V
4. 50 kV
EXPLANATION
HereV = W/QBy putting valuesV = 50×10e-3 / 1×10e-6V = 50 kV

21) The specific resistance of a metallic conductor ________ with the rise in temperature

1. Increases
2. Decreases
3. Remains same
4. Decreases first than start increases
EXPLANATION
As the temperature of a metallic conductor increases, the kinetic energy of the electrons of the conductor also increases ,due to which more obstruction is offered to the flowing electrons and hence the current.Therefore,as more obstruction is offered to the flowing electrons, the resistance is increased.
22) Insulators have ________ temperature coefficient of resistance

1. Zero
2. Negative
3. Positive
4. None of the above
EXPLANATION
In insulator, there is a large energy gap between the valance band and the conduction band. So, if the temperature rise is high, the electrons will go to the conduction band. But since it is not crowded due to less number of electrons, so the conductance will increase due to availability of free electrons in the conduction band. So, the resistance decreases with increase in temperature in insulator.
23) Eureka has __________ temperature coefficient of resistance

1. Positive
2. Negative
3. Almost zero
4. None of the above
EXPLANATION
Eureka has 0.1 x 10-3 temperature coefficient of resistance i.e. almost equals to zero.
24) What quantity of charge must be delivered by a battery with a potential difference of  100 V to do 500 J of work

1. 5 C
2. 0.5 C
3. 50 C
4. 500 C
EXPLANATION
Here,Q = W / VQ = 500 / 100Q = 5 C
25) A piece of aluminium wire is stretched to reduce its diameter to half of its original value. Its resistance will become

1. Two times
2. Half
3. Sixteen times
4. Eight times
EXPLANATION
Let,R1 = ρ× (L1/A1) & R2 = ρ× (L2/A2)R2/R1 = (L2/L1) × (A1/A2)L1×A1 = L2×A2orL1×A1 = (L2)×(A1/4)L2=4×L1R2/R1 = (4×L1/L1 ) × (A1/(A1/4)) = 16orR2=16×R1
26) Semiconductor have ______ temperature coefficient of resistance

1. Positive
2. Negative
3. Zero
4. None of the above
EXPLANATION
The resistivity of semi conductors decreases with the increase of temperature.
27) The thermal speed of electrons is of the order of

1. 1 ms-1
2. 10 ms-3
3. 106 ms-1
4. 3×108 ms-1
EXPLANATION
In the absence of an electric field, electrons buzz about with ‘thermal velocities’ of order 106 m/s in completely random directions.
28) The resistance of a material 2 m long and 2 m2 in area of cross-section is   1.6×10-8 Ω. Its specific resistance will be

1. 1.6×10-8 Ωm
2. 3.2×10-8 Ωm
3. 6.4×10-8 Ωm
4. 0.16×10-8 Ωm
EXPLANATION
Specific resistance for materials are given by ρ = RA/L. By solving we get 1.6×10-8 Ωm.
29) The value of α ( temperature coefficient of resistance ) depends on

1. Length of the material
2. Cross-sectional area of the material
3. Volume of the material
4. Nature of the material and temperature
EXPLANATION
The “alpha” (α) constant is known as the temperature coefficient of resistance and symbolizes the resistance change factor per degree of temperature change.
30) A copper wire of resistance Ro is stretched till its length is increased ‘n’ times of its original length. Its resistance will now be

1. n2 Ro
2. Ro / n2
3. nRo
4. n3 Ro
EXPLANATION
Solve this using relation R = ρ(L/A)
31) The resistivity of a material 2 × 10-8 Ωm. What will be the resistance of a hollow pipe of the material of length  1 m and having inner and outer radii of 10 cm and 20 cm respectively.

1. 2.0 × 10-4 Ω
2. 2.1 × 10-7 Ω
3. 2.0 × 10-5 Ω
4. 2.0 × 10-3 Ω
EXPLANATION
Solve this using relation R = ρ(L/A) Here,ρ = 2 × 10e-8 Ωm;L = 1 m; A = π ((R22) – ((R12));A = π ((0.22) – ((0.12)) = 9.42 × 10e-2 m2 ;R = (2 × 10e-8) × (1/9.42 × 10e-2) = 2.1× 10e-7 Ω
32) The example of non ohmic resistance is

1. Copper wire
2. Carbon resistance
3. Aluminium wire
4. Tungsten wire
5. All of the above

EXPLANATION
Non-ohmic conductors or resistances are those conductors or resistances which do not follow Ohms law.
33) A copper wire is stretched so that its length is increased by 0.1 %. The change in its resistance is

1. 0.1 %
2. 0.2 %
3. 0.3 %
4. 0.4 %
EXPLANATION
R = ρ(L/AR’ = ρ(L’/A’);L’ = L + (0.1/100)×L;As volume remains same. A×L = A’×L’ ;ORA’ = A(L/L’) = (A/1.001);R’/R = (L’/L)×(A/A’);R’/R = (1.001)×(1.001) = 1.002;OR(R’-R)/R = 0.002;Percentage increase =( (R’-R)/R )×100 = 0.002 %
34) If an electric current is passed through a nerve, the man

1. Begins to laugh
2. Gets shocked
3. Is excited
4. Becomes insensitive to pain
EXPLANATION
An electric current passes through us because our body is a very good conductor of electricity and we are connected to earth. Hence, our body becomes a medium for current to pass from high potential to low potential. Our body is made out of 70% fluid. When a current passes through a conducting fluid, it ionises the salts. The same happens to our blood. Depending on the intensity of current, the ionisation is at different levels and so the effects. The effect of ionisation of blood is conveyed to the brain as a shock.
35) The temperature coefficient of resistance of a wire is 0.00125 per °C. at 300 K, its resistance is  1 Ω. The resistance of the wire will be 2 Ω at

1. 1154 K
2. 1100 K
3. 1400 K
4. 1127 K
EXPLANATION
The resistance of wire at 300 K is R1 = 1 Ω.Let at T °C, the resistance of the wire be 2 Ω. i.e. R1 =2 Ω.R2 = R1×(1+(T-27))OR2 = 1×(1+0.00125(T-27))On Solving T=827 °CTemperature in Kelvin = 827+273 = 1100 K
36) Ampere-Hour is the unit of

1. Quantity of electricity
2. Energy
3. Power
4. Conductivity
EXPLANATION
An ampere hour is the amount of energy charge in a battery that will allow one ampere of current to flow for one hour.
37) The resistance of a conductor is 5 Ω at 50 °C and 6 Ω at 100 °C. What is the resistance at 0 °C.

1. 1 Ω
2. 2 Ω
3. 3 Ω
4. 4 Ω
EXPLANATION
No answer description required for this question.
38) The value of α0 of a conductor is 1/236 per °C. The value of α18 will be

1. 1/218 per °C
2. 1/244 per °C
3. 1/254 per °C
4. 1/264 per °C
EXPLANATION
α18 = α0/(1+(α0×18)) = 1/((1/α0)+18)α18 = 1/(236+18) = 1/254 per C
39) The value of α0 of a conductor is 1/255 per °C. The value of α20 will be

1. 1/300 per °C
2. 1/230 per °C
3. 1/250 per °C
4. 1/265 per °C
EXPLANATION
See the above question explanation, by using the same method one can solve this question.
40) The value of α50 of a conductor is 1/230 per °C. The value of α0 will be

1. 1/180 per °C
2. 1/280 per °C
3. 1/380 per °C
4. 1/250 per °C
EXPLANATION
See the above question explanation, by using the same method one can solve this question.

41) The number of free electrons passing through the filament of an electric bulb in one hour when the current through the current through it is 0.32 A filament will

1. 3 × 1022
2. 2× 1026
3. 7.2 × 1019
4. 3 × 1021
EXPLANATION
I = Q/T; I = Ne/T;N = IT/E;N = (0.32 × 3600)/1.6×10^-19N = 7.2×10^21
42) In which of the following substances the resistance decreases with increase in temperature

1. Carbon
2. Constantan
3. Copper
4. Silver
EXPLANATION
No answer description required for this question.
43) A wire having very high value of conductance is said to be

1. Very good conductor
2. Moderately good conductor
3. Insulator
4. None of the above
EXPLANATION
No answer description required for this question.
44) An electric heater is marked 2000 W, 200 V. The resistance of the coil is

1. 0.1 Ω
2. 20 Ω
3. 1/20 Ω
4. 200 Ω
EXPLANATION
By applying power formula, we get its resistance 20 Ω.
45) The current in a circuit having constant resistance is tripled. The power increases

1. 1/9 times
2. 3 times
3. 9 times
4. 1/3 times
EXPLANATION
Apply Power formula.
46) What voltage drop will be there across a&nbsp;1 KW electric heater whose resistance when hot is 40 Ω.

1. 100 V
2. 50 V
3. 150 V
4. 200 V
EXPLANATION
By applying power formula (P=V2/R, we get 200 V.
47) A resistor R1 dissipates the power P when connected to a certain generator. If resistance R2 is put in series with R1, the power dissipated by R1

1. Decreases
2. Increases
3. Remains the same
4. Any of the above depending upon the values of R1 and R2
EXPLANATION
When we are add a new resistor, we are decreasing current (I) whereas increasing resistance R. Here, which one changes by a bigger factor decides the fate of power. Suppose we double the existing resistance this makes the current flow half. From equation (P=I2R), we can see due to square term half reduction in current results in reduction in power and R term causes multiplication by 2, so in effect power reduces by factor of 2. This conclusion is applicable for any value of R2
48) In case of liquids, Ohm’s law is

1. Fully obeyed
2. Partially obeyed
3. There is no relation there between current and P.D
4. None of the above
EXPLANATION
Electrolytic solutions are good conductors of electricity and usually obey’s Ohm law.
49) Two electric bulbs rated for the same voltage have powers of 200 W and 100 W. If their resistances are respectively R1 and R2 then

1. R1 = 2 R2
2. R2 = 2 R1
3. R2 = 4 R1
4. R1 = 4 R2
EXPLANATION
One can derive equation for it by comparasion using power relation P = V2/R.
50) A copper wire a resistance of 10 Ω. It is stretched by one-tenth of its original length. Then its resistance will be

1. 10 Ω
2. 12.1 Ω
3. 9 Ω
4. 11 Ω
EXPLANATION
As we know that, resistance of a conductor is given by:R = ρL/AWhere ρ is the resistivity of the material, L is the length and A is the cross sectional area.Let’s assume the wire is cylindrical.Volume of a cylinder = V = πr2 x hWhere r is the radius and h is the height.Here, A = πr2 and h = LSo, V = ALWhen you are stretching the volume will remain the same.So V = A’L’ = A’ x (1.1)L (stretching by 10%)Thus A’ = V / (1.1L) = 0.909ASo new resistance = ρL’/A’ = ρ x 1.1L / (0.909xA) = 1.21R = 12.1 Ω
51) 1020 electrons each having a charge of 1.6×10-19 C pass from a point X to a point Y in 0.1 seconds. The current flowing is

1. 16 A
2. 1.6 A
3. 160 A
4. 1 A
EXPLANATION
Given,n= 1020electrons, t= 0.1 sece=1.6 x 10-19We know,Q=ne=1020x 1.6 x 10-19 Q=16 CAgain,I=Q/TI = 16/0.1I = 160 A
52) No current flows between two charged bodies if they had same

1. Capacity
2. Potential
3. Charge
4. None of the above

EXPLANATION
Current flows due to the potential difference between two charged bodies.
53) A carbon electrode has a resistance of 0.125 Ω at 20 °C. The temperature coefficient of carbon is -0.0005 at 20 °C. What will be the resistance of electrode at 85 °C

1. 4 Ω
2. 0.5 Ω
3. 1.2 Ω
4. 0.121 Ω
EXPLANATION
R85 = R20×(1+α20(T2-T1))R85 = 0.125×(1+(-0.0005×65))R85 = 0.121 Ω
54) The specific resistance of a wire 1.1 m long, 0.4 mm diameter having a resistance off 4.2 Ω will be

1. 48 × 10-8 Ωm
2. 40 × 108 Ωm
3. 4 × 10-6 Ωm
4. 4.8 × 10-8 Ωm
EXPLANATION
Using the general formula: R = ρ L / Aρ = RA / L= [4.2 Ω × π (0.4/2 × 10-3 m)2] / (1.1 m)≈ 4.8 × 10^-7 Ω mSpecific Resistance of wire is approximately 48 × 10-8 Ω m
55) Ampere-second is the unit of

1. Power
2. Energy
3. E.M.F
4. Charge
EXPLANATION
Ampere-second is the SI unit of electric charge (measured in coulomb).
56) The percentage by which the incandescence of lamp decreases due to drop of current by 3% is

1. 6%
2. 3%
3. 9%
4. 12%
EXPLANATION
P = I2 × RDelta ( P )= 2×I×(Delta( I )) × RDelta ( P )/P = 2×(Delta( I )) / I = 2×3% = 6%
57) Two resistance A and B have resistance RA and RB respectively with RA<RB . The resistivity of their materials are ρA and ρB. Then

1. ρA < ρB
2. ρA = ρB
3. ρA < ρB
4. Insufficient information
EXPLANATION
From the given information, we cannot derive the correct relation between the resistance. We also need to know the cross-section areas and the lengths of both the conductors before concluding about their resistivities.
58) As the temperature of a metallic resistor is increased, the product of its resistivity and conductivity

1. Increased
2. Decreased
3. Remains constant
4. May increased or decreased
EXPLANATION
As the temperature of a metallic resistor is increased the product of its resistivity and conductivity remains constant. This is because they are inversely proportional.
59) A resistor develops 400 J of thermal energy in seconds when a current of 2 A is passed through it. The resistance of the resistor is

1. 10 Ω
2. 20 Ω
3. 5 Ω
4. 40 Ω
EXPLANATION
Energy = I2×R×T;R = 400 / (4)×10R = 10 Ω
60) An electron moves in a circle of radius 10 cm with a constant speed of 4×106 m/s. The electric current of a point on the circle is

1. 1×10-12 A
2. 4×10-7 A
3. 2×10-6 A
4. None of the above
EXPLANATION
The no. of revolution made by electron in one second.= V/2πr = 4×106 / 2×π×10×10-2;= (2/π)×(1.6×10-19)= 1×10-12 A

61) The resistance of a human body is about

1. 200 Ω
2. 1000 Ω
3. 25 Ω
4. 10 Ω
EXPLANATION
A rough value for the internal resistance of the human body is 300-1,000 Ohms.
62) The hot resistance of an electric bulb filament is higher than its cold resistance because the temperature coefficient of filament is

1. Zero
2. Positive
3. Negative
4. None of the above
EXPLANATION
The temperature coefficient of a resistance of an electric bulb filament is positive i.e increases with the increase in temperature.
63) A nichrome wire used as a heater coil has a resistance of 2 Ω/m. For a heater of 1 kW at 200 V, the length of the wire required is

1. 80 m
2. 40 m
3. 20 m
4. 24 m
EXPLANATION
Resistance of heater wire = V2 / P = (200)2 / 1×1000 = 4 ΩThe resistance of the wire is 2 Ω per meter.Therefore,Length of heater wire = 40/2 = 20 Ω
64) When 1 V is applied in a circuit, a current of 1 μ A flows through it. The conductance of the circuit is

1. 1 μ mho
2. 10 μ mho
3. 100 μ mho
4. None of the above
EXPLANATION
One can solve it using Ohm law. Where conductance is the reciprocal of reistance.
65) The maximum possible conductance of a 75 kΩ , 10% resistor is

1. 20 μS
2. 14.81 μS
3. 18.71 μS
4. 24.5 μS
EXPLANATION
As we know that conductance G=1/R and conductance is maximum when resistance is minimum.Minimum Value of R = 75 – ( 75 × 10/100) = 67.5 kΩ.Now Maximum Conductance = 1/67.5 kΩ = 14.81 S
66) How much voltage is necessary to create a flow of 0.24 C in 0.8 seconds through a resistance of 500 Ω.

1. 50 V
2. 300 V
3. 150 V
4. 75 V
EXPLANATION
Current I = Q/TI = 0.24/0.8I = 0.3 ANow, Voltage is equal toV = IRV = 0.3 × 500 = 150 V
67) A certain electrical appliance consumes energy at a rate of 540 J/S. How many kWh of energy does it consumes in 3.5 hour

1. 5.4 kWh
2. 2.89 kWh
3. 1.89 kWh
4. 10.89 kWh
EXPLANATION
Power P = 540 J/s = 540 wattEnergy = 540 × 3.5 = 1890 Wh = 1890 kWh
68) The power dissipated in a resistor in term of its conductance G and the voltage V across it is

1. V2G
2. V2 / G
3. G2 V
4. G2 / V
EXPLANATION
Power dissipated P = V2 / RG = 1 / RP = (V2)×G
69) On the voltage versus current graph of a 10 kΩ resistor, what is the total change in voltage when the current changes from 1 mA to 5.5 mA

1. 55 V
2. 110 V
3. 45 V
4. 50 V
EXPLANATION
Graph between current and voltage remains linear. When I = 1 mA, V1 = IR = (1×10-3) × 10 = 10 VNow when I = (5.5×10-3)V2 = (5.5×10-3) × (10×103) = 55 VTherefore, change in voltage = V2 – V1 = 55 – 10 = 45 V
70) The current entering the positive terminal of an element is given by

i = 5 cos πt A

The power delivered to the element at t = 3 ms when the voltage across the element is 3t is given by

1. 18.43 W
2. 26.4 W
3. 53.48 W
4. 63.28 W
EXPLANATION
Power P = VI = (15 cos 60πt)×(5 cos 60πt) = 75 cos2 (60πt)) WAt t=3 msP = 75 cos2 (60π(3×10-3)) = 75 cos2 0.18π = 53.48 W
71) A copper wire of area of X-section 1 mm2 is carrying a current of 10 A. If the number density of conduction electrons is 10-28 m-3. The drift velocity of conduction electrons is

1. 1 / 160 ms-1
2. 1 / 20 ms-1
3. 1 / 30 ms-1
4. 1 / 80 ms-1
EXPLANATION
Drift velocity = I / nAe= 10 / 1028 × (4×10-6) × 1.6 ×10-19= 1/160
72) A potential difference of 3 V is applied across a conductor of resistance 1.5 Ω. The number of electrons flowing in 1 second is

1. 2 × 10-6 Ω
2. 0.8 × 10-6 Ω
3. 1.6 × 10 Ω
4. 4 × 10-6 Ω
EXPLANATION
Volume of cube = Mass/Density = 84/10.5 = 8 cm3Therefore, each side of the cube is (8)1/3 = 2 cm3R = ρL/A= (1.6×10-6) × (2/2×2) = 0.8 × 10-6 Ω
73) The temperature at which the resistance of a conductor becomes double to that at 0 °C is

1. 1/3 αo °C
2. 1/αo °C
3. 1/2 αo °C
4. 1/6 αo °C
EXPLANATION
Rt = Ro (1 + αoT)It is given that Rt = 2Ro2Ro = Ro (1 + αoT)OR= 1/αo °C
74) If a wire is melted and recast to half of its length, then the new resistance is

1. R/4
2. R/2
3. R
4. 2R
EXPLANATION
Since volume of the wire remains same,IA = I’A’orA/A’ = l’/l= 0.5l/l=l/2Now,R’/R = (l’/l) × (A/A’)=1/2 × 1/2= 1/4
75) Two wires A and B have the same cross-section and are made of the same material. RA = 600 Ω and RB = 100 Ω. The number of times A is longer than B is

1. 6
2. 2
3. 4
4. 5
EXPLANATION
Here, RA= 600 Ω and RB=100 Ω. We know that R=ρ L/A. As cross-section of both material are same. So RA ∝ L. Thus, LA is 6 times longer than LB.
76) The resistivity of a wire depends on

1. Length
2. Material
3. Cross-sectional area
4. None of the above
EXPLANATION
The resistivity is a property of the material in the wire that depends on the chemical composition of the material but not on the amount of material or the shape (length, cross-sectional area) of the material.
77) Kirchhoff’s second law is based on law of conservation of

1. Charge
2. Energy
3. Momentum
4. Mass
EXPLANATION
Kirchhoff’s voltage law (2nd Law) states that the sum of all voltages around any closed loop in a circuit must equal zero. This is a consequence of charge conservation and also conservation of energy.
78) The mass of the proton is roughly how many times the mass of an electron?

1. 184000
2. 18400
3. 1840
4. 184
EXPLANATION
A proton is about 1835 times more massive than an electron.
79) Which of the following is not same as watt ?

1. Joule / Second
2. Ampere / Volt
3. Ampere x Volts
4. (Ampere)2 x Ohm
EXPLANATION
All are the relation’s of power except the second.
80) One Kilowatt hour of electrical energy is the same as

1. 36 x 105 watt
2. 36 x 105 ergs
3. 36 x 105 joules
4. 36 x 105 B.T.U
EXPLANATION
The kilowatt hour (kWh) is a unit of energy equal to 3.6 megajoules.
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