CBSE X Science Effects of Current - EducareHut (Next Gen Knowledge Point)

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#1. An electric bulb is connected to a 220V generator. The current is 0.50 A. What is the power of the bulb?

440 W

110 W

55 W

0.0023 W

Explanation: Here, V = 220 V, I = 0.50 A

∴ Power (P) = VI = 220 x 0.50 = 110 W

#2. A cylindrical conductor of length l and uniform area of cross section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross-section.

A/2

3A/2

Resistivity of first conductior = (RA/l)
Resistivity of second conductor = (RA’/2l)

Resistivity of both material is same
so (RA)/(l) = (RA’/2l) => A’=2A

#3. The proper representation of series combination of cells obtaining maximum potential is

iii

Explanation: Maximum potential is obtained when cells are connected in series such that, negative terminal of the cell is connected to the positive terminal of the second cell and so on, as shown in the following diagram.

#4. The effective resistance between A and B is

4Ω

6Ω

May be 10 Ω

Must be 10 Ω

Answer: a

Explanation: (a) 6 Ω is shorted so effective resistance is 4 Ω.

#5. The temperature of a conductor is increased. The graph best showing the variation of its resistance is

Explanation: (a) Resistance is directly proportional to temperature of the conductor.

#6. In an electrical circuit three incandescent bulbs. A, B and C of rating 40W, 60 W and 100 W, respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

Brightness of all the bulbs will be the same

Brightness of bulb A will be the maximum

Brightness of bulb B will be more than that of A

Brightness of bulb C will be less than that of B

Explanation: We know that power is defined as rate of doing work. A bulb consumes electric energy and produces heat and light. Now, bulb with more power rating will produce more heat and light or we can say that

Power rating of bulb is directly proportional to the brightness produced by bulb.

Therefore, brightness of bulb B with power rating 60 W will be more than the brightness of bulb A having power rating as 40W.

#7. When electric current is passed, electrons move from:

high potential to low potential.

low potential to high potential.

in the direction of the current.

against the direction of the current

#8. Unit of electric power may also be expressed as

volt ampere

kilowatt hour

watt second

joule second

Explanation:

Electric power = voltage × current

SI Unit of voltage = Volt

SI Unit of current = Ampere

#9. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

1/5 Ω

10 Ω

5 Ω

1 Ω

Explanation: The maximum resistance is obtained when resistors are connected in series combination.

Thus equivalent resistance obtained by connecting five resistors of resistance 1/5 Ω each, in series = (1/5 1/5 1/5 1/5) = 1 Ω

#10. Two devices are connected between two points say A and B in parallel. The physical quantity that will remain the same between the two points is

Current

Voltage

Resistance

None of these

Explanation:(b) In parallel combination, voltage remains same across two points.

#11. If R1 and R2 be the resistance of the filament of 40 W and 60 W respectively operating 220 V, then

R1 < R2

R2 < R1

R1 = R2

R1 ≥ R2

Explanation: (b) Using power, P = V^2/R or R = V^2/P

For the same voltage, R ∝ 1/P

More the power, lesser the resistance.

Accordingly, R2 < R1

#12. A boy records that 4000 joule of work is required to transfer 10 coulomb of charge between two points of a resistor of 50 Ω. The current passing through it is

2 A

4 A

8 A

16 A

Explanation: (c) Work done in transferring the charge

W= qV = qlR …….. (V = IR)

#13. An Electric kettle Consumes 1 KW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

1 A

2 A

4 A

5 A

Explanation: Here, power = P = 1 KW = 1000 W

Voltage = V = 220 V

Current = I = ?

Now, I = P/V = 1000/220 = 4.5 A

Now rating of fuse wire must be slightly greater than 4.5 A i.e., 5 A.

#14. What is the commercial unit of electrical energy?

Joules

Kilojoules

Kilowatt-hour

Watt-hour

#15. A cell, a resistor, a key and an ammeter are arranged as shown in the circuit diagrams of figure. The current recorded in the ammeter will be

Maximum in (i)

Maximum in (ii)

Maximum in (iii)

The same in all the cases

Explanation: In series connections the order of elements in the circuit will not affect the amount of current flowing in the circuit.

#16. The resistivity does not change if

the material is changed

the temperature is changed

the shape of the resistor is changed

both material and temperature are changed

Explanation: The resistivity depends on the nature of the material and the temperature. It does not depends on dimension of resistor.

#17. A wire of length /, made of material resistivity ρ is cut into two equal parts. The resistivity of the two parts are equal to,

ρ2

2 ρ

4 ρ

Explanation: (a) Resistivity of the material depends only on the nature of material not dimensions

#18. The resistivity of insulators is of the order of

10^-8 Ω-m

10^1 Ω-m

10^-6 Ω-m

10^6 Ω-m

#19. A student carries out an experiment and plots the V ‒ I graph of three samples of nichrome wire with resistances R1, R2 and R3 respectively as shown in figure. Which of the following is true?

R1 = R2 = R3

R1 > R2 >R3

R3 > R2 > R1

R2 > R3 > R1

Explanation: Slope of VI graph is proportional to 1/ Resistance.

It means when slope will be maximum, then resistance will be minimum.

From the figure, we can see that, slope of R1 is maximum; hence its resistance will be minimum.

As, slope of R3 is minimum so, its resistance will be maximum.

Therefore, R3 > R2 > R1

#20. 1 kWh = ……….. J

3.6 × 10^-6 J

13.6 × 10^6 J

3.6 × 10^6 J

13.6 × 10^-6 J

#21. The heating element of an electric iron is made up of:

copper

nichrome

aluminium

iron

#22. Calculate the current flows through the 10 Ω resistor in the following circuit

1.2 A

0.6 A

0.2 A

2.0 A

Explanation: (b) In parallel, potential difference across each resistor will remain same. So, current through 10-Ω resistor

I = V/R=6/10 = 0.6 A

#23. Coulomb is the SI unit of:

Charge

Current

Potential difference

Resistance

#24. The electrical resistance of insulators is

high

low

zero

infinitely high

#25. Two wires of same length and area made of two materials of resistivity ρ1 and ρ2 are connected in series to a source of potential V. The equivalent resistivity for the same area is

Explanation: (a) For same length and area of cross-section R ∝ p.

For series combination, equivalent resistance is

Rs = R₁ R₂

⇒ Ps = ρ₁ ρ₂

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