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

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#1. If P and V are the power and potential of device, the power consumed with a supply potential V1 is

For Device P=VI=V*V/R=> R=V²/P

Resistance of device R=V²/P

Power with Potential V’ is P₁=V₁²/R = (V₁²/V²)P

#2. 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

#3. The resistance whose V-I graph is given below is

Explanation: (b) Resistance = slope line of V-I graph =

(9-0)/(15-0)=9/15=3/5

#4. 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)

#5. A cooler of 1500 W, 200 volt and a fan of 500 W, 200 volt are to be used from a household supply. The rating of fuse to be used is

2.5 A

5.0 A

7.5 A

10 A

Explanation: (d) Total power used, P = P₁ P₁ = 1500 500 = 2000 W.

Current drawn from the supply,I=P/V=2000/200=10A

#6. A battery of 10 volt carries 20,000 C of charge through a resistance of 20 Ω. The work done in 10 seconds is

2 × 10^3 joule

2 × 10^5 joule

2 × 10^4 joule

2 × 10^2 joule

Explanation: (b) W= qV= 20000 × 10 = 2,00,000 = 2 × 10⁵ J

#7. 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.

#8. 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

#9. 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.

#10. 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

#11. To get 2 Ω resistance using only 6 Ω resistors, the number of them required is

Explanation: (b) Three resistors of 2 Ω is required to get 6 Ω because resultant is more than individual so they all must be connected in series.

#12. 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.

#13. 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

#14. 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.

#15. 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

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

copper

nichrome

aluminium

iron

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

maximum in (i)

maximum in (ii)

maximum in (iii)

same in all the cases

Explanation :(d) Ammeter is always connected in series with in the circuit. The reading is independent from its location.

#18. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

100%

200%

300%

400%

If I is current and R is resistance then,

Power, P = I²R

Power in first case, P₁ = I²R

100% increase in current means that current becomes 2I

Power in second case, P₂ = (2I)²R = 4I²R

Now, increase in dissipated power = P₂– P₁ = 4I²R – I²R = 3I²R

Percentage increase in dissipated power = 3P₁/ P₁ × 100 = 300%

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

Explanation: (b) Equivalent resistance in parallel combination is

1/R_P =1/R₁ 1/R₂

For the same length and area of cross-section, R ∝ p (resistivity)

#20. Which of the following represents voltage?

Work done /(Current x Time)

Word done x Charge

(Work done x Time) / Current

Work done x Charge x Time

#21. Electrical resistivity of a given metallic wire depends upon

Its length

Its thickness

Its shape

Nature of the material

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

#22. In the following circuits, heat produced in the resistor or combination of resistors connected to a 12 V battery will be

Same in all the cases

Maximum in case (i)

Maximum in case (ii)

Maximum in case (iii)

#23. The least resistance obtained by using 2 Ω, 4 Ω, 1 Ω and 100 Ω is

< 100 Ω

< 4 Ω

< 1 Ω

> 2 Ω

Explanation: (c) In parallel combination, the equivalent resistance is smaller than the least resistance used in the circuit.

#24. Which of the following is not correctly matched?

None

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

1/5

1/25

1/10

Explanation: The minimum resistance is obtained when resistors are connected in parallel combination.
Thus equivalent resistance obtained by connecting five resistors of resistance 1/5 Ω each, parallel to each other =

1/R= 1/R₁1/R₂1/R₃1/R₄1/R₅

R=1/25

Finish