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

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#1. 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%

#2. Electric power is inversely proportional to

resistance

voltage

current

temperature

#3. Which of the following gases are filled in electric bulbs?

Helium and Neon

Neon and Argon

Argon and Hydrogen

Argon and Nitrogen

#4. 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)

#5. 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

#6. In an electrical circuit, two resistors of 2 and 4 respectively are connected in series to a 6V battery. The heat dissipated by the 4 resistor in 5s will be

10J

20J

30J

Explanation:

Here, first resistor, R1 = 2Ω

And second resistor, R2 = 4Ω

Voltage of cell, V = 6V

Time taken = t = 5s

Total resistance of the circuit = R = R1 R2 = 2 4 = 6 Ω

Current, I = V/R = 6/6 = 1A

Heat dissipated by the 4Ω resistor in 5s is given as,

H = I²Rt

⟹ H = 1 x 1 × 4 × 5 = 20J

#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. Which of the following is not correctly matched?

None

#9. A current of 1 A is drawn by a filament of an electric bulb. Number of electron passing through a cross-section of the filament in 16 seconds would be roughly

10^20

10^16

10^18

10^23

Current i= 1 A, time t=16s
Nuber of electrons=n
i=Q/t=ne/t
n=it/e=(1X16)/ (1.6*10^-19)=10^20

#10. Identify the circuit in which the electrical components have been properly connected.

iii

Explanation: Essential conditions are necessary when electrical components are connected

Voltmeter should be connected in parallel.
Ammeter is always connected in series.
Positive terminals of voltmeter and ammeter should be connected to positive terminal of the cell and their negative terminals should be joined to the negative terminal of the cell.

Thus, the above conditions are satisfied in case (ii).

#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. 1 mV is equal to:

10 volt

1000 volt

10^-3 volt

10^-6 volt

#13. 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.

#14. 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

#15. The electrical resistance of insulators is

high

low

zero

infinitely high

#16. 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)

#17. 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.

#18. 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

#19. 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.

#20. 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

#21. Two resistors of resistance 2 and 4 when connected to a battery will have

same current flowing through them when connected in parallel

same current flowing through them when connected in series

same potential difference across them when connected in series

different potential difference across them when connected in parallel

Explanation: In series combination of resistor, the current through both the resistor are same but potential difference across each will be different.

In parallel combination, current across each resistor will be different but the potential difference will be same.

#22. 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

#23. 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

#24. 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

#25. 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

Finish