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

Results

#1. 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.

#2. Electric power is inversely proportional to

resistance

voltage

current

temperature

#3. A fuse wire repeatedly gets burnt when used with a good heater. It is advised to use a fuse wire of

more length

less radius

less length

more radius

Explanation: (d) In order to get the working of heater properly, fused wire of higher rating must be used.

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

Helium and Neon

Neon and Argon

Argon and Hydrogen

Argon and Nitrogen

#5. 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

#6. 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.

#7. 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)

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

10^20

10^16

10^18

10^23

Explanation: (a) Q = ne and Q = It

∴ ne = It

n=IT/e= (1×16)/(1.6×10^-19)

10²⁰

#10. 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.

#11. Coulomb is the SI unit of:

Charge

Current

Potential difference

Resistance

#12. 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.

#13. 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

#14. 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

#15. Electrical resistivity of any given metallic wire depends upon

its thickness

its shape

nature of the material

its length

#16. 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.

#17. 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

#18. The electrical resistance of insulators is

high

low

zero

infinitely high

#19. 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.

#20. A coil in the heater consume power P on passing current. If it is cut into halves and joined in parallel, it will consume power

P/2

#21. Three resistors of 1 Ω, 2 ft and 3 Ω are connected in parallel. The combined resistance of the three resistors should be

greater than 3 Ω

less than 1 Ω

equal to 2 Ω

between 1 Ω and 3 Ω

#22. Electric potential is a:

scalar quantity

vector quantity

neither scalar nor vector

sometimes scalar and sometimes vector

#23. 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

#24. 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

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

copper

nichrome

aluminium

iron

Finish