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

Results

#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. 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

#3. Coulomb is the SI unit of:

Charge

Current

Potential difference

Resistance

#4. 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

#5. 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

#6. 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.

#7. 1 mV is equal to:

10 volt

1000 volt

10^-3 volt

10^-6 volt

#8. 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).

#9. 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

#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 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

#12. The resistance of hot filament of the bulb is about 10 times the cold resistance. What will be the resistance of 100 W-220 V lamp, when not in use?

48 Ω

400 Ω

484 Ω

48.4 Ω

R= V^2/P

#13. 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

#14. 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)

#15. 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.

#16. Electric potential is a:

scalar quantity

vector quantity

neither scalar nor vector

sometimes scalar and sometimes vector

#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. 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.

#19. 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)

#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. 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

#22. 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

#23. 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

#24. 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.

#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