Answer: P=VI= V

^{2}/R For the same V, R is inversely proportional to P.

Therefore, the bulb 60 W, 220 V has a greater resistance.

2. Question: A torch bulb has a resistance of 1 Ω when cold. It draws a current of 0.2 A from a source of 2 V and glows. Calculate

(i) the resistance of the bulb when glowing and

(ii) explain the reason for the difference in resistance.

(ii) explain the reason for the difference in resistance.

Answer:

(i) When the bulb glows:

(i) When the bulb glows:

V = I R ---- Ohm's law R = V/I = 2/.2 =10 Ω

(ii) Resistance of the filament of the bulb increases with increase in temperature. Hence when it glows its resistances is greater than when it is cold.

3. Question: Calculate the resistance of 1 km long copper wire of radius 1 mm. (Resistivity of copper = 1.72 x 1 0

^{-8}

^{}

Answer: L = 1 km = 1000 m

R = 1 mm = 1 x 1 0

p = 1.72 x 1 0-8 W m^{-3}^{-3 x }1 0

^{-3 }=

^{ }3.14 x 1 0

^{-6}

^{}

^{-6 }= 5.5 W

4. Question: When a potential difference of 2 V is applied across the ends of a wire of 5 m length, a current of 1 A is found to flow through it. Calculate:

(i) the resistance per unit length of the wire

(ii) the resistance of 2 m length of this wire

(ii) the resistance of 2 m length of this wire

(iii) the resistance across the ends of the wire if it is doubled on itself.

Answer: (i) V = I R ----- Ohm's law R=V/I=2/1= 2 Ohm

Resistance per unit length: 2/5= 0.4 Ohm/m

(ii) Resistance of 2 m length of the wire = 0.4 x 2=0.8 ohm

(iii) When the wire is doubled on itself:

(a) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A.

(b) The length becomes half i.e.L/2

Resistance of this wire =R' = p (l/2)/(2A) = 1/4(p(L/A)

But p(L/A) = 2 ohm

R' = 1/4 x 2=0.5 Ohm

5. How much work is done in moving 4 C across two point having pd. 10 v

Solution : W = VQ = 10 x 4 = 40J

6. How much energy is given to each coulomb of charge passing through a 9 v battery?

Solution: Potential difference = Work done = Potential difference × charge

Where, Charge = 1 C and Potential difference = 6 VWork done = 9×1 = 9 Joule.

7. 100 j of work is done in moving a charge of 5 C from one terminal of battery to another . What is the potential difference of battery?

Solution: V = W/Q = 100j/5C = 20 V

8. If 4 x 10

^{-3}J of work is done in moving a particles carrying a charge of 16 x 10^{- 6 }C from infinity to point P .What will be the potential at a point?
Solution: the potential at a point is work done to carry unit from one point to another

= (4 x 10

^{-3}) /(16 x 10^{- 6 }C) = 250 V9. Calculate the current and resistance of a 100 W ,200V electric bulb.

Solution:Power,P = 100W and Voltage,V = 200V

Power P = VI

So, Current I = P/v = 100/200 = 0.5A

Resistance R = V/I = 200/0.5 = 400W.

Resistance R = V/I = 200/0.5 = 400W.

Solution:

Voltage ,V = 220V and Current ,I = 5A,

Voltage ,V = 220V and Current ,I = 5A,

Power,P = VI = 220 × 5 = 1100W = 1.1 KW.

11.A lamp can work on a 50 volt mains taking 2 amps.What value of the resistance must be connected in series with it so that it can be operated from 200 volt mains giving the same power.

Solution: Lamp voltage ,V = 50V and Current ,I = 2 amps.

Resistance of the lamp = V/I = 50 / 2 = 25 Ω

Resistance connected in series with lamp = r.

Supply voltage = 200 volt. and Circuit current I = 2 A

Resistance connected in series with lamp = r.

Supply voltage = 200 volt. and Circuit current I = 2 A

Total resistance Rt= V/I = 200/2 = 100Ω

Rt = R + r => 100 = 25 + r => r = 75Ω

12. Calculate the work done in moving a charge of 5 coulombs from a point at a
potential of 210 volts to another point at 240 volts

Solution: Potential diffrence = 210 - 240 = 30 V

So, W.= V x Q = 30V x 5C = 150 Joules

13. How many electrons pass through a lamp in one minute if the current be 220 mA?

Solution:

I = 220 mA = 0.22 A

I = Q/T

0.22 = Q/60

Q= 0.22 x 60 = 13.2 C

No of electron carry 1 C charge = 6 x 10

^{18}
No of electron carry 13.2 C charge = 6 x 10

^{18}x 13.2 C = 79.2 x 10^{18}^{}

Solution:

I = Q/t = 12/4 = 3A

15. A 2 Volt cell is connected to a 1 Ω resistor. How many electrons come out of the negative terminal of the cell in 2 minutes?

Solution: V = IR => I = V/R = 2/1 = 2 A

I = Q/t => Q = It = 2 x 2 x 20 = 80 C

No of electron carry 1 C charge = 6 x 10

^{18}
No of electron carry 80 C charge = 6 x 10

^{18}x 80 C = 108 x 10^{18 }= 1. 08 x 10^{20}16. (a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?

Solution

(a) We are given V = 220 V; R = 1200 Ω.

we have the current I = V/R = 220 V/1200 Ω = 0.18 A.

we have the current I = V/R = 220 V/1200 Ω = 0.18 A.

(b) We are given, V = 220 V, R = 100 Ω.

we have the current I = V/R = 220 V/100 Ω = 2.2 A.

we have the current I = V/R = 220 V/100 Ω = 2.2 A.

17. The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?

Solution

We are given, potential difference V = 60 V, current I = 4 A.

According to Ohm’s law, R = V/I = 60/4 =15Ω

When the potential difference is increased to 120 V

the current is given by current = V/R = 120V/15 = 8A

The current through the heater becomes 8 A.

18. A 4 Ω resistance wire is doubled on it. Calculate the new resistance of the wire.

Solution

We are given, potential difference V = 60 V, current I = 4 A.

According to Ohm’s law, R = V/I = 60/4 =15Ω

When the potential difference is increased to 120 V

the current is given by current = V/R = 120V/15 = 8A

The current through the heater becomes 8 A.

18. A 4 Ω resistance wire is doubled on it. Calculate the new resistance of the wire.

Solution

We are given, R = 4 Ω.

When a wire is doubled on it, its length would become half and area of cross-section would double. T

So, a wire of length l and area of cross-section A becomes of length l/2 and area of cross section 2A. we have R = ρ(l/A)

R1 = ρ((l/A) / 2A) where R1 is the new resistance.

Therefore, R1/R = ρ((l/A)/2A) / ρ(l/A) = 1/4

Or, R1 = R/4 = 4Ω/4 = 1Ω

The new resistance of the wire is 1 Ω.

R1 = ρ((l/A) / 2A) where R1 is the new resistance.

Therefore, R1/R = ρ((l/A)/2A) / ρ(l/A) = 1/4

Or, R1 = R/4 = 4Ω/4 = 1Ω

The new resistance of the wire is 1 Ω.

Solution: Resistance of the wire = 0.4Ω

Resistance of bulb = 150Ω

Resistance of rheostat = 120Ω

In series, Total resistance ,R = 0.4 + 150 +120 = 270.4Ω

Resistance of rheostat = 120Ω

In series, Total resistance ,R = 0.4 + 150 +120 = 270.4Ω

Solution:

The potential difference at the ends of the conductor. = V = IR = 0.2 x 4.5 = 0.9 V

21. A lamp has a resistance of 96 ohms. How much current flows through the lamp when it is connected to 120 volts?

Solution: I = V/R = 120/96 = 1.25 A [V = IR]

The current through the lamp equals 1.25 A.'

22. The manufacturer specifies that a certain lamp will allow 0.8 ampere of current when 120 volts is applied to it. RRWhat is the resistance of the lamp?

Solution: V = IR So, R = V/I = 120/0.8 = 150 W

23. How much voltage is required to cause 1.6 amperes in a device that has 30 ohms of resistance?

Given: V = IR = 1.6 x 30 = 48 V

24. How much power is dissipated when 0.2 ampere of current flows through a 100-ohm resistor?

Ans: P = V I = IR x I = I2 R = 0.2 x 0.2 x 100 = 4 W

25, How much energy is converted by a device that draws 1.5 amperes from a 12-volt battery for 2 hours?

W = Pt, P = V I So, W = VIt = 12 x 1.5 x 2 = 36 Wh

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21. A lamp has a resistance of 96 ohms. How much current flows through the lamp when it is connected to 120 volts?

Solution: I = V/R = 120/96 = 1.25 A [V = IR]

The current through the lamp equals 1.25 A.'

22. The manufacturer specifies that a certain lamp will allow 0.8 ampere of current when 120 volts is applied to it. RRWhat is the resistance of the lamp?

Solution: V = IR So, R = V/I = 120/0.8 = 150 W

23. How much voltage is required to cause 1.6 amperes in a device that has 30 ohms of resistance?

Given: V = IR = 1.6 x 30 = 48 V

24. How much power is dissipated when 0.2 ampere of current flows through a 100-ohm resistor?

Ans: P = V I = IR x I = I2 R = 0.2 x 0.2 x 100 = 4 W

25, How much energy is converted by a device that draws 1.5 amperes from a 12-volt battery for 2 hours?

W = Pt, P = V I So, W = VIt = 12 x 1.5 x 2 = 36 Wh

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ReplyDeleteI want to know why r u taking 6.0×10^18 n not the charge of electrons1.67×10^-19 electrons..

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ReplyDeleteit is given that :

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