30 problems with solution of quantization and ohms law
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1. Question: Two bulbs have ratings 100 W, 220 V and 60 W, 220 V respectively. Which one has a greater resistance?
Answer: P=VI= V2/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.
Answer:
(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 0-3
p = 1.72 x 1 0-8 W m
Area of cross section = p r2 = 3.14 x 1 0-3 x 1 0-3 = 3.14 x 1 0-6
R = pl/A = (1.72 x 1 0-8 x 1000 ) / 3.14 x 1 0-6 = 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
(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 V
Work 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 V
9. 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.
10.Calculate the power rating of the heater coil when used on 220V supply taking 5 Amps.
Solution:
Voltage ,V = 220V and Current ,I = 5A,
Power,P = VI = 220 × 5 = 1100W = 1.1 KW.
Answer: P=VI= V2/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.
Answer:
(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 0-3
p = 1.72 x 1 0-8 W m
Area of cross section = p r2 = 3.14 x 1 0-3 x 1 0-3 = 3.14 x 1 0-6
R = pl/A = (1.72 x 1 0-8 x 1000 ) / 3.14 x 1 0-6 = 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
(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 V
Work 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 V
9. 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.
10.Calculate the power rating of the heater coil when used on 220V supply taking 5 Amps.
Solution:
Voltage ,V = 220V and Current ,I = 5A,
Power,P = VI = 220 × 5 = 1100W = 1.1 KW.
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