(a) Though same current flows through the electric line wires and the filament of bulb, yet only the filament glows. Why ? (b) The temperature of the filament of bulb is 2700 0 C when it glows. Why does it not get burnt up at such high temperature? (c) The filament of an electric lamp, which draws a current of 0.25 A is used for four hours. Calculate the amount of charge flowing through the circuit. (d) An electric iron is rated 2 kW at 220V. Calculate the capacity of the fuse that should be used for the electric iron .
Answers
Answer:
a) The resistance of electric line wires is very, very less than that of the filament of the bulb and as the wires are the good conductors of electricity which cannot stop electricity so they cannot heat up if the circuit is proper. Therefore, the current through high resistance filament produces more heat which makes it glow.
b) The temperature of the filament of the bulb is 2700 degrees celsius when it glows it not get burnt up at such high temperature because it has the melting point higher that it melting point of tungsten filament is 3422 °C so it does not burn or melt.
c) We know that,
Current = Charge / Time
=> I = Q / T
Also,
Charge = Number of electrons × Charge of each electron
Q = n × e
So, we get,
=> I = ne / T
Here,
I = 0.25 A
T = 4 hours = 4 × 3600 = 14400 seconds
e = 1.602 × 10^(-19)
By applying values, we get,
=> 0.25 = n × 1.602 × 10^(-19) / 14400
=> 0.25 × 14400 = n × 1.602 × 10^(-19)
=> 25 × 144 / 1.602 × 10^(-19) = n
=> n = 25 × 144 × 10^19 / 1.602
=> n = 2247.19101124 × 10^19
=> n = 224719101124 × 10^11 electrons
Now, amount of charge = Q
=> Q = I × t
=> Q = 0.25 × 14400
=> Q = 3600
Hence, amount of charge = 3600 coulomb
d) Power (P) = 2 kW
Since 1 kW = 1000 W
Therefore P = 2 kW = 2000 W
Potential difference (V) = 220 V
Current, = I
Since,
P = V × I
=> I = P / V
By applying values, we get,
I = 2000/220 = 9.09 A
Hence the capacity of the fuse that should be used for the electric iron rated 2kw at 220V will be withstand with 10 A capacity (by rounding off).
Answer:
Explanation:
a) The resistance of electric line wires is very, very less than that of the filament of the bulb and as the wires are the good conductors of electricity which cannot stop electricity so they cannot heat up if the circuit is proper. Therefore, the current through high resistance filament produces more heat which makes it glow.
b) The temperature of the filament of the bulb is 2700 degrees celsius when it glows it not get burnt up at such high temperature because it has the melting point higher that it melting point of tungsten filament is 3422 °C so it does not burn or melt.
c) We know that,
Current = Charge / Time
=> I = Q / T
Also,
Charge = Number of electrons × Charge of each electron
Q = n × e
So, we get,
=> I = ne / T
Here,
I = 0.25 A
T = 4 hours = 4 × 3600 = 14400 seconds
e = 1.602 × 10^(-19)
By applying values, we get,
=> 0.25 = n × 1.602 × 10^(-19) / 14400
=> 0.25 × 14400 = n × 1.602 × 10^(-19)
=> 25 × 144 / 1.602 × 10^(-19) = n
=> n = 25 × 144 × 10^19 / 1.602
=> n = 2247.19101124 × 10^19
=> n = 224719101124 × 10^11 electrons
Now, amount of charge = Q
=> Q = I × t
=> Q = 0.25 × 14400
=> Q = 3600
Hence, amount of charge = 3600 coulomb
d) Power (P) = 2 kW
Since 1 kW = 1000 W
Therefore P = 2 kW = 2000 W
Potential difference (V) = 220 V
Current, = I
Since,
P = V × I
=> I = P / V
By applying values, we get,
I = 2000/220 = 9.09 A