F a light bulb produces 60.0 J of electrical energy and is 45% efficient, how much energy does it use?
2 answers:
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Answer: 3 A
Explanation:
According to<u> Ohm's law</u>:
Where:
is the voltage
is the resistance of the resistor
is the electric current (the value we want to find)
Isolating :
Finally:
Answer:
a)n= 3.125 x electrons.
b)J= 1.515 x A/m²
c) =1.114 x
m/s
d) see explanation
Explanation:
Current 'I' = 5A =>5C/s
diameter 'd'= 2.05 x m
radius 'r' = d/2 => 1.025 x m
no. of electrons 'n'= 8.5 x
a) the amount of electrons pass through the light bulb each second can be determined by:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
As we know that: Q= ne
where e is the charge of electron i.e 1.6 x C
n= Q/e => 5/ 1.6 x
n= 3.125 x electrons.
b) the current density 'J' in the wire is given by
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x )²)
J= 1.515 x A/m²
c) The typical speed'' of an electron is given by:
=
=1.515 x / 8.5 x
x |-1.6 x
|
=1.114 x
m/s
d) According to these equations,
J= I/A
=
=
If you were to use wire of twice the diameter, the current density and drift speed will change
Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.
Also drift velocity will decrease as it is inversely proportional to the area
Answer:
Explanation:
The energy of a photon is given by:
where
h is the Planck constant
c is the speed of light
is the wavelength of the photon
In this problem, we have a microwave photon with wavelength
Substituting into the equation, we find its energy: