A light-bulb and a resistor is connected in a serie. The resistor is 100 Ohm. And the voltage over the bulb is described as U=I^2/(0.8*10^-3).
If i connect the serie to a 100 Volt DC source, how do i calculate the current and the resistance over the bulb?
Well Let us first analyze the situation. Your problem is illustrated in a picture given below.
It is always better to have a picture to solve the problems,you know.
Here is the representation...
Hope that this picture completely match with your problem.
Now let us start...........
We have applied a 100V dc across the series combination. Let us assume the voltage drop inside the resistor is V
R and voltage drop across the bulb is V
B.
So the total voltage will be V
DC=100V= V
R + V
B
But from Ohm's Law. V
R = I* R = I*100
also from your problem V
B =
There fore we will get an equation given below
I
2/(0.8*10
-3) + I* 100 = 100 V
Simplifying this, we get
I
2 + 0.08 I = 0.08 V
Solving this for I, we get
I = .2456571371 & I = - 0.3256571371
Here , it is not wise to assume a negative value for current.
so let us take
I = 0.24566 A
This is the current that flows through the series combination.........
Now your question
how do i calculate the current and the resistance over the bulb?
see, we got the current, but to find resistance we need the voltage drop across the Bulb.
so
V
B =
V
B = .24566
2/(0.8*10
-3) = 75.4360445 V
V
R = I*R = .24566*100=24.566V (no need, but just calculated for fun!)
now , By ohm's Law, V=I*R
therefore, R = V/I = 75.4360445/0.24566 = 307.075 ohms.... the resistance offered by the bulb.
thus, your final answers are
Current, I = 0.24566 A
Resistance, R=307.075 ohms