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[SOLVED] Basic circuit analysis

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Abhishek_Anand

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Let's say we have a light bulb of the following rating : 20W, 100V, 20 Ohms.
Let's say that, 20W of power is being supplied to the bulb.
By Ohm's Law current through the bulb will be 100/20 A.
Now if I use the power formula P=VI, I=20/100 A.
Again if I use the heat loss formula P=I^2*R, I will be different.
I want to know where exactly are these formulas used. There must be particular cases where we use them.
Am I interpreting the power formulas wrong?
Now what if the supply is not able to supply the 20W that the load needs. How will it affect the current in the load.
 

Let's say we have a light bulb of the following rating : 20W, 100V, 20 Ohms.
Let's say that, 20W of power is being supplied to the bulb.
By Ohm's Law current through the bulb will be 100/20 A.
Now if I use the power formula P=VI, I=20/100 A.
Again if I use the heat loss formula P=I^2*R, I will be different.
I want to know where exactly are these formulas used. There must be particular cases where we use them.
Am I interpreting the power formulas wrong?
Now what if the supply is not able to supply the 20W that the load needs. How will it affect the current in the load.
Click to expand...
Hi Abhishek_Anand
The answer is simple ! the impedance of bulb is not what you are going to measure like that ! when it is going to be warmer the impedance is going to be much higher ! on the other hands a bulb which has a filament , can be considered as something like PTC !
Best Wishes
Goldsmith
 
I guess you have measured a 20W bulb and found the 20 ohms? Filament bulbs have a resistance when hot which is maybe 15 times the cold resistance. So, a 100V 20W bulb would be 500 ohms when hot - it must be otherwise it wouldn't be a 20W/100V bulb. However, when cold you might measure only 30 ohms.

Keith.
 
Let's say we have a light bulb of the following rating : 20W, 100V, 20 Ohms
Click to expand...
There is an error here P= V^2 /R if the bulb is 100v 20w then its resistance is 500ohms (with 100v applied)
By Ohm's Law current through the bulb will be 100/20 A.
Click to expand...
I=V/R .2A= 100/500 .
P=I^2 * R 20=.2^2 * 500
Everything comes right when you correct your assumption of 20 ohms.
Now what if the supply is not able to supply the 20W that the load needs. How will it affect the current in the load.
Click to expand...
A light bulb is a case of nonlinear resistance. If you supply the bulb with something different than 100v then its resistance will be lower (as has been said before). I=V/R will not hold out for the R that you calculated before.
 
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