Simple Electricity Calculation Question. HELP!?

Question

What is the voltage drop across the tungsten filament in a 100-W light bulb? The resistance of the filament is 144 (Ohm) and a current of 0.833 (amps) is flowing through it.

The formulas I've learned so far are:
I= Q/t
V= E/Q
R= V/I

Answer 1

Voltage = Current * Resistance (which is a simple rearrangement of your third equation) will work fine.

Many other options here such as

Voltage = Power / Current

Take your pick.

Answer 2

First of all,what is the voltage?Assuming that the voltage is 120 volts,the formula electricians use is..VD=IxR This translates simply to....voltage drop is equal to the amperage times the resistance.Again assuming 120 volts,the equation should read as follows.... 0.833 x 144 = 119.952 This seems like a lot of voltage drop for just 1 lightbulb.It also means that there is only.048 volts left over for the circuit.This cannot be correct because the National Electrical Code only allows for a recommendation of 3% on a branch circuit.Therefore,I think that the resistance of the filament should be .144 This would translate as follows 0.833 x 0.144 = 0.120 VD would then be 0.120 or basically 12% of 1 volt. I am sure this is what the question is looking for,otherwise,it does not make sense from an electrical code standpoint.Good Luck.....PS.....Check out this book for help in the future....OHM'S LAW,ELECTRICAL MATH AND VOLTAGE DROP CALCULATIONS by TOM HENRY......Plain english calculations for electric....

Answer 3

Let's learn a few more formulas
Power is a product of voltage (V) and Current (I)
P=VI
So V= P/I= 100/ .833=120V

Also since R=V/I then I=V/R so
P=V^2 / R

V= sqrt(P R)
V= sqrt( 100 x 144) = 120 v