What is the resistance and current through a 62 W lightbulb if it is connected to its proper source voltage of

Question

in ohms (resistance) and amps (current)

Answer 1

Basically, power = current^2 * resistance

So you have 62 = I^2 * R

And also V = I * R

if V = 120 volts (like it is for most of the U.S.A), then
120 = I * R
62 = I*I * R
62 / 120 =
0.5167 amps = I

And also 62 = (0.5167^2) * R
So R = 232.2 ohms

Answer 2

We have P = E^2/R, so if it is run off a 120 volt source, the resistance can be determined as R = 120(120)/62, or roughly 240 ohms. The current can be determined using P = EI, so it will be about half an ampere.

Answer 3

V= I R and P = V * I , so 62= 120* I so I = .5166 amps

v= volts, I= current in amps, and R= resistance in ohmns

120v = .5166 * R R= 120/.5166 = 232.28 ohms

Note: the Resistance changes when the bulb heats up. That is why you can not put a ohm metter on it and get an accurate reading.

Answer 4

Ohms law: P = IV and V = IR

So, P = I(IR) = Isquared * R

If you know the power (in watts) and the voltage, then you have it.

Answer 5

assumed v=120volt
r=v*v/w=120*120/6.2=?ohm
i=w/v=6.2/120=?amp

Answer 6

challenging step. seek with bing and yahoo. that may help!

Answer 7

0.5636363636363 amps
195.161290322580 ohms
at 110V