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Draw a circuit diagram, and find the actual power of the bulb in the circuit described.?
A lightbulb marked “75 W [at] 120 V” is screwed into a socket at one end of a long extension cord in which each of the two conductors has a resistance of 0.800 ohms. The other end of the extension cord is plugged into a 120-V outlet. Draw a circuit diagram, and find the actual power of the bulb in the circuit described.
- billrussell42Lv 71 month ago
Assuming, and it is a big assumption, and not true, that the bulb is linear.... and assuming that it is actually exacty 75 watts at that voltage, another false assumption.
P = E²/R
R = E²/P = 120²/75 = 192 Ω
assuming the "120 volt outlet" is actually exactly 120 volts...
voltage divider, V = 120•192 / (192+0.8+0.8) = 119 volts
P = E²/R = 119²/192 = 73.8 watts
diagram is 3 resistors in series, the middle the bulb, the other two wire resistance
yet another silly problem, written by someone who has no idea of what this is about.
- qrkLv 71 month ago
In reality, this problem can't be solved unless you have the I-V curve of the bulb. Filament resistance is voltage dependent. As usual, a really poorly presented homework problem.
- oldschoolLv 71 month ago
V²/R = P so R = V²/P = 120²/75 = 192Ω
Assume the bulb's resistance is constant = 192Ω
The power consumed by that resistance in the ckt described:
P = i²*R = [120/(0.8Ω+192Ω+0.8]² *192Ω = 73.77 or about 74W
- UserLv 71 month ago
voltage supply = 120 V
light bulb = 192 ohms (see below for calculation)
R2 = 0.800 ohms
light bulb resistance calculation
with 120V drop across the light bulb
(no other resistance in the circuit)
the light bulb dissipates 75 W
we know the equation
W = IV
and in this case we know W and V
but want to calculate R for the lightbulb
since V = IR
I = V/R
plugging that into our W=IV wattage equation
(eq x) W = Vsquared/R
solving for R
R = Vsquared/W
Rlightbulb = 120 * 120 / 75 = 192 ohms
You'll need to calculate the voltage across the light bulb in your circuit
and from that and from the resistance of the light bulb
we can use the equation we already figured out
and solve for the power dissipated by the light bulb in that circuit.