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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.

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• 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.

• 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.

• 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

• voltage supply = 120 V

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R1=0.800 ohms

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light bulb = 192 ohms (see below for calculation)

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R2 = 0.800 ohms

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ground

light bulb resistance calculation

75W @120V

that means

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

so

since V = IR

I = V/R

and so

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

(eq x)

and solve for the power dissipated by the light bulb in that circuit.

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