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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.

4 Answers

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  • 1 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.

  • qrk
    Lv 7
    1 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.

  • 1 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

  • User
    Lv 7
    1 month ago

    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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