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A 75 watts110 volts bulb is connected in parallel with a 40 watts110 volts bulb. What is its net resistance?
Wiki User
∙ 15y ago
We have two bulb in parallel debiting 75 + 40 = 115 watts under 110 volts.
I -current amperes
V -potential volts
W -power watts
R -resistance ohms
knowing
W = V*I
V = I*R
W = R*I2
Then:
115 watts = 110 volts * I
=> I = 115/110 = 1,045 amperes
R = 115/(115/110)2 = 1102 / 115 = 105,217 ohms
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Another way:
First get the resistance of each bulb.
Then we know that Rparallel = 1/(1/R1 + 1/R2 )
75 watts = 110 volts * I => I = 75/110 ampere.
R1 = 75/(75/110)2 = 1102/ 75 = 161,333 ohms.
for the other bulb
40watts = 110 volts * I => I = 40/110 ampere.
R2 = 40/(40/110)2 = 1102/ 40 = 302,5 ohms.
meaning
Rparallel = 1/(1/161,333+1/302,5) = 105,217 ohms
That it's
Wiki User
∙ 15y ago
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