A light bulb is wired in series with a 144-Ω resistor, and they are connected across a 120-V source. The power delivered to the light bulb is 23.0 W. What are the two possible resistances of the light bulb?
lower value Ω
higher value Ω
lower value Ω
higher value Ω
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According to Kirchoffs Voltage Law: Sum of Voltage Drops across bulb and resistor must equal to 120
Hence using, V=P/I , V=IR and R=P/I^2
23/I + 144I = 120
Multiply all terms by I
23 + 144I^2 = 120I
144I^2-120I+23=0
Using quadratic formula you get two currents i-e: I=0.534A and I=0.298A
Now using R=P/I^2
R(lower)=23/0.534^2 = 80.6ohms
R(Upper)=23/0.298^2 = 259ohms
Hence using, V=P/I , V=IR and R=P/I^2
23/I + 144I = 120
Multiply all terms by I
23 + 144I^2 = 120I
144I^2-120I+23=0
Using quadratic formula you get two currents i-e: I=0.534A and I=0.298A
Now using R=P/I^2
R(lower)=23/0.534^2 = 80.6ohms
R(Upper)=23/0.298^2 = 259ohms
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80.5 Ω
257.6 Ω
257.6 Ω