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A 950 W bulb and a 475 W bulb are both designed to operate at standard household voltage of 120 V. Determine which bulb has the lower filament resistance and then calculate the value of its resistance. Answer in units of Ω.

Answer :

Explanation:

Given that,

Power of bulb 1, [tex]P_1=950\ W[/tex]

Power of bulb 2, [tex]P_2=475\ W[/tex]

Household voltage, V = 120 V

We know that the power of any electrical equipment is given by :

[tex]P=V\times I[/tex]

Also,

[tex]P=\dfrac{V^2}{R}[/tex]

For bulb 1,

[tex]R_1=\dfrac{V^2}{P_1}[/tex]

[tex]R_1=\dfrac{(120)^2}{950}[/tex]

[tex]R_1=15.15\ \Omega[/tex]

For bulb 2,

[tex]R_2=\dfrac{V^2}{P_2}[/tex]

[tex]R_2=\dfrac{(120)^2}{475}[/tex]

[tex]R_2=30.31\ \Omega[/tex]

Bulb 1 has lower filament resistance having 15.15 ohms resistance. Hence, this is the required solution.

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