Answer:
[tex]62.5 \Omega[/tex]
Explanation:
First of all, we calculate the resistance of the lamp. The power of the lamp is given by
[tex]P=\frac{V^2}{R'}[/tex]
Where
P = 100 W is the power
V = 250 V is the voltage
R' is the resistance of the lamp
Solving for R',
[tex]R'=\frac{V^2}{P}=\frac{250^2}{100}=625 \Omega[/tex]
Now this resistance is connected in parallel to the other resistance R, so their equivalent resistance is
[tex]\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R'}[/tex] (1)
Now the total power in the circuit is
P = 1100 W
And the voltage is
V = 250 V
So we can write
[tex]P=\frac{V^2}{R_{eq}}[/tex]
to find the equivalent resistance:
[tex]R_{eq}=\frac{V^2}{P}=\frac{250^2}{1100}=56.8 \Omega[/tex]
Therefore, rearranging (1), we find R:
[tex]\frac{1}{R}=\frac{1}{R_{eq}}-\frac{1}{R'}=\frac{1}{56.8}-\frac{1}{625}=0.016 \Omega^{-1} \rightarrow R=62.5 \Omega[/tex]