To solve this problem we will start considering the relationship between rms voltage and the maximum voltage. Once the RMS voltage is obtained, we will proceed to find the system current through the given power and the voltage found. Finally by Ohm's law we will find the resistance of the system
The relation of RMS Voltage is,
[tex]V_{rms} = \frac{1}{\sqrt{2}}*V_{max}[/tex]
Note that this conversion is independent from the Frequency.
[tex]V_{rms} = \frac{1}{\sqrt{2}}*(170)[/tex]
[tex]V_{rms}= 120.20V[/tex]
Now the current is,
[tex]I =\frac{P}{V}[/tex]
[tex]I = \frac{(25 W)}{(120.20V)}[/tex]
[tex]I = 0.2079A[/tex]
By Ohm's Law
[tex]V = IR \rightarrow R = \frac{V}{I}[/tex]
Replacing,
[tex]R = \frac{ (120.20 V)}{ ( 0.2079 A)}[/tex]
[tex]R = 578.16 \Omega[/tex]
Therefore the resistance of this lightbulb is[tex]578.16 \Omega[/tex]
