Answer:
The power consumed by air filter is 10.08 W
Explanation:
Given:
Primary voltage [tex]V_{p} = 120[/tex] V
Current in secondary coil [tex]I_{s} = 2 \times 10^{-3}[/tex] A
Turn ratio [tex]\frac{N_{s} }{N_{p} } = \frac{42}{1}[/tex]
For finding power consumed by filter,
[tex]P = V_{p} I_{p}[/tex]
We know that the relation between current and number of turns,
[tex]\frac{I_{s} }{I_{p} } = \frac{N_{p} }{N_{s} }[/tex]
[tex]P = V_{p} I_{s} \frac{N_{s} }{N_{p} }[/tex]
[tex]P = 120 \times 2 \times 10^{-3} \times 42[/tex]
[tex]P = 10.08[/tex] W
Therefore, the power consumed by air filter is 10.08 W
