Question

The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is 45:1. The primary coil is plugged into a standard 120-V outlet. The current in the secondary coil is 1.2 x 10-3 A. Find the power consumed by the air filter.

Answer 1

Answer:

The power consumed by the air filter is 6.48 watt.

Explanation:

 Since

The power used by filter is

$P = I_{p} V_{p}$

and we know that

$\frac{I_{s} }{I_{p} } = \frac{N_{p} }{N_{s} }$

but we need current in primary coil that is

$I_{p} =I_{s} (\frac{N_{s} }{N_{p} })$

Substituting this into the equation of the power we get

$P = I_{s} (\frac{N_{s} }{N_{p} }) V_{p}$

P  = ($1.2*10^-3$)(45/1)(120)

we get

P = 6.48 watt

This the power consumed by the air filter.

Answer 2

Answer:

6.48 W

Explanation:

P = VsIs.................... Equation 1

Where P = Power consumed in the air filter, Is = current in the secondary coil, Vs = Voltage in the secondary coil

But,

Ns/Np = Vs/Vp..................... Equation 2

Where Ns = Number of turns in the secondary coil, Np = Number of turns in the primary coil, Vp = Voltage in the secondary coil, Vs = Voltage in the primary coil.

Make Vs the subject of the equation

Vs = NsVp/Np................... Equation 3

Substitute equation 3 into equation 1

P = NsVpIs/Np................. Equation 4

Given: Ns:Np =45/1, Vp = 120 V, Is = 1.2×10⁻³ A

Substitute into equation 4

P = 120×45/1×1.2×10⁻³

P = 6480×10⁻³ W.

P = 6.48 W

Hence the power consumed by the air filter = 6.48 W.