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 46:1. The primary coil is plugged into a standard 120-V outlet. The current in the secondary coil is 1.8 x 10-3 A. Find the power consumed by the air filter.

Answer 1

Answer:

9.936 Watt

Explanation:

Ns / Np = 46 : 1

Vp = 120 V

Is = 1.8 x 10^-3 A

Ns / Np = Vs / Vp

46 / 1 = Vs / 120

Vs = 5520 V

Power consumed by the air filter = Vs x Is = 5520 x 1.8 x 10^-3 = 9.936 Watt

Answer 2

Answer:

The power consumed by the air filter is 9.936 watts      

Explanation:

It is given that, the secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter.

Turn ratio of the transformer, $\dfrac{N_s}{N_p}=\dfrac{46}{1}$

Voltage of primary coil, $V_p=120\ V$

Current in the secondary coil, $I_s=1.8\times 10^{-3}\ A$

The power consumed by the air filter is :

$P_s=I_s\times V_s$...........(1)

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

So, $P_s=I_s\times (\dfrac{N_s}{N_p})\times V_p$

$P_s=1.8\times 10^{-3}\times (\dfrac{46}{1})\times 120$

$P_s=9.936\ Watts$

So, the power consumed by the air filter is 9.936 watts. Hence, this is the required solution.