Respuesta :
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, [tex]\dfrac{N_s}{N_p}=\dfrac{46}{1}[/tex]
Voltage of primary coil, [tex]V_p=120\ V[/tex]
Current in the secondary coil, [tex]I_s=1.8\times 10^{-3}\ A[/tex]
The power consumed by the air filter is :
[tex]P_s=I_s\times V_s[/tex]...........(1)
For a transformer, [tex]\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}[/tex]
So, [tex]P_s=I_s\times (\dfrac{N_s}{N_p})\times V_p[/tex]
[tex]P_s=1.8\times 10^{-3}\times (\dfrac{46}{1})\times 120[/tex]
[tex]P_s=9.936\ Watts[/tex]
So, the power consumed by the air filter is 9.936 watts. Hence, this is the required solution.
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
