the question has a transformer, the hv side of which is connector to a pot that is said to be an attenuator in the question with attenuation factor = .8?
the turns ratio of transformer is 1:1.125.
I have to find HV side voltage after the attenuation...i know attenuator reduces voltage but how does the factor come into picture?
transformer HV Voltage = n1/n2*LV Voltage = 1.125/1*230 = 258.75
The Attempt at a Solution
I considered attenuator as resistor with .8 value of total value. like a pot and took output voltage as V=V(HV)*.8/(.8+.2)=V *.8 = 258.75*.8 = 207V
I dont think i used the attenuatior factor rightly. Google suggested a log formula. Like 10 raised to A/20. where A is factor.
I'm not sure how to proceed.