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## Homework Statement

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?

## Homework Equations

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.

Any help?