Question

What is the power at the load of a series equalizer?

(a) [Vmax^2/(Ro^2+X1^2)]Ro

(b) [Vmax^2/(2(Ro^2+X1^2))]Ro

(c) [Vmax^2/(3(Ro^2+X1^2))]Ro

(d) [Vmax^2/(4(Ro^2+X1^2))]Ro

I had been asked this question during an interview.

My question is based upon Series Equalizer in portion Filters and Attenuators of Network Theory

Answer 1

The correct choice is (d) [Vmax^2/(4(Ro^2+X1^2))]Ro

To elaborate: The power at the load of a series equalizer is P=(Vmax/√((2Ro)^2+(2X1)^2))^2 Ro = [Vmax^2/(4(Ro^2+X1^2))]Ro.