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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))]RoI had been asked this question during an interview.My question is based upon Series Equalizer in portion Filters and Attenuators of Network Theory

Answer»

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.



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