The document describes a circuit with a 5Vrms voltage source in series with a 3kOhm internal resistance connected through an ideal transformer to a 150Ohm load resistance. It asks to calculate the power delivered to the load in milliwatts as the transformer ratio varies from 1.0 to 10.0, and to determine the ratio that delivers maximum power. The solution shows the circuit equations and takes the derivative of power with respect to transformer ratio to find that maximum power is delivered at a ratio of 6/5.

1. A source which can be represented by a voltage source of 5Vrms in series with an internal
resistance of 3kOhm is connected to a 150Ohm load resistance through an ideal transformer.
Calculate the power in milliwatts supplied to the load as a function of the transformer ratio,
covering ratios from 1.0 to 10.0. For what turns ratio is the maximum power transferred to the
load? (Hint: power is max at dPL/da = 0)
Solution
v1/v2 = i2/i1 = a (turns ratio)Â Â Â Â Â Â Â Â Â Â Â Â Â Â i1 primary current i2 is
secondary current
5 - 3i1 = v1
power transfer to load = v2*i2
= v2*i1*a
= (5 - 3i1 )*i1
P =(5-3 i2/a)*i2/a
MAX POWER dp/da =0
a = 6/5 i2