An a.c. source of internal resistance `9000 Omega` is to supply current to a load resistor of 10 ohm. How should the source be matched to the load and what is the ratio of the currents passing through the load and the source ?

Explanation

The a.c. source must be connected to primary of transformer and load should be connected of secondary of the transformer.
As `I_(s)^(2) R_(s) = I_(P)^(2) R_(P)`
`(I_(s))/(I_(P)) = sqrt((R_(P))/(R_(s))) = sqrt((900)/(10)) = 30`
Also, `(I_(s))/(I_(P)) = (n_(P))/(n_(s)) = 30 :. (n_(s))/(n_(P)) = (1)/(30)`