A non-ideal voltage source supplies current of 250mA to load resistance of 3.3kOhms. What is the magnitude of the source voltage if the power efficiency is 94%. Also calculate the value of the additional resistance which must be added to the circuit to limit the current drawn from the supply to 200mA and calculate the new power efficiency (PL/PS). So far I think I have worked out the supply voltage: V=IR, (250mA)*(3300)Ohms, To get 825V (825/94%)*100=877.6V How to I carry on with this problem to find internal resistance and the new power effieciency. Please help. Thanks.