# Power Efficiency Type Question

1. Nov 10, 2007

### Kobayashi

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.

2. Nov 10, 2007

### dontharmanimals

internal resistance = (877.6 - 825 ) / 250 = 0.21 K =210 ohms

3. Nov 10, 2007

### dlgoff

"So far I think I have worked out the supply voltage:
V=IR, (250mA)*(3300)Ohms, To get 825V"

In your equation, what units should I and R be inorder to get V in volts?

Now what resistance would you need to add to get I=200mA?

4. Nov 11, 2007

### dontharmanimals

If current I is in milliamperes, resistance R must be in kilo-ohms to get Volts.
or
If current I is in amperes, resistance R must be in ohms to get Volts.

Total resistance R, required in the circuit to limit current to 200 mA is given by R = V / I.

R ( total ) = 877.6 volts / 200 mA = 4.388 kilo-ohms = 4388 ohms.

Additional resistance required = R ( total ) - internal resistance - resistance already
available in the circuit

= 4388 - 210 - 3300 = 878 ohms

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