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Old Apr 13th 2009, 05:38 AM   #1
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Generator question



Require some pointers, thanks.



A generator has an open circuit voltage of 12V and an internal resistance of 40Ω. Calculate;
a)The load resistance for maximum power transfer
b)The on-load terminal voltage
c) The power supplied to the load
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Old Apr 13th 2009, 09:35 AM   #2
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(a) The maximum power is transferred when external resistance is equal to internal one. Hence the load resistance required in this case is 40 ohm.

(b) The current through the resistance is V/R = 12/80 A. ( 80 ohm is the total resistance of the circuit ). Hence the potential difference accros the external resistance (terminal voltage) = IR = 12*40/80 = 6 V.

(c) The power is then VI = 6*12/80 W
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Old Apr 14th 2009, 02:58 PM   #3
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thanks, much appreciated.
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Old Apr 14th 2009, 09:44 PM   #4
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The max power transfer occurs when the internal resistance and the load are equal.
In this case Ri = 40Ω
Total resistance = 80Ω. The current thru the ckt = 12/80.
The voltage at the terminlas = (3/20) x 40 = 6V
The power = (V^2) /R = 36/40 = 0.9 W.
Thus the resistor used should be of at least the same or greater wattage. Standard 1 W resistors are available and may be used.
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