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 Apr 13th 2009, 06:38 AM #1 Junior Member   Join Date: Apr 2009 Posts: 6 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
 Apr 13th 2009, 10:35 AM #2 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 (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
 Apr 14th 2009, 03:58 PM #3 Junior Member   Join Date: Apr 2009 Posts: 6 thanks, much appreciated.
 Apr 14th 2009, 10:44 PM #4 Physics Team   Join Date: Feb 2009 Posts: 1,425 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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