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Determine the resistance of a resistor which must be placed in series with a 75 ohm resistor across 120V source in order to limit the power dissipation in the 75 ohm resistor to 90 watts...

Answer»

t ADDED resistance be xCurrent in the circuit if x is added=>I = V/RtotalI = 120 / 75+xPower accros 75 ohm resistorP = I2R90 = (120/75+x)2 x 75x2 + 5625 + 150X= 12000x2 + 150x - 6375 = 0x=34.6



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