Question

Q5/// The maximum power dissipated in a 4 load is 100 w when connected to a d.c. voltage V and internal resistance R. Calculate: the current in the load, internal * .resistance R, and voltage V 4 A, 3 Ohm, 30 v o 3 A, 5 Ohm, 12 v 0 2 A, 6 ohm, 24v O 6 A, 2.5 ohm, 220 v O 5 A, 4 Ohm, 40 v O 10 A 4.5 ohm. 60v 0

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Answer 1

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Given data! Load Besistance(RU= 42 Maximum Power dissipated in lood is (Pmax) = 100w I R = 100 T R = 42 IL M 4 I 2 = 100 I 2 = Load current I = 25 I = 5 Amps To get maximum power to the load RL=B Internal Besistance(R)=42 DC voltage (V) = I (R+R) 5(4+4) = 5x8 DC voltage (V). 40/volts 5th option Answer - 5A, 4ohm, 400