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physics -- three resistors of 8, 12 and 24 Watts are in parallel

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physics -- three resistors of 8, 12 and 24 Watts are in parallel and a current of 20 amps is drawn by the combination. Determine (a) the voltage across the combo (b) the current through each resistor. thx

Ok. For parallel circuits, total power P = P1 + P2 + P3.

P1=8, P2 =12, P3 =24. P = 8 + 12 + 24 =44 Watts.

a) Voltage across the combo V= P/I. By plugging I=20A and P=44 Watts, we get V= 44/20 = 2.2V. b) Current through each resistor can be measured as I1 = P1/V, I2 =P2/V, and I3 = P3/V. By plugging the values, we get I1 = 8/2.2 =3.63A, I2 =12/2.2 =5.45A, and I3 =24/2.2 =10.9A.

Chris, quick follow up question -- previously you mentioned "For parallel circuits, total power P = P1 + P2 + P3" -- Is this also true for Power in series or is the equation different? Thx