Consider three identical resistors, R. The maximum power each can dissipate is P. Two of the three resistors are connected in series, and the third is connected in parallel with the other two. What is the maximum power this network dissipate?
2007-10-30 09:44:34 · 5 answers · asked by littlet9393 1 in Science & Mathematics ➔ Engineering
If voltage V is applied to the network, the power dissipated by the single resistor will be V^2/R. Each resistor of the pair of series connected resistor will have V/2 across it and dissipate (V/2)^2/R. To avoid exceeding the power rating of the single resistor, V must be adjusted so that V^2/R = P. Therefore the power dissipated each of the other two resistors will be P/4 and the total power will be 1.5 X P.
If a balanced three phase AC voltage is applied to the three nodes of the circuit, including the point between the series connected resistors, the voltage can be adjusted so that the maximum power, P is dissipated by each of the three resistors.
2007-10-30 11:54:44 · answer #1 · answered by EE68PE 6 · 1⤊ 1⤋
The above solutions are astounding. regardless of if it truly is many times counseled to apply a resistor with a wattage score equivalent to 2 times the dissipation interior the resistor. A 5 watt resistor, at the same time because it is going to attend to the dissipation, gets extremely heat and could in all possibility replace over the years. stable prepare says that a ten watt resistor could be used.
2016-11-09 20:57:16 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Power = Current x Voltage and using ohm's law
Power = current^2 x Resistance or (Voltage^2)/Resistance
For your problem you have do find the equivalent resistance of the network...
Resistors in series, R total = R1 + R2
Reistors in parallel, 1/(Rtotal) = (1/R1) + (1/R2)
Use those formulas to come up with the equivalent resistance
2007-10-30 09:51:31 · answer #3 · answered by Dave C 7 · 0⤊ 0⤋
you have R connected across 2R (although you say 3 resistors in series)
so the Voltage is halved for the 2 in series
Power is proporttional to voltage squared
(V^2/R)
So the single Resistor sees V
and the other two see V/2
so the power is V^2/R for the single one
and (V/2)^2/R for the other 2
total power is V^2/R and V^2/2R
or 1.5V^2/R
or 1.5P
2007-10-30 12:29:50 · answer #4 · answered by Anonymous · 0⤊ 0⤋
All you want to know and more:
http://www.watlow.com/reference/equations/0103.cfm
2007-10-30 10:07:39 · answer #5 · answered by Anonymous · 0⤊ 2⤋