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Introduction to power dissipation

# Introduction to power dissipation

Anonymous
#1 Sep 12, 2012
I'm new to the field, and I'm taking my first course through EDX, it's free and need a little help on one questions.

If I have three resistors with values of 4, 4, and 6 ohms respectively and given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up? I'd prefer knowing how to solve it rather than just the answer.
krcs

Vaggeryd, Sweden

#2 Sep 13, 2012
2.67
SIMPLE!!
4/1.5=2.67
anon

Philippines

#3 Sep 13, 2012
How could you come up with that solution???
krcs

Vaggeryd, Sweden

#4 Sep 13, 2012
anon:the smallest value of 4,4,6 ohm res. is 1.5 there for 2.67
sam

#5 Sep 13, 2012
ok, but 4?
comendante

Kraków, Poland

#6 Sep 13, 2012
the resistance of the smallest of three resistors is 4 ohm. The voltage which should be put on the terminals of this resistor to get power 1 W is 2 volts. (P = (U^2)/R). When you change R value from 4 ohm, to resistance of whole circuit you get P = 4/1.5)
snorlax

Europe

#7 Sep 14, 2012
great comendante!
sam

#9 Sep 14, 2012
Sorry, but that is not possible, you made ohm/ohm, you need to have, amp2/ohm.
anon

Philippines

#10 Sep 14, 2012
comendante

Kraków, Poland

#11 Sep 15, 2012
sam ->
P = U*I [V*A = W](1)
R = U/I [V/A = Ohm]=> I = U/R [V/Ohm = A](2)

P =(U^2)/R [V^2 / Ohm](3)

because of (2)[V/Ohm = A],

[V^2 / Ohm]=[V *(V/Ohm)]=[V*A]=[W]
Siddharth Prakash

Trivandrum, India

#12 Nov 16, 2015
It is not stated in the question that the smallest resistor consumes 1 watt only that the individual resistors consumes 1 watt..how can that be right because the individual resistors are of diffeent value
Calvs

Dasmarińas, Philippines

#13 Apr 29, 2016
The total power P in Watts that can be consumed by the smallest-valued composite resistor is found by calculating the voltage at which the first of the three resistors burns up. As the composite resistor consists of three resistors in parallel, all three resistors share the same potential V. It follows that the voltage at which the smallest resistor burns up is the maximum voltage one can apply to the resistor network. Given that the smallest resistor is Rmin=4&#937;, we can find the maximum voltage the network can sustain before burning up:

Using P=V^2/Rmin=1 W, the maximum voltage is V=2 V.

At a potential of 2V both 4&#937; resistors dissipate 1W. The 6&#937; resistor dissipates:
(2^2)/6=2/3W
The total power consumed is therefore 1W+1W+2/3W=2.6667W

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