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Circuits with Series and Parallel Wiring

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ehild

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Correct .

ehild
Ok! So now part b how much power dissipated through a 10 ohm resistor. The power equation.
P=IV
or
P=I2(r)
or
P= v2/R

I tried a few answers but none of them work.
 
Last edited:

ehild

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What is the voltage across the 10 ohm resistor?

ehild
 

NascentOxygen

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Ok! So now part b how much power dissipated through a 10 ohm resistor. The power equation.
P=IV
or
P=I2(r)
or
P= v2/R

I tried a few answers but none of them work.
They don't? http://img690.imageshack.us/img690/3484/hidden17.gif [Broken]

How much current is flowing in just one of the 10Ω resistors?
 
Last edited by a moderator:
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They don't? http://img690.imageshack.us/img690/3484/hidden17.gif [Broken]

How much current is flowing in just one of the 10Ω resistors?
.8 amps?
 
Last edited by a moderator:
65
0
Then I2·R =
and what makes you think this isn't right?
It would be 6.4, and it's wrong because I plugged in the answer and it said it was wrong.
 
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questions like these can be solved in your head!
first, for the current, the net resistance of the topmost two resistors is 5 ohm. now, the reciprocal of 5 is .2 and the reciprocal of 20 is .05 as you should be able to do immediately. they add up to be .25 - which is the reciprocal of the net resistance of the equivalent resistor containing the 20 ohm resistor and also the reciprocal of 4. so, to date, the net resistance is 4 ohms and this adds to the 6 ohms resistor in series. now, clearly, the net current through the battery is 20/(6+4) A = 2 A. this 2 A current divides in two sections in the ratio 4:1 (the current varies inversely as the resistance with p.d. constant). so, the current going through the ammeter is 2[itex]\times[/itex]4/5 = 16/10 = 1.6 A.
for the power, the current 1.6 A just divides in two equal parts, so it's .8 A through each 10 A resistor. so, the power dissipated is 10[itex]\times[/itex].82 = 6.4 W.

so, don't use the loop and the junction rules unless you really need them. they'll mess up everything. try to solve this kinds of problems intuitively or using the method i've described - they are really easy.
 

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