Circuits with Series and Parallel Wiring

ehild

Homework Helper
No? So it would be 1.6 A?

Correct .

ehild

Snape1830

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.

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ehild

Homework Helper
What is the voltage across the 10 ohm resistor?

ehild

Snape1830

What is the voltage across the 10 ohm resistor?

ehild
8? It seems wrong for some reason

NascentOxygen

Mentor
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:

Snape1830

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:

Mentor

Snape1830

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.

NascentOxygen

Mentor
It would be 6.4, and it's wrong because I plugged in the answer and it said it was wrong.
6.4 watts is correct.

Snape1830

6.4 watts is correct.
Oh, right yeah, I just didn't put the decimal point in. Silly me.

Thanks!

Rajatmo

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$\times$4/5 = 16/10 = 1.6 A.
for the power, the current 1.6 A just divides in two equal parts, so it's .8 A through each 10 A resistor. so, the power dissipated is 10$\times$.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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