# find the voltage across

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• Oct 11th 2010, 08:09 PM
bigwave
find the voltage across
If a source of $200V$ is connected across the circut of (see attached image)
(a) find the voltage across the $8\Omega$ resister
(b) find the power dissipated in the $1\Omega$ resistence.
Ans. (a) $40V$ (b) $25W$

Attachment 19284

I know that $I=\frac{V}{R}$ but i don't see how they got the answers from this also there is parallel and series resistence in the diagram.... just want to see how they got these answers. thanks ahead.(Cool)
• Oct 12th 2010, 07:37 AM
Ackbeet
(a) What is the current through the $1\Omega$ resistor? I'm not sure I agree with the answer of $40V$ through the $1\Omega$ resistor, incidentally.

(b) For a resistor, $P=VI=(IR)I=I^{2}R.$ I'm not sure I agree with the book's answer here, either.

Just to be clear, the voltage is applied across ae, correct?
• Oct 12th 2010, 11:44 AM
bigwave
what happens to the voltage when it goes thru this circut
I see what you mean, however I am new to this so easy to get stuck

actually I don't know exactly what happens to the voltage when it goes thru this circut, but I am assuming the the current thru

$8\Omega = \frac{200v}{8\Omega} = 25A$

but then we don't get the 40V from this.

so something has to be happening as the circut goes into the series and parallel configuration. yes it appears that we concerned with the $R_{ae}$
• Oct 12th 2010, 12:02 PM
Ackbeet
Whoa! I think you changed the OP. That's generally discouraged in this forum, because it makes subsequent posts seem irrelevant. If you need to change the problem, do so in a later post. Generally, don't edit posts that aren't the last post in the thread. So, think of posts not in pencil, but in pen. The last post is drying ink, and all posts before the last post are dry, and can't be changed. (Ok, pretend there's no such thing as a pen erasor!)

If you want the voltage across the $8\Omega$ resistor, that'll be the same as across the network of resistors that has the equivalent $8\Omega$ resistance. Why? Because the termination points are the same.

If the negative side of the 200V is at e, and the positive side is at a, then the voltage across the $8\Omega$ resistor is just 200V, the same as the network. Resistors in parallel have the same voltage across them, and resistors in series have the same current through them.
• Oct 12th 2010, 12:31 PM
bigwave
so thats what I thot
ok, I repent no more OP sins.

so thats what I thot $200v$ across the $8\Omega$resistor not $40v.$

I'll be posting more ??? as they come.... hope you reply(Cool)

I attached another image as an example trying to see how they got 40v

Attachment 19297
• Oct 12th 2010, 12:48 PM
Ackbeet
Ok. I'll look for them. Have a good one!
• Oct 12th 2010, 01:18 PM
Archie Meade
Quote:

Originally Posted by bigwave
If a source of $200V$ is connected across the circut of (see attached image)
(a) find the voltage across the $8\Omega$ resister
(b) find the power dissipated in the $1\Omega$ resistence.
Ans. (a) $40V$ (b) $25W$

Attachment 19284

I know that $I=\frac{V}{R}$ but i don't see how they got the answers from this also there is parallel and series resistence in the diagram.... just want to see how they got these answers. thanks ahead.(Cool)

Hi bigwave,

your point "b" is clearly to the right of the 16 Ohm resistor.
If the 200V source is placed across ab rather than ae, then you will get the answers quoted,
if the question is asking for the voltage across the 8 Ohm resistor
and the power dissipated in the 1 Ohm resistor.

The effective series resistance seen by the source is then 20 Ohms,
giving a circuit current of 10 Amps.

5 Amps will pass through each of the 8 Ohm parallel arms.
That's 40V across the 8 Ohm resistor and 5V across the 1 Ohm resistor.

The power dissipated in the 1 Ohm resistor is VI=25 Watts.
• Oct 12th 2010, 01:21 PM
Ackbeet
Ah, very nice. I was always wondering in the back of my mind what that 16 Ohm resistor was doing, but I successfully repressed it.