Can anyone give me a hand?
John Creighton
dennis wrote:
> please help me solve this problem, thank you!!
>
> [Image]
Just use Kirchoves (SP?) Voltage Law around the outside loop.
John Creighton
dennis wrote:
> please help me solve this problem, thank you!!
>
> [Image]
Form a third loop connecting the one amp current going out to the one
amp current going in. This loop will act as a current source. Then apply
mesh analysis to the three loops.
Jon Bell
>M&#$C)1TH.C,]/#DS.#= 2%Q.0$1713<X4&U15U]B9VAG/DUQ>7!D>%QE9V/_
[snip]
If your problem is that a noisy phone line is messing up your Internet
connection, you'd better contact your telephone company. ;-)
--
Jon Bell <jtb...@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
Old Man
dennis <den...@hongkong.com> wrote in message
news:9f5nek$48...@imsp212.netvigator.com...
> please help me solve this problem, thank you!!
Since they both have the same (unknown) voltage drop, the 10 ohm and 40 ohm
resistors can be treated as if they are in parallel:
1 / Rp = 1 / 10 + 1/40
Rp = 8 ohm
Vp = (1 amp) * Rp = 8 volt
I (upper branch) = (8 volt) / (10 ohm) = 0.8 amp
I (lower branch) = (8 volt) / (40 ohm) = 0.2 amp
The battery and the resistor, R, also have the same voltage across them:
R = (6 volt) / (0.2 amp) = 30 ohms
With battery internal resistance of 2 ohms
Vbat = (6 volt) + (2 ohms)*( 0.8 amps) = 7.6 volts
R = (7.6 volt) / (0.2 amp) = 38 ohms
You should never trust strangers. So you are still required to understand
the principles of solving the circuit before you can trust the answer The
main principle here is that current is separately conserved in the upper
circuit branch and the lower branch. [Old Man]
dennis
i am such an idiot
"Old Man" <o...@bg.net> wrote in message
news:2RER6.569$G91....@newsfeed.slurp.net...
jmfb...@aol.com
<grin> We all have those. Your first line indicates
it's a momentary condition that has been cured ;-).
/BAH
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