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Studyinghard · « **on:** December 30, 2009, 06:46:08 pm »

Can someone help me with 4d and 5b thanks

image uploaded below

kriptik · « **Reply #1 on:** December 30, 2009, 09:41:17 pm »

4d is parallel so use

$(\frac{1}{r(1)}+\frac{1}{r(2)}+\frac{1}{r(3)})^{-1}$

which gives

$(\frac{1}{12\Omega }+\frac{1}{8\Omega }+\frac{1}{24\Omega })^{-1}$

which then = $4\Omega$

5b

a series circuit so

$Rtotal =R1 + R2 + R3 ....$

$R= 10\Omega + 5\Omega + 40\Omega$

which gives

$60\Omega$

Studyinghard · « **Reply #2 on:** December 30, 2009, 10:10:25 pm »

Quote from: kriptik on December 30, 2009, 09:41:17 pm

4d is parallel so use

$(\frac{1}{r(1)}+\frac{1}{r(2)}+\frac{1}{r(3)})^{-1}$

which gives

$(\frac{1}{12\Omega }+\frac{1}{8\Omega }+\frac{1}{24\Omega })^{-1}$

which then = $4\Omega$

5b

a series circuit so

$Rtotal =R1 + R2 + R3 ....$

$R= 10\Omega + 5\Omega + 40\Omega$

which gives

$60\Omega$

for 5b the nelson physics answer gives $13.3/omega$

kriptik · « **Reply #3 on:** December 30, 2009, 10:27:26 pm »

oops sorry my bad

the two resistors at the top are in series, so add them together you get 20 ohms.

then add the 20 with 40 using the parallel formula, i didn't see it was a parallel circuit i made a mistake and assumed it was all in series.

so its

$(\frac {1}{20} + \frac {1} {40})^-1$

and it should give 13.3 $omega$

Studyinghard · « **Reply #4 on:** December 30, 2009, 10:50:27 pm »

oh sweet ..i get it now. thanks

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