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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Studyinghard on December 30, 2009, 06:46:08 pm

Title: Resistance question
Post by: Studyinghard on December 30, 2009, 06:46:08 pm: Can someone help me with 4d and 5b thanks

image uploaded below
Title: Re: Resistance question
Post by: 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

$60\Omega$
Title: Re: Resistance question
Post by: Studyinghard 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 $13.3/omega$
Title: Re: Resistance question
Post by: kriptik 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 $omega$
Title: Re: Resistance question
Post by: Studyinghard on December 30, 2009, 10:50:27 pm: oh sweet ..i get it now. thanks