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Old Sep 24th 2016, 03:47 AM   #1
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Join Date: Sep 2016
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problem with circuit


I got a problem of this circuit diagram, where a and b are open circuit
I need to find the voltage across R4 but i feel confused with finding equivalent resistance
Here my way:
R1 || R2 and R3
R2 is in series with R3
R2 || R4
therefore:
1/R2 + 1/R4 = 0.1= 10k ohms
10k + R3 = 10k + 20k = 30k ohms
1/10k + 1/30k = 7.5k ohms

by V=IR
I=4/7.5k=5.3333x10^-4
the voltage across R4=(5.33x10^-4 )(20k)=10.667V

am i correct in this way since i dont have the answer
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Old Sep 24th 2016, 12:26 PM   #2
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You can't solve this circuit reducing it like that. You have to use Kirchoff's laws to solve for the currents in the circuit and the the voltage across $ab$ is the sum of the voltage across $R_3$ and the voltage across $R_4$. I would set this up as

$v=(i_1-i_2)R_2 + i_1 R_4$

$0 = i_2(R_1+R_3) + (i_2-i_1)R_2$

or

$\begin{pmatrix}R_2+R_4 &-R_2 \\ -R_2 &R_1+R_2+R_3 \end{pmatrix} \begin{pmatrix}i_1 \\ i_2 \end{pmatrix} = \begin{pmatrix}v \\ 0 \end{pmatrix}$

solving this you obtain

$\begin{pmatrix}i_1 \\ i_2 \end{pmatrix} = \begin{pmatrix}

\dfrac{v(R_1+R_2+R_3)}{R_2^2-(R_2+R_4)(R_1+R_2+R_3)} \\

\dfrac{v R_2 }{R_1 R_2 + R_1 R_4+R_2 R_3 + R_2 R_4 + R_3 R_4}

\end{pmatrix}$

and $V_{ab} = i_1 R_4 + i_2 R_3 = \dfrac{v(R_1+R_2+R_3)}{R_2^2-(R_2+R_4)(R_1+R_2+R_3)}R_4 + \dfrac{v R_2 }{R_1 R_2 + R_1 R_4+R_2 R_3 + R_2 R_4 + R_3 R_4}R_3$
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Last edited by Romsek; Sep 25th 2016 at 09:25 AM.
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