problem with circuit

Sep 2016

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
1/R2 + 1/R4 = 0.1= 10k ohms
10k + R3 = 10k + 20k = 30k ohms
1/10k + 1/30k = 7.5k ohms

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

am i correct in this way since i dont have the answer
Dec 2013
Encinitas, CA
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

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

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


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

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}


and $V_{ab} = i_1 R_4 + i_2 R_3 = \dfrac{v(R_1+R_2+R_3)}{R_2^2-(R_2+R_4)(R_1+R_2+R_3)}R_4 + \dfrac{v R_2 }{R_1 R_2 + R_1 R_4+R_2 R_3 + R_2 R_4 + R_3 R_4}R_3$


Last edited:
  • Like
Reactions: 1 person