# What is the current flowing through the resistor in this circuit?

I have to find the current flowing through the resistor connecting node a to d. I know it's a current divider problem, but I'm having a hard time finding the correct expression. I presume R5 is obsolete, since it's connected to an open circuit. Since $R_{1}$, $R_{2}$, $R_{4}$ are connected in parallel, their equivalent resistance $\frac{1}{R_{T}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{4}}$. Since we have to find the current through through $R_{3}$, the current $i$ should be:

$$i = I_{0} \frac{R_{T}}{R_{3}+R_{T}}$$.

Is this correct? Am I doing something wrong?

• R1,R2 and R4 are _not _ in parallel. R3,R6 and R4 are in series. – Bob Jacobsen Mar 30 '18 at 5:51
• R1 is in parallel with R3+R6+R4, it is not in parallel with R3, R4. – Harry Svensson Mar 30 '18 at 5:51

$R1$, $R2$ and $R4$ are not in parallel, they are connected in a star or T configuration.
The current $I_0$ is devided between $R_1$ and $R_{346}$ in the first part of circuit, where $R_{346}=R_3+R_4+R_6$.
From the basic current divider formula: $$I_{346}=\frac{R1}{R1+R_{346}}I_0$$