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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 I presume R5 is obsolete, since it's connected to an open circuit.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?

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  • \$\begingroup\$ R1,R2 and R4 are _not _ in parallel. R3,R6 and R4 are in series. \$\endgroup\$
    – Bob Jacobsen
    Commented Mar 30, 2018 at 5:51
  • \$\begingroup\$ R1 is in parallel with R3+R6+R4, it is not in parallel with R3, R4. \$\endgroup\$
    – Harry Svensson
    Commented Mar 30, 2018 at 5:51

1 Answer 1

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\$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: $$I_{346}=\frac{R1}{R1+R_{346}}I_0$$

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