0
\$\begingroup\$

enter image description here

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 I presume R5 is obsolete, since it's connected to an open circuit.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


\$\endgroup\$
  • \$\begingroup\$ R1,R2 and R4 are _not _ in parallel. R3,R6 and R4 are in series. \$\endgroup\$ – Bob Jacobsen Mar 30 '18 at 5:51
  • \$\begingroup\$ R1 is in parallel with R3+R6+R4, it is not in parallel with R3, R4. \$\endgroup\$ – Harry Svensson Mar 30 '18 at 5:51
1
\$\begingroup\$

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

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.