For this circuit (a and b are connected by a battery),

circuit diagram

Will I be able to find the total resistance of the circuit by adding resistors that are in series and combining resistors that are in parallel without using the method of Kirchhoff's Voltage and Current Law?

I tried adding R3+R4 and this forms one resistor, which is in parallel with R2 and R1. And R2 is in parallel with R1 and R3, and R1 is in parallel with R2 and R3. This can be verified by Kirchhoff's Voltage laws. The problem is, I get stuck here. I know that if R1,R2 and R3+R4 can be combined into one resistor, then parallel-series would solve the problem neatly. But I can't see how these resistors be combined.

  • $\begingroup$ Step by step: R3 and R4 are in series, so replace them with a single resistor R7. Now R2 and R7 are in parallel, then... You can get there this way $\endgroup$ – Ross Millikan Feb 5 '14 at 23:25
  • $\begingroup$ @RossMillikan That's what I did, but then I get stuck. $\endgroup$ – user29157 Feb 5 '14 at 23:41
  • $\begingroup$ Can hold be released? $\endgroup$ – user29157 Feb 5 '14 at 23:48
  • $\begingroup$ You can combine three parallel resistors just fine. One way is just to combine two of them (they are in parallel), then combine that with the third. Now you have one resistor in place of R1, R2, R3, R4. $\endgroup$ – Ross Millikan Feb 5 '14 at 23:51
  • $\begingroup$ but then where would the combined resistor be? Would it be in series with R6? Or Would it be in series with R5? $\endgroup$ – user29157 Feb 5 '14 at 23:53

The net resistance depends on the point where you have applied the potential difference. Indicating the direction of current is very useful.

You can Use $[$ $($ $R_3$ series $R_4$ $)$ parallel $R_2$ $]$ to be $R_7$

Now Redraw the circuit and by naming the point with same potential as one point as I have done in my diagram. Helps a lot:

Seems like I have made an error in direction of current in $R_7$ . It should be opposite. But it matters not as if you apply Kirchoff's law the current will come out to be negative.

enter image description here


Many times this kind of problems becomes very simple just by redrawing the circuit in a more standard way. Have a look at this:

These are all equivalent circuits!

Now I think you will have no problem to solve it!


R_total=((((R_1+R_2)/(R_1 R_2 )+1/(R_3+R_4 ))^(-1)+R_5 )^(-1)+1/R_6 )^(-1)


protected by Qmechanic Oct 5 '17 at 18:17

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