I know that finding the equivalent resistance depend on whether resistors are in serie (R1+R2+...+Rn) or parallel (1/Req=1/R1+1/R2+...+1/Rn), but i don't know how to apply that to find the equivalent resistance of this circuit (not including Ra):


simulate this circuit – Schematic created using CircuitLab

My solution:


simulate this circuit

R3, R1 and R2 are in parallel so: 1/Req=1/R3+1/R1+1/R2 so Req=3.299 Ohm.

I assumed that R3 isn't in parallel with R2 and R1, so I calculated Req in different ways: Req=(R1//R2)+R3+ Or: Req=(R1//R3)+R2 Or: Req=(R3//R2)+R1

I know this may be the most ridiculous question you have seen today, but i'm very beginner in electronics. Any help would be apprieciated.

Thanks for reading my question.

  • \$\begingroup\$ From your description, it's not clear exactly what you're trying to do. Your description states that you want the equivalent resistance but the solution diagram looks more like a Thevenin transformation. \$\endgroup\$
    – vini_i
    Sep 27, 2017 at 18:27
  • \$\begingroup\$ If you don't include Ra, then you get (10+1 = 11) and 9.7 in parallel. Thus, 1/Re = 1/11 + 1/9.7. \$\endgroup\$ Sep 27, 2017 at 18:32
  • \$\begingroup\$ @vini_i I want to replace R1, R2 and R3 with a single resistance Req in serie with V1, my question is how to find the value of Re?. \$\endgroup\$
    – Mehdi
    Sep 27, 2017 at 18:32
  • \$\begingroup\$ No, \$R_1\$, \$R_2\$, and \$R_3\$ are not in parallel with each other. \$\endgroup\$
    – jonk
    Sep 27, 2017 at 18:38
  • \$\begingroup\$ @BartekBanachewicz Then Req=5.15Ohm, and the simulation would be: link, but Ra receive in the original circuit 90.8uA not 995uA. \$\endgroup\$
    – Mehdi
    Sep 27, 2017 at 19:03

2 Answers 2


The two possible solutions are as follows.

The first diagram is Ra in parallel with R2. Then R3 is in series with the result. Finally, R1 is in parallel with the result.

Ra || R2 = 0.99

0.99 + R3 = 10.99

10.99 || R1 = 5.15

Notice that in the second diagram the voltage of the supply is lower. This is called a Thevenin equivalent.

For a Thevenin equivalent first, remove the load and calculate the voltage across where the load was. In this case, R2 and R3 form a resistive divider.

V1 * (R2 / (R3 + R2)) = 0.090V

Then second, short the supply (turn it off or reduce it to zero for a voltage supply and calculate the equivalent resistance. With the supply sorted R3 is completely bypassed. This leaves R2 in parallel with R1.

R1 || R2 = 0.9 ohms.

Finally, take the voltage and resistance and set it up in the manner shown.


simulate this circuit – Schematic created using CircuitLab

To keep current the same though Ra first you need to calculate what the current in Ra is. Then find the resistance needed to keep the same current. First Ra and R2 are in parallel.

R2 || Ra = .99

Then the result is in series with R3. (R1 is ignored because it is directly across the supply and does not affect Ra)

R3 + 0.99 = 10.99

The current is then V1 / 10.99 = 0.090A

Then you do a current divider to see what part of the current is flowing through Ra and what part is flowing through R2.

0.090 * (R2 / (Ra + R2)) = 9.09e-5

Now set up an equasion with X equal to the needed resistance. The total voltage drop in the circuit should be 1V.

9.09e-5 * 1000 + 9.09e-5 * X = 1

Solve for X to get 10121 ohms.


simulate this circuit

  • \$\begingroup\$ -1 because you don't explain at all how you got to the solution. \$\endgroup\$
    – The Photon
    Sep 27, 2017 at 18:41
  • \$\begingroup\$ @jonk I thought it was 1k not 1ohm, I fixed it. \$\endgroup\$
    – vini_i
    Sep 27, 2017 at 18:42
  • \$\begingroup\$ @vini_i Yeah. I just noticed and deleted my comment. The update wasn't present, then. \$\endgroup\$
    – jonk
    Sep 27, 2017 at 18:43
  • \$\begingroup\$ @vini_i But V1 and Ra shouldn't be touched, i just want to replace R1, R2 and R3 with a single resistance Req in serie with V1, Ra must receive the same current as in the original circuit. \$\endgroup\$
    – Mehdi
    Sep 27, 2017 at 18:46
  • \$\begingroup\$ @Mehdi Then the second circuit is what you want. \$\endgroup\$
    – vini_i
    Sep 27, 2017 at 18:48
  1. Ra and R2 is in parallel so it will yield 1/2
  2. that 1/2 is in series with R3 so add and will be 10.5
  3. finally the 10.5 is in parallel with R1 so it will yield 5

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