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Suppose we have a circuit like this where \$R_1 = R_2 = R_3 = 1\$

schematic

simulate this circuit – Schematic created using CircuitLab

Why can we not divide this into two branches, \$ B_1, B_2 \$ and then do the following to find total current?

$$ I_t = \frac{V_1}{R_1+R_2}+\frac{V_1}{R_1+R_3} = 1 A $$

However, when we get the equivalent resistance and use it to find the current in the usual way like so, we get a different answer.

$$ I_t = \frac{V_1}{R_1 + \frac{1}{\frac{1}{R_2} + \frac{1}{R_3}}} = 2/3 A $$

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    \$\begingroup\$ Because the voltage across R1 (and thus the current through it) doesn't just depend on R2 or just depend on R3. It depends on the parallel combination of R2 and R3. \$\endgroup\$ – The Photon Feb 19 '15 at 21:21
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The circuit you analyse with your formula is essentially this one, with the switch open: two independent braches, each with two resistors in series.

schematic

simulate this circuit – Schematic created using CircuitLab

The current in the two braches is independennt, each current is V / ( R + R ).

Note that the potential (voltage) at both sides of the switch is the same, so we can close the switch, without effect on the circuit. Now we have your circuit, except that R1 is represented by TWO parallel resistors, each R, so the equivalent is R/2.

To summarize, you analyzed your circuit as if it were the one I show, which is different from your circuit in the value of R1.

The comment from The Photon gives another view, which amounts to the same thing.

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  • \$\begingroup\$ I believe you meant R1 and R2 to be 2 ohm resistors. \$\endgroup\$ – Andy aka Feb 19 '15 at 21:40
  • \$\begingroup\$ Might be easier for OP to follow if you renamed R1 and R2 to R1a and R1b, and R4 to R2. Took me a bit to realize what you had done because I assumed R2,R3 in your schematic was the same as in OPs. @Andyaka - He's showing the equivalent circuit of what OP calculated in OPs first equation, in which OP created two branches, each with an R1 of 1ohm, so it's correct as shown \$\endgroup\$ – I. Wolfe Feb 19 '15 at 21:40
  • \$\begingroup\$ @Andyaka Note the line in the answer: "To summarize, you analyzed your circuit as if it were the one I show, which is equivalent to your circuit with R1 = 0.5 Ohm." \$\endgroup\$ – Tut Feb 19 '15 at 21:42
  • \$\begingroup\$ @tut - yep sorry! \$\endgroup\$ – Andy aka Feb 19 '15 at 21:49
  • \$\begingroup\$ I tried the show the circuit that corresponds to TS's calculations, and then show that it is not the circuit he shows in his diagram. \$\endgroup\$ – Wouter van Ooijen Feb 19 '15 at 22:14
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In a way you can do just that but you have to use more symmetry. For example the common 1 ohm resistor (R1) could be replaced by two 2 ohm resistors in parallel. This leaves two branches with of 3 ohms each. Current in each is one-third an amp. Total current is 2/3 amps.

You have to exchange the common resistor (R1) into 2 resistors that are proportional to the two independant resistors (R2 and R3) and together reduce to a value the same as R1.

For instance, if R2 was 1 ohm and R3 was 3 ohms, we know the net resistance of R2 and R3 is 0.75 ohms. If R1 is 2 ohms, the total resistance is 2.75 ohms and the current is 0.3636 amps. Now create two resistors from R1 having the same proportions as R2 and R3 but forming 2 ohms in parallel. Call it R1a and R1b.

I'll leave you to do the algebra. Then calculate the current with R2 in series with the lower of R1a and R1b. Then add this to the current taken by R3 in series with the higher of R1A and R1B and you'll get the right answer.

I don't think it's particularly useful in practise though.

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Your method does look appealing at first glance. It even resembles superposition, and we know that works. The problem is that your method obeys KCL but violates KVL. By definition, two branches in parallel must share the same voltage, but in your method, they don't. This is easiest to see if we replace R2 with a short circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

(Does anyone know how to make schematics smaller?)

Let's try your method:

$$I_1 = \frac{1 V}{1 \Omega + 0 \Omega} + \frac{1 V}{1 \Omega + 1 \Omega} = 1 A + 0.5A = 1.5A$$

Now look at the voltages across the "resistors":

$$V_{R1} = 1.5A \times 1 \Omega = 1.5 V$$ $$V_{R2} = 1A \times 0 \Omega = 0 V$$ $$V_{R3} = 0.5A \times 1 \Omega = 0.5 V$$ $$V_{mid} = 1V - V_{R1} = V_{R2} = V_{R3}$$ $$V_{mid} = -0.5 V? = 0 V? = 0.5 V? \; (contradiction)$$

The problem, of course, is that no current should flow through R3 when it's shorted out.

Another obviously pathological case is having a million parallel resistors instead of two. The current should converge to \$V_1 / R_1\$, but instead your method gives \$I_1 = 500,000 A\$ and a possible \$V_{mid} = -499,999V\$! This is actually similar to what goes wrong in your specific example. You calculate \$I_1 = 1 A\$, but by Ohm's Law that makes \$V_{mid} = V_1 - I_1 \times R_1 = 0V\$. There can't be 0V across R2 and R3 if there's current flowing through them.

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