Electronic – Why can’t you divide a series + parallel circuit into branches to get total current

circuit analysisparallelseries

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.

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

Best Answer

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.