I've a simple circuit with two parallel resistors, but my derivation of the circuit equations gives a sign the opposite of what I know to occur, so I'm looking for help fixing the error as well as a general strategy for keeping the signs correct in the future. Consider the circuit:

DC circuit with parallel resistors

The voltage law on the left gives

$$v=v_1 \Longrightarrow v=i_1 R_1$$

and the circuit law at the top node gives

$$i = i_1 + i_2.$$

Now, I know that the voltage drop across the parallel resisitors should be the same and equal to v. However, when I consider the voltages on the rightmost loop, I see

$$v_1 + v_2 = 0 \Longrightarrow i_1 R_1 + i_2 R_2=0$$

To me, if I have polarity labeled, even if it doesn't matter for resistors, I place positive flow into + as the left and positive flow into - on the right. This is reversed for negative flow. That said, this leads to the incorrect equation above. If we do the outer loop, this gives the correct sign with

$$ v = v_2 \Longrightarrow v = i_2 R_2. $$

which would imply, correctly, that

$$ i_1 R_1 = i_2 R_2. $$

As such, what is the correct sign convention in this case in order to give the correct result? What is a general strategy for determining the sign and maintaining consistency when the current is divided at a node?

  • \$\begingroup\$ Welcome. Since voltage stays the same, it is only a current issue. \$ i1 + i2\$. Not the same as R1 || R2. If R1 = R2 then R = R1/2. \$\endgroup\$
    – user105652
    Commented Jul 31, 2020 at 21:35

1 Answer 1


Your problem is here:

However, when I consider the voltages on the rightmost loop, I see $$v_1 + v_2 = 0 \Longrightarrow i_1 R_1 + i_2 R_2=0$$

As you go around the loop you're entering one resistor at the "+" terminal and the other resistor at the "-" terminal, so you need to give one of these a negative sign.

Then you'd have

$$v_1 + v_2 = 0 \Longrightarrow i_1 R_1 - i_2 R_2=0$$

which is exactly what you expect.


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