# Correctling a sign in a simple circuit with parallel resistors

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: 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?

• 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. – user105652 Jul 31 '20 at 21:35

## 1 Answer

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.