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Consider a simple circuit with a battery of $\theta\ \text V$s, and two resistors of $R_1 \ \Omega$s and $R_2\ \Omega$s connected in series. Let us assume that $R_1$ is connected nearer to the positive terminal of the battery. We are required to find the power $P_1$ and $P_2$, dissipated in these resistors.

Now, here if we take the direction of electric current from negative to positive, we get the following result:

$P_2 = I_2^2R_2$. Since $I_2 = \frac{\theta}{R_2}$, $P_2 = \frac{\theta^2}{R_2}$

Now, similarly, $P_1 = \frac{\theta^2}{R_1}$. But, there is a voltage drop across $R_2$, which equals $IR_2 = R_2\frac{\theta}{R_1 + R_2}$. Therefore, the voltage just before $R_1$ equals $\theta\left(1 -\frac{R_2}{R_1 + R_2}\right)$ and $P_1 = \frac{\theta^2\left(1 -\frac{R_2}{R_1 + R_2}\right)^2}{R_1}$.

But, consider the case in which the direction of electric current is taken from positive to negative. We encounter $R_1$ first. In this case $P_1 = I_1^2R_1 = \frac{\theta^2}{R_1}$. There will be voltage drop between $R_1$ and $R_2$ which will equal $\theta\left(1 - \frac{R_1}{R_1 + R_2}\right)$, so $P_2 = \frac{\theta^2\left(1 - \frac{R_1}{R_1 + R_2}\right)^2}{R_2}$.

These are two different results. Which direction should the electric current be taken in this case?

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If the resistors are in series, then $I_2 = I_1 = \frac{\theta}{R_1 + R_2}$. – Ali Aug 4 '13 at 10:10
Yes, that is true, but the numbers should work out the other way too. – Gerard Aug 4 '13 at 10:14
And if you choose the wrong direction, you will perhaps find: $I_1 = I_2 = - \frac{\theta}{R_1 + R_2}$. The minus sign means you have chosen the wrong direction. – Ali Aug 4 '13 at 10:18
up vote 2 down vote accepted

Actually, there is a mix-up in your calculations. While calculating the power of a resistor, you can use the formula $P = \frac{V^2}{R}$, however the voltage in this formula is the voltage drop between the resistor's terminals.

In the circuit that you described, in order to find the power consumption of the resistors you have to first calculate the voltage drops across both resistors, and in such a simple circuit $V_1 = \frac{\theta \times R_1}{R_1 + R_2}$ and $V_2 = \frac{\theta \times R_2}{R_1 + R_2}$. This is a very simple voltage division but if it is not obvious at first glance, you can first calculate the current through the resistors and the battery, $I = \frac{\theta}{R_1+R_2}$ and then multiply this with resistance values of each resistor to get the voltage drop across the terminals of the resistor being investigated.

If you do the same calculation from an inverse point of view, you will find the current and the voltage values same in magnitude, opposite in sign. However, since both power formulas (i.e. $P = \frac{V^2}{R}$ and $P = \frac{I^2}{R}$) include the square of either the voltage or the current (or a third and more fundamental and generally applicable formula, $P = V \times I$ includes the multiplication of both terms), the power is the same in both cases.

All in all, this confusion only arises because of a mistake in your calculations and all circuits obey the laws of Kirchhoff(Kirchhoff's Current Law and Kirchhoff's Voltage Law) and the results are always consistent no matter which direction you perform your calculations.

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In circuit analysis, the reference direction of a current variable is arbitrary; there is no wrong direction. The same holds true for the reference polarity of a voltage variable.

If, for the same current, I choose the reference direction to be to the right and you choose the reference direction to be to the left, our answers will differ by a sign but both answers give the correct direction for the actual electric current.

However, in choosing the reference direction and reference polarity, it is wise to use the passive sign convention where the current through a resistor enters the positive labelled terminal:

enter image description here

Now, consider a circuit like you described:

enter image description here

Since this is a series circuit, there is only one electric current, $I = \dfrac{V_{in}}{R_1 + R_2}$, that circulates clockwise (the battery delivers power to the circuit so the electric current exits the positive terminal).

The power associated with each resistor is then:

$P_1 = I^2 R_1$

$P_2 = I^2 R_2$

If we had chosen the reference direction for the current variable to be counter-clockwise, the series current would be $I' = -\dfrac{V_{in}}{R_1 + R_2}$

But note that:

$P_1 = I'^2 R_1 = I^2 R_1$

so, regardless, the power calculation gives the same result.

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It doesn't matter the current direction. Take it allways the same direction, from - to + or + to -. The calculations results will allways be the same. I prefer take that current flows as from - to + (cathode to anode). This is the electrons flow. In a cathode ray tube, electrons leave cathode (negative) to hit the positive phosphorescent screen and produce the light.

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