I recently came across a question, for which I saw two possible methods of finding the solution. I was required to calculate the "current flowing" $I$ when given the voltage $V$, frequency $f$, total resistance $R$ and impedance of the circuit $Z$. In the book that these questions were written, there were two methods (Note - AC Current):

First Method:

$$V=|I\cdot Z|\ \Longrightarrow\ \frac{V}{|Z|}=|I|$$

Second Method:

$$V=I\cdot R\ \Longrightarrow\ \frac{V}{R}=I$$

(I use all the time for DC Currents)

As I found out, through trial and error, both of these give different solutions. Personally, I tended towards the first solution, but I couldn't help noticing the second method, as it gave a nice round answer, as opposed to the fractional answer the first gave. Which of these two methods is the correct one to use?

  • 2
    $\begingroup$ Well, when it's oscillating and the quantities have phases - so that you talk about the impedance instead of a resistance at all - the first method is clearly right by the definition of impedance (definition that is a straight generalization of resistance from Ohm's law, method 2, which is only for DC). Nicer numbers don't mean anything. They can be chosen by the authors of exercises to please the solvers who get the nice numbers but they may also be picked to deliberately mislead them. ;-) Numerical values of things such as voltage in the real world problems have no reason to be nice numbers. $\endgroup$ – Luboš Motl Feb 26 '13 at 11:02

Method 1 is universal, method 2 works only for dc.

Method 2 is for DC supply, whereas the Method 1 is for AC supply.In AC circuit the Inductor (a coil) and a capacitor pose resistance which is calculated as impedence of circuit. .

Whereas in DC , capacitor fully blocks the supply and thus capacitive reactance is infinite and the inductive reactance is zero. thus the current in a capacitive circuit is zero. z^2=R^2 + (xl -xc )^2 ; z(dc)=R; Also the difference of Xl and Xc is small and thus answer is approximately the same by the method 2 in AC circuits too.

  • $\begingroup$ @asryael review my ans. $\endgroup$ – ABC Mar 19 '13 at 12:22

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