# Potential difference between A.C. circuit

I the following A.C circuit . I could not understand why the V$_{AB}$ does not depend on non zero value of R$_0$.

According to me it should depend on it . As it will decide the amount of current in that part of wire which will decide the potential at B.

• Who says it doesn't? – The Photon Jan 11 '17 at 19:09
• @ThePhoton in my book it is written – Koolman Jan 12 '17 at 2:31
• Can you share the exact quote and the context? – The Photon Jan 12 '17 at 3:01
• @ThePhoton dc.allenbpms.in/testpaper/solution/… – Koolman Jan 12 '17 at 3:09

$$V_{AB} = v_0\left(\frac{R_0}{j\omega{}L+R_0}-\frac{1}{2}\right)$$

$$\frac{V_{AB}}{v_0} = \frac{2R_0 - j\omega{}L-R_0}{2(j\omega{}L+R_0)}$$

$$\frac{V_{AB}}{v_0} = \frac{1}{2}\frac{R_0-j\omega{}L}{R_0+j\omega{}L}$$

Now consider the term $\frac{R_0-j\omega{}L}{R_0+j\omega{}L}$. Since the numerator and the denominator have equal magnitude, the magnitude of the quotient is always 1, so the magnitude of $V_{AB}$ is always $\frac{v_0}{2}$.

The phase of the quotient is $$2\tan^{-1}\left(\frac{R_0}{\omega{}L}\right)$$ but the original question only asks about the r.m.s. value (which you didn't correctly relate in your post), so this variation doesn't matter.

• Here What is $j$ – Koolman Jan 12 '17 at 3:34
• Sorry, that's Electrical Engineering speak for $i$. – The Photon Jan 12 '17 at 5:20
• Can you expalin it in simple words for a high school student – Koolman Jan 12 '17 at 7:48
• If you haven't learned enough math to follow what I wrote, then you aren't ready to answer this question yet. – The Photon Jan 12 '17 at 17:04
• @Koolman It is a mathematical trick: hyperphysics.phy-astr.gsu.edu/hbase/electric/impcom.html – Yashas Apr 6 '17 at 6:24