Potential Difference - A Classical Circuit - Kirchhoff´s Law

This article treats about a classical standard approach on potential difference.

Consider the following basic circuit

In which we will admit the following polarities and direction of current, thus like the electricals nets I and II, and junctions A, A', A'', B, B' and B''.

We will attack the problem considering the Kirchhoff´s Law about "voltages" and currents.

In the Net I:

$-{\Large\epsilon}+R{ i }_{ 1 }+2R{ i }_{ 1 }=0\Leftrightarrow{ i }_{ 1 }=\dfrac {{\Large\epsilon}}{ 3R }$

In the Net II:

$2R{ i }_{ 2 }+R{ i }_{ 2 }-2R{ i }_{ 1 }-R{ i }_{ 1 }=0\Leftrightarrow3R{ i }_{ 2 }=3R{ i }_{ 1 }\Leftrightarrow{i}_{2}={i}_{1}$

In the A' junction(Point):

$i={i}_{1}+{i}_{2}\Leftrightarrow{i}_{1}={i}_{2}=\dfrac{i}{2}$

That Way:

${U}_{A'A}=R.{i}_{1}=R.\dfrac {{\Large\epsilon}}{ 3R }=\dfrac {{\Large\epsilon}}{ 3 }$

So,

${V}_{A}={V}_{A'}+{U}_{A'A}={\Large\epsilon}-\dfrac {{\Large\epsilon}}{3}=\dfrac {{2\Large\epsilon}}{ 3 }$

In the Same Way:

${U}_{B'B}=2R.{i}_{2}=2R.\dfrac {{\Large\epsilon}}{ 3R }=\dfrac {{2\Large\epsilon}}{ 3 }$

So,

${V}_{B}={V}_{B'}+{U}_{B'B}={\Large\epsilon}-\dfrac {{2\Large\epsilon}}{3}=\dfrac {{\Large\epsilon}}{ 3 }$

Because,

${V}_{A'}={V}_{B'}={\Large\epsilon}$

It can be this values any potential. Including ${\Large\epsilon}$. Verify!!!

Then,

${V}_{A}-{V}_{B}=\dfrac {{2\Large\epsilon}}{ 3 }-\dfrac {{\Large\epsilon}}{ 3 }=\dfrac {{\Large\epsilon}}{ 3 }$

$\Huge \boxed{!!!}$

Take a look that,

${V}_{B}-{V}_{A}=-\dfrac {{\Large\epsilon}}{ 3 }$

And

${V}_{A''}={V}_{B''}=0$

These ones are dependents of ${V}_{A'}$ e ${V}_{B'}$.