Question

The potential difference between the points A and B in figure will be:

A. $\dfrac{2}{3}\,{\text{V}}$
B. $\dfrac{8}{9}\,{\text{V}}$
C. $\dfrac{4}{3}\,{\text{V}}$
D. $2\,{\text{V}}$

Answer 1

Consider the potential difference in the lower terminal of the circuit diagram is zero. Hence, derive the equation for the potential at points A and B. Determine the value of the electric current flowing through the two terminals of the circuit. Determine the value of the potential through the resistors and hence determine the potential difference between the points A and B.

Complete step by step answer:
Let us suppose that the potential in the lower terminal of the battery is zero. Hence, the potential at point A becomes
${V_A} = \left( {2\,{\text{V}}} \right) - V$ …… (1)
Here, $V$ is the potential through each resistor.
And the potential at point A becomes
${V_B} = \left( {0\,{\text{V}}} \right) + V$ …… (2)
In both the parts of the above circuit diagram, the same electric current $i$ flows according to Kirchhoff’s current law.

Let us determine this electric current ${V_A} = \left( {2\,{\text{V}}} \right) - V$ using Ohm’s law.
$i = \dfrac{V}{R}$
$\Rightarrow i = \dfrac{{{\text{2 V}}}}{{5\,\Omega + 5\,\Omega + 5\,\Omega }}$
$\Rightarrow i = \dfrac{{\text{2}}}{{15}}\,{\text{A}}$
Hence, the current flowing through both the parts of the circuit diagram is $\dfrac{{\text{2}}}{{15}}\,{\text{A}}$.

Let us determine the value of the potential using Ohm’s law.
$V = iR$
Substitute $\dfrac{{\text{2}}}{{15}}\,{\text{A}}$ for $i$ and $5\,\Omega$ for $R$ in the above equation.
$V = \left( {\dfrac{{\text{2}}}{{15}}\,{\text{A}}} \right)\left( {5\,\Omega } \right)$
$\Rightarrow V = \dfrac{{\text{2}}}{3}\,{\text{V}}$
Hence, the potential for each resistor is $\dfrac{{\text{2}}}{3}\,{\text{V}}$.

Substitute $\dfrac{{\text{2}}}{3}\,{\text{V}}$ for $V$ in equation (1).
${V_A} = \left( {2\,{\text{V}}} \right) - \left( {\dfrac{{\text{2}}}{3}\,{\text{V}}} \right)$
$\Rightarrow {V_A} = \dfrac{4}{3}\,{\text{V}}$
Substitute $\dfrac{{\text{2}}}{3}\,{\text{V}}$ for $V$ in equation (2).
${V_B} = \left( {0\,{\text{V}}} \right) + \left( {\dfrac{{\text{2}}}{3}\,{\text{V}}} \right)$
$\Rightarrow {V_B} = \dfrac{2}{3}\,{\text{V}}$
The potential difference between points A and B is
$\Rightarrow {V_A} - {V_B} = \dfrac{4}{3}\,{\text{V}} - \dfrac{2}{3}\,{\text{V}}$
$\therefore {V_A} - {V_B} = \dfrac{2}{3}\,{\text{V}}$

Therefore, the potential difference between the points A and B is $\dfrac{2}{3}\,{\text{V}}$.

Note: One can also solve the same question by another method. One can determine the net resistance in two loops of the circuit individually and the current through these two loops which is the same for both the loops due to the same arrangement of the resistors. Then one can determine the potential at point A and point B of the circuit using the proper value of resistance and then determine the potential difference between the points A and B.