In Fig. 27-8a, calculate the potential difference between a and c by considering a path that contains R, r₁, and${\mathit{\epsilon }}_{\mathbf{1}}$.

• a) From the given figure, the emf values of the batteries ${\epsilon }_1=4.4V,{\epsilon }_2=2.1V$.
• b) The values of the given resistances $r_1=1.8\Omega ,r_2=2.3\Omega ,R=5.5\Omega$

For the pair of resistors in a parallel connection, the potential difference at two junctions remains the same. Thus, using the loop rule, the potential drop between the points is related, and thus, the value of the required potential drop is calculated using the calculated current value.

Formulae:

The voltage equation using Ohm’s law,

V=IR (i)

Here I is the current, and R is the resistance.

Kirchhoff’s voltage law,

$\sum _{closedloop}V=0$ (ii)

Step 3: Calculation of the potential difference across a and c

From the given figure, we can see that the current flowing through the junction is the same, which can be given using equation (i) as follow:

(a)

localid="1662212141571" $$\begin{gathered} i=\frac{Netemf}{Totalresis\tan ce} \\ =\frac{{\epsilon }_1-{\epsilon }_2}{R+r_1+r_2} \end{gathered}$$

Now, the potential drop across point a is equal to that at point c, thus using equations (i) and (ii), we get the potential difference across a and c as follows:

localid="1662212150021" $$\begin{gathered} V_a-{\epsilon }_1=V_c-i{r}_1-iR \\ {V}_a-V_c={\epsilon }_1-i{r}_1-iR \\ ={\epsilon }_1-i(r_1+R) \end{gathered}$$

Substitute the value from expression a in the above expression, and we get,

Substitute the values in the above expression, and we get,

$$\begin{gathered} V_a-V_c=44.4V-\left(\frac{4.4V-2.1V}{5.5\Omega +1.8\Omega +2.3\Omega }\right)\left(2.3\Omega +5.5\Omega \right) \\ =2.5V \end{gathered}$$

Hence, the value of the potential difference is 2.5V .