Question

Find the potential difference V_a – V_b in the circuits shown in the figure (32-E12).

Answer 1

Applying KVL in loop 1, we get:
$$\begin{gathered} i_1{R}_2-E_2+\left(i_1+i_2\right)R_3=0 \\ (R_2+R_3)i_1+R_3{i}_2=E_2...\left(1\right) \end{gathered}$$
Applying KVL in loop 2, we get:
$$\begin{gathered} i_2{R}_1-E_1+\left(i_1+i_2\right)R_3=0 \\ \left(R_1+R_3\right)i_2+R_3{i}_1=E_1...\left(2\right) \end{gathered}$$
Multiplying equation (1) by (R₁+R₃) and (2) by R₃ and then subtracting (2) from (1), we get:
$i_1=\frac{E_2\left(R_1+R_3\right)-E_1{R}_3}{\left(R_1{R}_2+R_2{R}_3+R_3{R}_1\right)}$
Similarly, multiplying equation (1) by R₃ and (2) by (R₁+R₃), and then subtracting (2) from (1), we get:
$i_2=\frac{E_1\left(R_2+R_3\right)-E_2{R}_3}{\left(R_1{R}_2+R_2{R}_3+R_3{R}_1\right)}$
From the figure,
$$\begin{gathered} V_{\mathrm{a}}-V_{\mathrm{b}}=\left(i_1+i_2\right)R_3 \\ \Rightarrow V_{\mathrm{a}}-V_{\mathrm{b}}=\left[\frac{E_1{R}_2+E_2{R}_1}{\left(R_1{R}_2+R_2{R}_3+R_3{R}_1\right)}\right]R_3 \\ \Rightarrow V_{\mathrm{a}}-V_{\mathrm{b}}=\frac{{\displaystyle \frac{E_1}{R_1}}+{\displaystyle \frac{E_2}{R_2}}}{{\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}+{\displaystyle \frac{1}{R_3}}} \end{gathered}$$

(b) The circuit in figure b can be redrawn as shown below:

We can see that it is similar to the circuit in figure a and, hence, the answer obtained will be same.