Physics - Current Electricity Question with Solution | TestHub

The formula for the equivalent emf of the parallel combination of batteries is ${\epsilon }_e=r_{\text{eq }}\left(\frac{e_1}{r_1}+\frac{c_2}{r_2}\right)$ Here, $r_{\mathrm{rq}}$ is the equivalent resistance $\begin{array}{rl}& \frac{1}{r_{\text{eq }}}=\frac{1}{r_1}+\frac{1}{r_2}\\ & \frac{1}{r_{\text{84 }}}=\frac{1}{2}+\frac{1}{1}\\ & \frac{1}{r_{\text{eq }}}=\frac{3}{2}\end{array}$ Substituting the values $\begin{array}{rl}{\epsilon }_e& =r_{\infty }\left(\frac{e_1}{r_1}+\frac{e_2}{r_2}\right)\\ {\epsilon }_{\mathrm{e}}& =\frac{2}{3}\left(\frac{12}{2}+\frac{6}{1}\right)\\ {\epsilon }_e& =\frac{2}{3}\times 12\\ \therefore {\epsilon }_e& =8\mathrm{V}\end{array}$