Two cells of E.M.F. 10 V and 2 V and internal resistances 10 Ω and 5 Ω
respectively, are connected in parallel as shown. Find the effective voltage
across R.?

Dear Student,

Please find below the solution to the asked query:

Given situation is drawn in the following diagram.

So, the circuit can be simplified to a state, where a single cell of emf and a single internal resistance can be placed between A and B. Here, the question is to calculate V_AB.
So, from effective resistance formula,
$\frac{1}{R_{eff}} = \frac{1}{R_1}+\frac{1}{R_2} = \frac{1}{10}+\frac{1}{5} = \frac{1+2}{10}\Rightarrow R_{eff} = \frac{10}{3} \Omega$
from effective emf formula,
$$\begin{gathered} I_{eff} = I_1 + I_2 \\ \Rightarrow \frac{V_{eff}}{R_{eff}} = \frac{V_1}{R_1} + \frac{V_2}{R_2} \Rightarrow \frac{V_{eff}}{{\displaystyle \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} = \frac{10}{10}+\frac{2}{5} \Rightarrow V_{eff} = \left(\frac{10}{3}\right)\left(\frac{7}{5}\right) \\ \Rightarrow V_{eff} = \frac{14}{3} = 4.67 V \end{gathered}$$
Therefore, $V_{AB} = 4.67 V$



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