# two cells of emf 10v and 2 v and internal resistance 10 n 5 respectively are connected in parallel.find effective voltage across R which is placed parallel to both

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,

${I}_{eff}={I}_{1}+{I}_{2}\phantom{\rule{0ex}{0ex}}\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)\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{eff}=\frac{14}{3}=4.67V$

Therefore, ${V}_{AB}=4.67V$

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