Two identical cells each of emf 1.5 ,V are co | vedclass

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(C) The two $20 \,\Omega$ resistors are connected in parallel,so the equivalent external resistance $R$ is given by $\frac{1}{R} = \frac{1}{20} + \fra

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$5$

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(C) The two $20 \,\Omega$ resistors are connected in parallel,so the equivalent external resistance $R$ is given by $\frac{1}{R} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$,which means $R = 10 \,\Omega$.The two identical cells of emf $E = 1.5 \,V$ and internal resistance $r$ are connected in parallel. The equivalent emf $E_{eq} = E = 1.5 \,V$ and equivalent internal resistance $r_{eq} = \frac{r}{2}$.The terminal voltage $V$ across the external resistance is given by $V = E_{eq} - I r_{eq}$,where $I$ is the total current.Given $V = 1.2 \,V$,we have $1.2 = 1.5 - I \left(\frac{r}{2}\right)$,which implies $I \left(\frac{r}{2}\right) = 0.3$.Also,the total current $I$ is given by $I = \frac{E_{eq}}{R + r_{eq}} = \frac{1.5}{10 + \frac{r}{2}}$.Substituting $I$ into the equation $I \left(\frac{r}{2}\right) = 0.3$:$\left(\frac{1.5}{10 + \frac{r}{2}}\right) \left(\frac{r}{2}\right) = 0.3$$1.5 \left(\frac{r}{2}\right) = 0.3 \left(10 + \frac{r}{2}\right)$$0.75r = 3 + 0.15r$$0.6r = 3$$r = \frac{3}{0.6} = 5 \,\Omega$.

Two identical cells each of emf $1.5 \,V$ are connected in parallel across an external resistance formed by two $20 \,\Omega$ resistors connected in parallel. $A$ voltmeter connected in the circuit measures $1.2 \,V$. The internal resistance of each cell is ................. $\Omega$.

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