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(a) Two cells of different emfs and internal resistances are connected in parallel with one another. Derive the expression for the equivalent emf and equivalent internal resistance of the combination.

(b) Two identical cells of emf 1·5 V and internal resistance r are each connected in parallel providing a supply to an external circuit consisting of two resistances of $17\; \Omega$ each joined in parallel. A very high resistance voltmeter reads the terminal voltage of the cell to be 1·4 V. Calculate the internal resistance of each cell.

Answers (1)

(a) Let the total amount $(I)=I_{1}+I_{2}$ ______(i)

$\because V=E_{1}-I_{1}r_{1}$

$\because I_{1}=\frac{E_{1}-V}{r_{1}}$ and

$\because V=E_{2}-I_{2}r_{2}$

$\because I_{2}=\frac{E_{2}-V}{r_{2}}$

On putting the values of $I_{1}$ and $I_{2}$ in equation (i) ,we get

$I=\frac{E_{1}-V}{r_{1}}+\frac{E_{1}-V}{r_{2}}$

$I=\left [ \frac{E_{1}}{r_{1}}+\frac{E_{2}}{r_{2}} \right ]-V\left [ \frac{1}{r_{1}}+\frac{1}{r_{2}} \right ]$

$V=\frac{E_{1}r_{2}+E_{2}r_{1}}{r_{1}+r_{2}}-\frac{Ir_{1}r_{2}}{r_{1}+r_{2}}$

Now, comparing with;-

$V=E_{eq}-Ir_{eq}$

Then, we get the values of $E_{eq}$ and $r_{eq}$

$E_{eq}=\frac{E_{1}r_{2}+E_{2}r_{1}}{r_{1}+r_{2}}$

$r_{eq}=\frac{r_{1}r_{2}}{r_{1}+r_{2}}$

(b)

$V=1.4 \; Volt$ and emf $(E_{eq})=1.5\; V$

$r_{eq}=\frac{17}{2}=8.5\; \Omega$

To calculate the internal resistance

We have,

$I=\frac{V}{r_{eq}}$

On putting the values, we get

$I=\frac{1.4}{8.5}\; Ampere$

Now, $V=E_{eq}-Ir_{eq}$

So, $1.4=1.5-\frac{1.4}{8.5}\times \frac{r}{2}\; \; \; \; \left ( \because r_{eq}=\frac{r}{2} \right )$

Hence, $r=1.21\; \Omega$ is the required internal resistance.

Posted by

Safeer PP

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