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(a) Two cells of emf $E_{1}$ and $E_{2}$ have their internal resistances $r_{1}$ and $r_{2}$ , respectively. Deduce an expression for the equivalent emf and internal resistance of their parallel combination when connected across an external resistance $R$ . Assume that the two cells are supporting each other.

(b) In case the two cells are identical, each of emf $E=5\; V$ and internal resistance $r=2\; \Omega$ , calculate the voltage across the external resistance $R=10\; \Omega$ .

Answers (1)

a)

Emf are $\varepsilon _{1}and \varepsilon _{2}$

Internal resistance are $r _{1}andr _{2}$

Here, $I=I _{1}+I_{2}$ ........(1)

Let V is the potential difference between B₁ and B₂

Then $V = \varepsilon _{1}-I_{1}r_{1}$

Therefore , $I_{1}= \frac{\varepsilon _{1}-V}{r_{1}}$ ......(2)

Similarly. $I_{2}= \frac{\varepsilon _{2}-V}{r_{2}}$ .......(3)

Putting equation (2) and (3) in (1)

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

$I=\left ( \frac{\varepsilon _{1}}{r_{1}}+\frac{\varepsilon _{2} }{r_{2}} \right )-V\left ( \frac{1}{r_{1}}+\frac{1}{r_{2}} \right )$

or $V = \left ( \frac{\varepsilon _{1}r_{2}+\varepsilon _{2}r_{1}}{r_{1}+r_{2}} \right )- I\left ( \frac{r_{1}r_{2}}{r_{1}+r_{2}} \right )$

If we replace this with a single cell it can be written as

$V = \varepsilon _{equivalent}-Ir_{eqivalent}$

Then,

$\varepsilon _{eqivalent} = \left ( \frac{\varepsilon _{1}r_{2}+\varepsilon _{2}r_{1}}{r_{1}+r_{2}} \right )$

$r_{equivalent} =\left ( \frac{r_{1}r_{2}}{r_{1}+r_{2}} \right )$

b)

$\\r_{eff}=\frac{r}{2}=\frac{2}{2}=1\Omega\\E_{ext}=IR\\\\I=\frac{E_{eff}}{R+r}=\frac{5}{10+1}=\frac{5}{11}A\\\\E_{ext}=\frac{5}{11}\times10=4.54\ Volts$

Posted by

Safeer PP

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