JEE MAIN - Physics (2005 - No. 6)
Two sources of equal $$emf$$ are connected to an external resistance $$R.$$ The internal resistance of the two sources are $${R_1}$$ and $${R_2}\left( {{R_1} > {R_1}} \right).$$ If the potential difference across the source having internal resistance $${R_2}$$ is zero, then
$$R = {R_2} - {R_1}$$
$$R = {R_2} \times \left( {{R_1} + {R_2}} \right)/\left( {{R_2} - {R_1}} \right)$$
$$R = {R_1}{R_2}/\left( {{R_2} - {R_1}} \right)$$
$$R = {R_1}{R_2}/\left( {{R_1} - {R_2}} \right)$$
Explanation
$${\rm I} = {{2\varepsilon } \over {R + {R_1} + {R_2}}}$$
Potential difference across second cell
$$ = V = \varepsilon - {\rm I}{R_2} = 0$$
$$\varepsilon - {{2\varepsilon } \over {R + {R_1} + {R_2}}}.{R_2} = 0$$
$$R + {R_1} + {R_2} - 2{R_2} = 0$$
$$R + {R_1} - {R_2} = 0$$
$$\therefore$$ $$R = {R_2} - {R_1}$$
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