JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 11)
Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are $$r_{1}$$ and $$r_{2}$$ $$\left(r_{1}>r_{2}\right)$$. If the potential difference across the source of internal resistance $$r_{1}$$ is zero, then the value of R will be :
$$r_{1}-r_{2}$$
$$\frac{r_{1} r_{2}}{r_{1}+r_{2}}$$
$$\frac{r_{1}+r_{2}}{2}$$
$$r_{2}-r_{1}$$
Explanation
_27th_July_Morning_Shift_en_11_1.png)
$$\Delta V = 0 \Rightarrow {{2\varepsilon } \over {{r_1} + {r_2} + R}}{r_1} = \varepsilon $$
$$ \Rightarrow R = {r_1} - {r_2}$$
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