A cell of constant e.m.f. first connected to a resistance {R_1} and then connected to a resistance {R

Q: A cell of constant $e.m.f.$ first connected to a resistance ${R_1}$ and then connected to a resistance ${R_2}$. If power delivered in both cases is then the internal resistance of the cell is

a (a) Power dissipated$ = {i^2}R = {\left( {\frac{E}{{R + r}}} \right)^2}R$ ${\left( {\frac{E}{{{R_1} + r}}} \right)^2}{R_1} = {\left( {\frac{E}{{{R_2} + r}}} \right)^2}{R_2}$ $==>$ ${R_1}(R_2^2 + {r_2} + 2{R_2}r) = {R_2}(R_1^2 + {r^2} + 2{R_1}r)$ $==>$ $R_2^2{R_1} + {R_1}{r^2} + 2{R_2}r = R_1^2{R_2} + {R_2}{r^2} + 2{R_1}{R_2}r$ $==>$ $({R_1} - {R_2}){r^2} = ({R_1} - {R_2}){r^2} = ({R_1} - {R_2}){R_1}{R_2}$ $==>$ $r = \sqrt {{R_1}{R_2}} $

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

A cell of constant $e.m.f.$ first connected to a resistance ${R_1}$ and then connected to a resistance ${R_2}$. If power delivered in both cases is then the internal resistance of the cell is

A$\sqrt {{R_1}{R_2}} $
B$\sqrt {\frac{{{R_1}}}{{{R_2}}}} $
C$\frac{{{R_1} - {R_2}}}{2}$
D$\frac{{{R_1} + {R_2}}}{2}$

Diffcult

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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