For ensuring dissipation of same energy in all three resistors $({R_1},\,{R_2},\,{R_3})$ connected as shown in figure, their values must be related as
A${R_1} = {R_2} = {R_3}$
B${R_2} = {R_3}$ and ${R_1} = 4{R_2}$
C${R_2} = {R_3}$ and ${R_1} = \frac{1}{4}{R_2}$
D${R_1} = {R_2} + {R_3}$
AIIMS 2005, Diffcult
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C${R_2} = {R_3}$ and ${R_1} = \frac{1}{4}{R_2}$
c (c) As the voltage in ${R_2}$ and ${R_3}$ is same therefore, according to,
$H = \frac{{{V^2}}}{R}.t,$ ${R_2} = {R_3}$
Also the energy in all resistance is same.
${i^2}{R_1}t = i_1^2{R_2}t$
Using ${i_1} = \frac{{{R_3}}}{{{R_2} + {R_3}}}i = \frac{{{R_3}}}{{{R_3} + {R_3}}}i = \frac{1}{2}i$
Thus ${i^2}{R_1}t = \frac{{{i^2}}}{4}{R_2}t$ or, ${R_1} = \frac{{{R_2}}}{4}$
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