Explanation:
The voltage across \(R_{2}\) and \(R_{3}\) is same.
Therefore, for same energy loss according to \(H=\dfrac{V^{2}}{R} t, R_{2}\) must be equal to \(R_{3}\).
\(i.e.\) \(R_{2}=R_{3}\)
Again, the energy is same in all resistances.
\(\therefore {i^2}{R_1}t = i_1^2{R_2}t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
and \({i_1} = \frac{{{R_3}}}{{{R_2} + {R_3}}}\;\;\;{\mkern 1mu} {\kern 1pt} i = \frac{{{R_3}}}{{{R_3} + {R_3}}}i = \frac{1}{2}i\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
\(\left[\because R_{2}=R_{3}\right]\)
Thus, \(i^{2} R_{1} t=\dfrac{i^{2}}{4} R_{2} t\) [From Eqs. (1) and (2)]
\(\therefore R_{1}=\dfrac{R_{2}}{4}\)
So, correct option is (3)