Suppose, heat energy $\mathrm{H} \propto \mathrm{I}^{a} \mathrm{R}^{b} t^{c}$

$\therefore \mathrm{H}=k \mathrm{I}^{a} \mathrm{R}^{b} \mathrm{t}^{c} \ldots$ $(i)$ (where $k$ is a dimensionless constant.)

Now, writing the dimensional formula on both sides,

$\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}=(\mathrm{A})^{a}\left(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right)^{b}(\mathrm{~T})^{c}$

$\therefore \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}=(\mathrm{A})^{a-2 b} \mathrm{M}^{b} \mathrm{~L}^{2 b} \mathrm{~T}^{c-3 b}$

${[\mathrm{H}]=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}}$

${[\mathrm{I}]=\mathrm{A}[t]=\mathrm{T}}$

${[\mathrm{R}]=\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}}$

Equating the indices on both sides,

$a-2 b=0, b=1 \therefore a-2(1)=0 \therefore a=2$

$-3 b+c=-2$

$\therefore-3+c=-2$

$\therefore c=-2+3=1$

Thus, $a=2, b=1$ and $c=1$.

Substituting these values in equation $(i)$, we get

$\mathrm{H}=k \mathrm{I}^{2} \mathrm{R} t$

$\mathrm{H}=k \mathrm{I}^{2} \mathrm{R} t \text { But, } k=1 \quad \therefore \mathrm{H}=\mathrm{I}^{2} \mathrm{R} t$