(N/A) Suppose,heat energy $H \propto I^{a} R^{b} t^{c}$.
$\therefore H = k I^{a} R^{b} t^{c} \dots (i)$ (where $k$ is a dimensionless constant).
Now,writing the dimensional formula on both sides:
$[H] = M^{1} L^{2} T^{-2}$
$[I] = A^{1}$
$[R] = M^{1} L^{2} T^{-3} A^{-2}$
$[t] = T^{1}$
Substituting dimensions in equation $(i)$:
$M^{1} L^{2} T^{-2} = (A^{1})^{a} (M^{1} L^{2} T^{-3} A^{-2})^{b} (T^{1})^{c}$
$M^{1} L^{2} T^{-2} = M^{b} L^{2b} T^{-3b+c} A^{a-2b}$
Equating the indices of $M, L, T,$ and $A$ on both sides:
For $M$: $b = 1$
For $L$: $2b = 2 \implies b = 1$
For $A$: $a - 2b = 0 \implies a - 2(1) = 0 \implies a = 2$
For $T$: $-3b + c = -2 \implies -3(1) + c = -2 \implies c = 1$
Thus,$a = 2, b = 1, c = 1$.
Substituting these values in equation $(i)$,we get:
$H = k I^{2} R t$
Experimentally,$k = 1$,therefore $H = I^{2} R t$.