In an electrical circuit, two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be?

Two resistances of $2\mathit{ \Omega }$ and $4\mathit{ \Omega }$ are given. Also, these resistances are connected in series.
${\mathrm{R}}_{\mathrm{s}}=2+4$
$\Rightarrow {\mathrm{R}}_{\mathrm{s}}=6\mathrm{ \Omega }$
Using Ohm’s law, the current can be calculated as,
$\mathrm{I}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{s}}}$
$\Rightarrow \mathrm{I}=\frac{6}{6}$
$\Rightarrow \mathrm{I}=1 \mathrm{A}$
Therefore, heat dissipation in the $4\mathit{ \Omega }$ resistor is equal to
$\mathrm{H}={\mathrm{I}}^2\mathrm{Rt}$
$\Rightarrow \mathrm{H}=1^2\times 4\times 5$
$\Rightarrow \mathrm{H}=20 \mathrm{J}$
Hence, the heat dissipation is equal to 20 J.