In an electrical circuit, two resistors of  $6\mathrm{\Omega }$ and $8\mathrm{\Omega }$ respectively are connected in series to a $12 \mathrm{V}$ battery. The heat dissipated by the $8\mathrm{\Omega }$ resistor in 5 s will be

 Calculate the equivalent resistance using the formula for a series combination of resistors.
$R_{\mathrm{s}}=R_1+R_2$
Then calculate the current flowing through the circuit using Ohm's law.
$I=\frac{V}{R_{\mathrm{qq}}}$

Finally, calculate the heat dissipated in the given resistor by following the formula.
$H=I^2{R}_{2}i$
Resistance $\left(R_1\right)=6\Omega$
Resistance $\left(R_2\right)=8\Omega$
Potential difference $\left(V\right)=12 \mathrm{V}$
Time $\left(t\right)=5 \mathrm{s}$
Resistors are connected in series and the total resistance is
Resistance of the circuit, $R_{\mathrm{eq}}=6+8=14\Omega$
The total current is calculated using Ohm's law as follows,
$I={\left(\frac{V}{R_{\mathrm{eq}}}\right)}_{\mathrm{I}}$
Current,
$I=\left(\frac{12}{14}\right)=0.85 \mathrm{A}$
Hence, the heat dissipated by $8\Omega$ in 5 seconds will be
$H=I^2{R}_{2}t=0.{85}^2\times 8\times 5=28.9 Joule$
Hence the heat dissipated by the $8\Omega$resistor in 5 seconds is $28.9 Joule$.