Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

Ans: Option C - 1:4

Equation of heat,

                                            $\mathrm{H} =\frac{{\mathrm{V}}^2}{\mathrm{R}}\mathrm{t}$

For the same voltage, V and time, t ; ratio of heat produced in series ${\mathrm{H}}_{\mathrm{S}}$ to heat produced in parallel ${\mathrm{H}}_{\mathrm{P}}$ is given by,

                                                      $\frac{{\mathrm{H}}_{\mathrm{S}}}{{\mathrm{H}}_{\mathrm{P}}}=\frac{{\displaystyle \left(\frac{{\mathrm{V}}^2}{{\mathrm{R}}_{\mathrm{S}}}\mathrm{t}\right)}}{{\displaystyle \left(\frac{{\mathrm{V}}^2}{{\mathrm{R}}_{\mathrm{P}}}\mathrm{t}\right)}}=\frac{{\mathrm{R}}_{\mathrm{P}}}{{\mathrm{R}}_{\mathrm{S}}} ------------ \left(1\right)$

Now, ${\mathrm{R}}_{\mathrm{S}}=\mathrm{R}+\mathrm{R}=2\mathrm{R} -----------\left(2\right)$

and ${\mathrm{R}}_{\mathrm{P}}=\frac{1}{\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)}=\frac{\mathrm{R}}{2} -----------\left(3\right)$

Substituting equations (2) and (3) in (1), we get

                                                       $\frac{{\mathrm{H}}_{\mathrm{S}}}{{\mathrm{H}}_{\mathrm{P}}}=\frac{\left({\displaystyle \frac{\mathrm{R}}{2}}\right)}{2\mathrm{R}} =\frac{1}{4}$

Therefore, the ratio of heat produced in series and parallel combinations is $1:4$.