Question

the coil of heater is cut into two equal halves and only one of them is used in the heater . what is the ratio of heat producedby half of the coil to the original coil?

Answer 1

When the coil is cut into two equal parts then the resistance of each part will be one-half of the resistance of the original coil. Since, the applied potential difference is the same, therefore, the current will be doubled.

$$\begin{gathered} HeatproducedbythehalfcoilH\text{'}=\left(2I{)}^2\right(\frac{R}{2})t \\ =4I^{2}X\frac{Rt}{2} \\ =2I^{2}Rt \\ \mathit{H}\mathit{e}\mathit{a}\mathit{t}\mathbf{}\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{e}\mathit{d}\mathbf{}\mathit{b}\mathit{y}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{h}\mathit{a}\mathit{l}\mathit{f}\mathbf{}\mathit{c}\mathit{o}\mathit{i}\mathit{l}\mathbf{}\mathbf{=}\mathbf{2}{\mathit{I}}^{\mathbf{2}}\mathit{R}\mathit{t}\mathbf{} \\ \mathit{H}\mathbf{\text{'}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}{\mathit{I}}^{\mathbf{2}}\mathit{R}\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{H}\mathbf{}\mathbf{}(H\hspace{0.17em}istheheatproducedbytheoriginalcoil) \\ Hence, \\ RatiooftheheatproducedbytheHalfcoiltowardsoriginalcoil=\frac{H\text{'}}{H}\mathbf{=}\mathbf{}\mathbf{2} \end{gathered}$$