Question
Two heater coils made of same material are connected in parallel across the mains. The length and diameter of the wire of one of coils are double that of the other. Which one of them will produce more heat?
✓
Answer
We have resistance of one wire,
$\text{R}_1=\frac{\rho\text{l}}{\text{A}}=\frac{\rho\text{l}}{\pi\text{r}^2}=\frac{4\rho\text{l}}{\pi\text{D}^2}$
Where l is length and D is diameter of the wire. The resistance of second wire of double the length and double the diameter is,
$\text{R}_2=\frac{4\rho(2\text{l})}{\pi(2\text{D})^2}=\frac{4\rho\text{l}}{\pi\text{D}^2}.\frac12,\text{ i.e.,}\text{ R}_2=\frac{\text{R}_1}{2}$
Heat produced per second,
$\text{H}=\frac{\text{V}^2}{\text{R}}\propto\frac{1}{\text{R}}$
As second coil has resistance equal to half of first coil, therefore heat produced in second coil is double than that in first coil.
$\text{R}_1=\frac{\rho\text{l}}{\text{A}}=\frac{\rho\text{l}}{\pi\text{r}^2}=\frac{4\rho\text{l}}{\pi\text{D}^2}$
Where l is length and D is diameter of the wire. The resistance of second wire of double the length and double the diameter is,
$\text{R}_2=\frac{4\rho(2\text{l})}{\pi(2\text{D})^2}=\frac{4\rho\text{l}}{\pi\text{D}^2}.\frac12,\text{ i.e.,}\text{ R}_2=\frac{\text{R}_1}{2}$
Heat produced per second,
$\text{H}=\frac{\text{V}^2}{\text{R}}\propto\frac{1}{\text{R}}$
As second coil has resistance equal to half of first coil, therefore heat produced in second coil is double than that in first coil.
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