Question
Two wires A and B of the same material and having same length, have their cross sectional areas in the ratio 1 : 6. What would be the ratio of heat produced in these wires when same voltage is applied across each?
✓
Answer
$\text{A}_\text{A}:\text{A}_\text{B}=1:6$
$\text{H}-\text{V}^2\frac{\text{t}}{\text{R}}$ and $\text{R}=\frac{\rho\text{l}}{\text{A}}$
$\text{H}_\text{A}=\frac{\text{V}^2\text{t}}{\frac{\rho\text{l}}{\text{A}_\text{A}}};\text{H}_\text{B}=\frac{\text{V}^2\text{t}}{\frac{\rho\text{l}}{\text{A}_\text{B}}}$
$=\frac{\text{H}_\text{A}}{\text{H}_\text{B}}=\frac{\text{V}^2\text{t}\times\text{A}_\text{A}}{\rho\text{l}}\times\frac{\rho\text{l}}{\text{V}^2\text{tA}_\text{H}}$
$\Rightarrow\frac{\text{H}_\text{A}}{\text{H}_\text{B}}=\frac{\text{A}_\text{A}}{\text{A}_\text{B}}=1:6$
$\text{H}-\text{V}^2\frac{\text{t}}{\text{R}}$ and $\text{R}=\frac{\rho\text{l}}{\text{A}}$
$\text{H}_\text{A}=\frac{\text{V}^2\text{t}}{\frac{\rho\text{l}}{\text{A}_\text{A}}};\text{H}_\text{B}=\frac{\text{V}^2\text{t}}{\frac{\rho\text{l}}{\text{A}_\text{B}}}$
$=\frac{\text{H}_\text{A}}{\text{H}_\text{B}}=\frac{\text{V}^2\text{t}\times\text{A}_\text{A}}{\rho\text{l}}\times\frac{\rho\text{l}}{\text{V}^2\text{tA}_\text{H}}$
$\Rightarrow\frac{\text{H}_\text{A}}{\text{H}_\text{B}}=\frac{\text{A}_\text{A}}{\text{A}_\text{B}}=1:6$
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