The same mass of copper is drawn into two wires of diameters 1 mm and 2 mm. If the two wires are connected in series and a current is passed through them, what is the ratio of the heat produced in the wires?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 16 : 1
In a series connection the same current flows through both wires. So by Joule’s law, \( H=I^{2}Rt \), the heat produced in each wire is proportional to its resistance. The same mass of copper has the same volume, so the thinner wire is also the longer one.
For equal volumes \( A_1l_1=A_2l_2 \), so \( l\propto1/A \). With \( R=\rho l/A \):
\( \displaystyle R\propto\frac{1}{A^{2}} \)
Since \( A=\pi D^{2}/4 \) for a wire of diameter \( D \):
\( \displaystyle R\propto\frac{1}{D^{4}} \)
The ratio of the heats is then:
\( \begin{aligned} \frac{H_1}{H_2} &= \frac{R_1}{R_2} \\ &= \left(\frac{D_2}{D_1}\right)^{4} \\ &= \left(\frac{2\,\mathrm{mm}}{1\,\mathrm{mm}}\right)^{4} \\ &= 16 \end{aligned} \)
Therefore, the heat produced in the 1 mm and 2 mm wires is in the ratio 16 : 1.