A 10 m long wire of resistance \( 20\,\Omega \) is connected in series with a battery of emf 3 V (negligible internal resistance) and a resistance of \( 10\,\Omega \). The potential drop per metre along the wire (in \( \mathrm{V\,m^{-1}} \)) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.2
The wire, the \( 10\,\Omega \) resistor and the battery are in series, so the same current flows through all of them. Along a uniform wire the potential falls by the same amount on every metre. So the drop per metre is the drop across the wire divided by its length.
\( \begin{aligned} I &= \frac{\varepsilon}{R_{\mathrm{w}}+R} \\ &= \frac{3\,\mathrm{V}}{20\,\Omega+10\,\Omega} \\ &= 0.1\,\mathrm{A} \end{aligned} \)
\( \begin{aligned} V_{\mathrm{w}} &= IR_{\mathrm{w}} \\ &= 0.1\,\mathrm{A}\times20\,\Omega \\ &= 2\,\mathrm{V} \end{aligned} \)
\( \begin{aligned} \frac{V_{\mathrm{w}}}{L} &= \frac{2\,\mathrm{V}}{10\,\mathrm{m}} \\ &= 0.2\,\mathrm{V\,m^{-1}} \end{aligned} \)