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A $10 \mathrm{~m}$ long wire of resistance $20 \Omega$ is connected in series with a battery of e.m.f. $3 \mathrm{~V}$ (negligible internal resistance) and a resistance of $10 \Omega$. The potential gradient along the wire is
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$0.2 \mathrm{~V} / \mathrm{m}$
Two resistances are in series
$\mathrm{R}_{\mathrm{net}}=20 \Omega+10 \Omega=30 \Omega$
Current $\mathrm{I}=\frac{3}{30}=0.1 \mathrm{~A}$
Potential through the wire is $\mathrm{V}=0.1 \times 20=2 \mathrm{~V}$
$\therefore \quad$ Potential gradient $=\frac{\mathrm{V}}{10}=\frac{2}{10}=0.2 \frac{\mathrm{V}}{\mathrm{m}}$
$\mathrm{R}_{\mathrm{net}}=20 \Omega+10 \Omega=30 \Omega$
Current $\mathrm{I}=\frac{3}{30}=0.1 \mathrm{~A}$
Potential through the wire is $\mathrm{V}=0.1 \times 20=2 \mathrm{~V}$
$\therefore \quad$ Potential gradient $=\frac{\mathrm{V}}{10}=\frac{2}{10}=0.2 \frac{\mathrm{V}}{\mathrm{m}}$
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