A 10 m long wire of resistance $20 \Omega$ is connected in series with a battery of e .m. f., $3 V$

Question

A 10 m long wire of resistance $20 \Omega$ is connected in series with a battery of e .m. f., $3 V$ (negligible internal resistance) and a resistance of $10 \Omega$. The potential gradient along the wire in volt per meter is\

Professor PC Thomas Classes · Answer

Total resistance $=20+10=30 \Omega$\ Current $=\frac{3}{30}=\frac{1}{10} \mathrm{~A}$\ P.d. across the wire\ $=$ current $	imes$ resistance of the wire $=\frac{1}{10} 	imes 20$\ $=2 \mathrm{~V}$\ This is the p.d over 10 m\ Therefore p.d over $1 \mathrm{~m}=\frac{2 \mathrm{~V}}{10 \mathrm{~m}}$\