Question 523: Current Electricity

Question

A $$10 \mathrm{~m}$$ long wire of resistance $$20 \Omega$$ is connected in series with a battery of emf $$3 \mathrm{~V}$$ (negligible internal resistance) and a resistance of $$10 \Omega$$. The potential gradient along the wire is

Accepted Answer

Total length of wire $$=10 \mathrm{~m}$$
The equivalent resistance of the circuit, $$R_{\mathrm{eq}}=10+20 =30 \Omega$$
The current flowing through the wire, $$i=V / R$$
$$=\frac{3 V}{30 \Omega}=0.1 \mathrm{~A}$$
$$	herefore$$ Potential gradient along the wire,
$$\begin{aligned}
& =\frac{	ext { potential difference across } 20 \Omega}{	ext { length of } 20 \Omega 	ext { wire }} \
& =\frac{0.1 	imes 20 \Omega}{10 \mathrm{~m}}=0.20 \mathrm{~V} / \mathrm{m}
\end{aligned}$$

Answer

$$3 \mathrm{~V} / \mathrm{m}$$

Answer

$$0.1 \mathrm{~V} / \mathrm{m}$$

Answer

$$0.2 \mathrm{~V} / \mathrm{m}$$

Answer

$$0.3 \mathrm{~V} / \mathrm{m}$$

Current Electricity

A $10 \mathrm{~m} long wire of resistance 20 \Omega is connected in series with a battery of emf 3 \mathrm{~V} (negligible internal resistance) and a resistance of 10 \Omega$. The potential gradient along the wire is

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