Question 558: Current Electricity

Question

Resistance of a potentiometer wire is $$2 \Omega / \mathrm{m}$$. A cell of e.m.f. $$1.5 \mathrm{~V}$$ balances at $$300 \mathrm{~cm}$$. The current through the wire is

Accepted Answer

$$l=300 \mathrm{~cm}=3 \mathrm{~m}$$
Total resistance of wire,
$$\mathrm{R}=3 	imes 2=6 \Omega$$
Since, the potentiometer is balanced. Voltage across wire segment $$=1.5 \mathrm{~V}$$
$$\begin{array}{ll}
	herefore & \mathrm{IR}=1.5 \mathrm{~V} \
	herefore & \mathrm{I}=\frac{1.5}{6}=250 \mathrm{~mA}
\end{array}$$

Answer

$$2.5 \mathrm{~mA}$$

Answer

$$7.5 \mathrm{~mA}$$

Answer

$$250 \mathrm{~mA}$$

Answer

$$750 \mathrm{~mA}$$

Current Electricity

Resistance of a potentiometer wire is $2 \Omega / \mathrm{m}. A cell of e.m.f. 1.5 \mathrm{~V} balances at 300 \mathrm{~cm}$. The current through the wire is

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