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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 volt and a resistance of $10 \Omega$. The potential gradient along the wire in volt/meter
is
Options:
3 volt and a resistance of $10 \Omega$. The potential gradient along the wire in volt/meter
is
Solution:
2822 Upvotes
Verified Answer
The correct answer is:
$0 \cdot 20$
(B)
Total resistance $\mathrm{R}=20+10=30 \Omega$
current I $=\frac{\mathrm{E}}{\mathrm{R}}=\frac{3}{30}=0.1 \mathrm{~A}$
p.d. acros wire $V=I R_{w}$
$=0.1 \times 20=2 \mathrm{~V}$
$\begin{aligned} \therefore \text { Potential gradient } &=\frac{\mathrm{V}}{\mathrm{L}} \\ &=\frac{2}{10}=0.2 \mathrm{~V} / \mathrm{m} \end{aligned}$
Total resistance $\mathrm{R}=20+10=30 \Omega$
current I $=\frac{\mathrm{E}}{\mathrm{R}}=\frac{3}{30}=0.1 \mathrm{~A}$
p.d. acros wire $V=I R_{w}$
$=0.1 \times 20=2 \mathrm{~V}$
$\begin{aligned} \therefore \text { Potential gradient } &=\frac{\mathrm{V}}{\mathrm{L}} \\ &=\frac{2}{10}=0.2 \mathrm{~V} / \mathrm{m} \end{aligned}$
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