A potentiometer is 10 m long. It has a resistance of 20 Ω. It is connected in series with a battery of 3 V and a resistance of 10 Ω. What is the potential gradient along with wire?
Solution:
In figure, AB is a potentiometer wire.
length of potentiometer wire, \(L\) = 10 m
resistance of potentiometer wire, \(R_{AB}\) = 20 Ω
resistance in series, \(R\) = 10 Ω
potential gradient , \(k\) = ?
total current flowing through the wire is,
\[\begin{align*} I&=\frac{E}{R+R_{AB}}\\ &=\frac{3}{30}\\ \therefore I&=0.1 \hspace{0.1cm} A\\ \end{align*}\] Now,
potential drop across AB is,
\[\begin{align*} V&=I R_{AB}\\ &=0.1 \times 20\\ \therefore V&=2 \hspace{0.1cm} V\\ \end{align*}\] So, the potential gradient along with the wire is, \[\frac{V}{L}=\frac{2}{10}=0.2 \hspace{0.1cm} V/m\]
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