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A potentiometer wire of length $100 \mathrm{~cm}$ and resistance $3 \Omega$ is connected in series
with resistance of $8 \Omega$ and an accumulator of 4 volt whose internal resistance is
$1 \Omega$
A cell of e.m.f. 'E' is balanced by $50 \mathrm{~cm}$ length of the wire. The e.m.f. of the
cell is
Options:
with resistance of $8 \Omega$ and an accumulator of 4 volt whose internal resistance is
$1 \Omega$
A cell of e.m.f. 'E' is balanced by $50 \mathrm{~cm}$ length of the wire. The e.m.f. of the
cell is
Solution:
1217 Upvotes
Verified Answer
The correct answer is:
$0 \cdot 50$ volt.
(C)
Total resistance $=3+8+1=12 \Omega$ current. $\quad I=\frac{E}{R}=\frac{4}{12}=\frac{1}{3} A$
P.D. across the wire $=I R_{\omega}=\frac{1}{3} \times 3=1 \mathrm{~V}$
potential gradient $\mathrm{K}=\frac{\mathrm{V}}{\mathrm{L}}=\frac{1}{1}=1 \mathrm{v} / \mathrm{m}$ $\mathrm{E}^{\prime}=\mathrm{K} \ell^{\prime}=1 \times 0.5=0.5 \mathrm{~V}$
Total resistance $=3+8+1=12 \Omega$ current. $\quad I=\frac{E}{R}=\frac{4}{12}=\frac{1}{3} A$
P.D. across the wire $=I R_{\omega}=\frac{1}{3} \times 3=1 \mathrm{~V}$
potential gradient $\mathrm{K}=\frac{\mathrm{V}}{\mathrm{L}}=\frac{1}{1}=1 \mathrm{v} / \mathrm{m}$ $\mathrm{E}^{\prime}=\mathrm{K} \ell^{\prime}=1 \times 0.5=0.5 \mathrm{~V}$
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