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Questions  >>  CBSE XII  >>  Physics  >>  Current Electricity
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Q)

The adjacent fig shows a $2.0 \;V$ potentiometer used for the determined of internal resistance of a cell $1.5 \;v$ cell. The balance point of the cell is open circuit is $76.3 \;cm$ . When a resistor of $9.5 \Omega$ is used in the external circuit of the cell, the balance point shifts to $6.48 \;cm$ length of the wire . Determine the internal resistance of the cell

$\begin{array}{1 1} 16.8\;\Omega \\ 0.168\;\Omega \\ 1.68\;\Omega \\ 168\;\Omega \end{array} $

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A)
Solution :
$l_1= 76.3 \;cm$
$l_2 = 64.8\;cm$
$R= 9.5 \; \Omega$
$r= \bigg(\large\frac{l_1}{l_2}$$-1 \bigg)$
Substituting the values
$r= \bigg( \large\frac{76.3}{64.8} $$-1\bigg) \times 9.5$
$\qquad= 1.68 \Omega$
Hence the internal resistance of the cell is $1.68 \Omega$
Answer : $1.68\;\Omega$
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