In the secondary circuit of a potentiometer a cell of internal resistance 1.5 gives balancing length of 52 cm. To get a balancing length of 40 cm, how much resistance is to be connected across the cell?

5 Ω

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6 Ω

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7 Ω

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4 Ω

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Solution

The correct option is A $5 Ω$
$r = 1.5 Ω, I_{1} = 52 c m, I_{2} = 40 c m \frac{r}{R} = \frac{I_{1} - I_{2}}{I_{2}}$
Resisteance R $= \frac{r I_{2}}{(I_{1} - I_{2})} = \frac{1.5 \times 40}{(52 - 40)} = \frac{1.5 \times 40}{12} = 5 Ω$

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In the secondary circuit of a potentiometer a cell of internal resistance 1.5 $Ω$ gives balancing length of 52 cm. To get a balancing length of 40 cm, how much resistance is to be connected across the cell?

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In the secondary circuit of a potentiometer a cell of internal resistance 1.5 gives balancing length of 52 cm. To get a balancing length of 40 cm, how much resistance is to be connected across the cell?

The correct option is A 5Ωr=1.5Ω,I1=52cm,I2=40cmrR=I1−I2I2 Resisteance R =rI2(I1−I2)=1.5×40(52−40)=1.5×4012=5Ω

The correct option is A $5 Ω$
$r = 1.5 Ω, I_{1} = 52 c m, I_{2} = 40 c m \frac{r}{R} = \frac{I_{1} - I_{2}}{I_{2}}$
Resisteance R $= \frac{r I_{2}}{(I_{1} - I_{2})} = \frac{1.5 \times 40}{(52 - 40)} = \frac{1.5 \times 40}{12} = 5 Ω$