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Two potentiometer wires W₁ and W₂ of equal length $l$ connected to a battery of emf E_P and internal resistance ‘r’ through two switches S₁ and S₂. A battery of emf E is balanced on these potentiometer wires. If potentiometer wire W₁ is of resistance 2r and balancing length on W₁ is $l/2$ when only switch S₁ is closed and S₂ is open. On closing S₂ and opening S₁ the balancing length on W₂ is found to be $\left ( \frac{2l}{3} \right )$ , the resistance of the potentiometer W₂ will be

Option: 1
$r$

Option: 2
$\frac{2r}{5}$

Option: 3
$3r$

Option: 4
$\frac{2r}{3}$

Answers (1)

best_answer

Answer (1)

When $S_{1}$ is closed and $S_{2}$ is opened

$I=\frac{E_{P}}{3r}$

$E=Ir_{1}$

$E=I\times\frac{1}{2}\times\frac{2r}{I}$

$E=\frac{E_{P}}{3}$

When $S_{2}$ is opened and $S_{1}$ is closed

$I=\frac{E_{P}}{r_{1}+r}$

$E= I \times r_{2}$

$=I \times \frac{2I}{3}\times \frac{r_{1}}{I}$

$=\frac{2E_{P}r_{1}}{3\left ( r_{1}+r \right )}$

$\frac{E_{P}}{3}=\frac{2E_{P}r_{1}}{3\left ( r_{1}+r \right )}$

$r_{1}+r=2r_{1}$

$r_{1}=r$

Posted by

HARSH KANKARIA

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