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# Need explanation for: - the resistance of meter bridge AB in given figure is . With the cell of emf and rehostat resistanc - Current Electricity - JEE Main

the resistance of meter bridge AB in given figure is $4\Omega$. With the cell of emf $\varepsilon = 0.5 V$and rehostat resistance Rh = $2\Omega$ the null point is obtained at some point J. When the cell is replaced by another one of emf $\varepsilon =\varepsilon _{2}$  the same null point J is found for $R_{h } = 6\Omega$. The emf $\varepsilon _{2}$ is:

• Option 1)

0.6 V

• Option 2)

0.4 V

• Option 3)

0.3 V

• Option 4)

0.5 V

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Meter bridge -

$\frac{P}{Q}=\frac{R}{S}\Rightarrow S= \frac{(100-l)}{l}R$

- wherein

$AB=l$

$BC=(100-l)$

When $R_{h}=2\Omega$

$\frac{dV}{dL}=(\frac{6}{4+2})\times \frac{4}{L}$

$L=100 cm$

Let null point be at l cm

$\varepsilon _{1}=0.5V=(\frac{6}{2+4})\frac{4}{L}l-(1)$

for $R_{n}=6\Omega$

$\varepsilon _{2}=(\frac{6}{4+6})\frac{4}{L}\times l$

$\frac{0.5}{\varepsilon _{2}}=\frac{10}{6}$

$\Rightarrow \varepsilon _{2}=0.3$

Option 1)

0.6 V

Option 2)

0.4 V

Option 3)

0.3 V

Option 4)

0.5 V

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