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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 Vand 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

Answers (1)
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A admin

 

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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