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Help me solve this Two sources of equal emf are connected to an external resistance R The internal resistances of he two sources are R1 and R2 ( R2 > R1 ). If the potential difference across the source having internal resistance R2 is zero, then

Two sources of equal emf are connected to an external resistance The internal resistances of he two sources are  R1 and R2   ( R> R). If the potential difference across the source having internal resistance  R2   is zero, then :

  • Option 1)

    R=\frac{R_{1}R_{2}}{R_{1}+R_{2}}

  • Option 2)

    R=\frac{R_{1}R_{2}}{R_{2}-R_{1}}

  • Option 3)

    R=R_{2}\frac{\left ( R_{1}+R_{2} \right )}{\left ( R_{2}-R_{1} \right )}

  • Option 4)

    R=R_{2}-R_{1}

 
Answers (1)
1117 Views
V Vakul

As we learnt in

Current given by the cell -

i=\frac{E}{R+r}

- wherein

R- External resistance

r-  internal resistance

 

 I =\frac{2E}{R_{1}+R_{2}+R}

\because E-IR_{2}=0 \ (given)

\therefore E=IR_{2}

E =\frac{2E \ R_{2} }{R_{1}+R_{2}+R} \Rightarrow R_{1}+R_{2}+R = 2R_{2}

R = R_{2} - R_{1} 

 


Option 1)

R=\frac{R_{1}R_{2}}{R_{1}+R_{2}}

This is incorrect.

Option 2)

R=\frac{R_{1}R_{2}}{R_{2}-R_{1}}

This is incorrect.

Option 3)

R=R_{2}\frac{\left ( R_{1}+R_{2} \right )}{\left ( R_{2}-R_{1} \right )}

This is incorrect.

Option 4)

R=R_{2}-R_{1}

This is correct.

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