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)
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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