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Two sources of equal emf are connected t...

Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are` R_(1) and R_(2) (R_(1) gt R_(1))`. If the potential difference across the source having internal resistance `R_(2)` is zero, then

A

`R=((R_2 times (R_1+R_2))/((R_2-R_1)`

B

`R=R_2-R_1`

C

`R=(R_1R_2)/((R_1+R_2)`

D

`R=(R_1R_2)/((R_2-R_1))`

Text Solution

Verified by Experts

As `R_1 , R_2` are R in series
`R_(eq)=R_1+R_2+R therefore` New current `I=(2E)/(R_1+R_2+R)`
According to the question `-(V_A-V_B)=E-IR_2`
`therefore 0=E-IR_2 or E=IR_2`
or `E=(2E)/(R_1+R_2+R) R_2 or R_1+R_2+R=2R_2 or R=R_2-R_1`
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