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Visualizing the Electric Current

A battery of ...

Question

A battery of emf E and internal resistance r sends current I1 and I2, when connected to external resistance R1 and R2 respectively. Determine the emf and internal resistance of the battery.

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Solution

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$w h e n r e s i s t o r R_{1} i s c o n n e c t e d w i t h t h e c e l l o f i n t e r n a l r e s i s \tan c e r . F r o m k i r c h o f f s l o o p r u l e c u r r e n t i n c i r c u i t i s I_{1} = \frac{E}{r + R_{1}} . . . 1 w h e n r e s i s t o r R_{2} i s c o n n e c t e d w i t h t h e c e l l o f i n t e r n a l r e s i s \tan c e r . F r o m k i r c h o f f s l o o p r u l e c u r r e n t i n c i r c u i t i s I_{2} = \frac{E}{r + R_{2}} . . . 2 E q u a t i n g E f r o m b o t h e q u a t i o n I_{1} (r + R_{1}) = I_{2} (r + R_{2}) s o l v i n g f o r r r = \frac{I_{2} R_{2} - I_{1} R_{1}}{I_{1} - I_{2}} P u t t i n g v a l u e o f r i n e q u a t i o n . . 1 r + R_{1} = \frac{E}{I_{1}} \frac{I_{2} R_{2} - I_{1} R_{1}}{I_{1} - I_{2}} + R_{1} = \frac{E}{I_{1}} E = I_{1} (\frac{I_{2} R_{2} - I_{1} R_{1}}{I_{1} - I_{2}} + R_{1}) E = \frac{I_{1} I_{2} (R_{2} - R_{1})}{I_{1} - I_{2}} R e g a r d s$

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