Two resistance `R_(1)` and `R_(2)` are joined as shown in figure to two batteries of emf `E_(1)` and `E_(2)`. If `E_(2)` is short circuited, what is the current through `R_(1)` ?
Text Solution
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When `E_(2)` is short circuited, it means positive pole of battery `E_(2)` is connected to negative pole of battery `E_(2)`. The battery gets discharged. The resistance of the arm become zero. The current from cell `E_(1)`, flowing through `R_(1)` will not pass through `R_(2)`, but will go through the short circuited path. Therefore, `R_(2)` becomes inffective. Now current through `R_(1)` is due to emf `E_(1)` with resistance `R_(1)` in circuit, i.e., `I= E_(1)//R_(1)`
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