The circuit in Fig. shows two cells connected in opposition to each other. Cell `E_(1)` is of emf `6V` and internal resistance `2Omega,` the cell `E_(2)` is of emf `4 V` sand internal resistance `8omega`. Find the potential difference between the points `A` and `B`.
Text Solution
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Here, `E_(1) = 6 V, r_(1)= 2Omega, E_(2)=4 V, r_(2) = 8Omega` Refer Fig.`2(EP)`.12, current in circuit, `l=(6-4)/(2+8) = 1/5A` Potential difference across `A` and `B` is the terminal voltage of cell `E_(1) = E_(1) - Ir_(1) = 6-1/5xx2 = 5.6 V`
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