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If E(1) = 10 V, E(2) = 5V(1),r(1) = 0.1 ...


If `E_(1) = 10 V, E_(2) = 5V_(1),r_(1) = 0.1 Omega , r_(2) = 0.1 Omega , R_(1) = 5 Omega and R_(2) = 4 Omega` . Find the potential difference between A and B .

Text Solution

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Let I current flows , in the circuit applying kirchhoff.s voltage law .
`-ir_(2) - E_(2) -iR_(2) + E_(1) -ir_(1)-I R_(1) = 0`
`E_(1) - E_(2) = ir_(1) + ir_(2) + iR_(2) + iR_(1)`
`E_(1) - E_(2) = i(r_(1)+r_(2)+R_(2) + R_(1))`
`10-5 = i (0.1 + 0.1+5+4)`
` 5 = i (9.2)`
`l=(5)/(9.2) = 0.54A`
Now , potential difference between A and B is `V_(A) - V_(B) = ir_(2) + E_(2)`
`V_(A) - V_(B) = (0.54 xx 0.1) + 5=5 + 0.1 = 5.1V`
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