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epsi(1) and epsi(2) are two batteries ha...

`epsi_(1) and epsi_(2)` are two batteries having e.m.f of 34V and 10V respectively and internal resistance of `1Omega and 2Omega` respectively. They are connected as shown in figure below. Using Kirchhoff.s Laws of electrical networks, calculate the current `I_(1) and I_(2)`

Text Solution

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Applying Kirchhoff.s second law in mesh ABEFA, `1I_(1) + 4I_(1) + 5(I_(1) + I_(2)) + 7I_(1) + 34`
or `17I_(1) + 5I_(2)= 34`…(i)
Applying Kirchhoff.s second law in mesh ACDFA, `1I_(1) + 4I_(1)- 4i_(2)- 2I_(2)- 7I_(2) + 7I_(1) = 34-10` or `12I_(1)- 13I_(2)= 24` ...(ii)

Multiply equation (i) by 13 and equation (ii) by 5, and add both the equations, `221I_(1) + 65I_(2)= 442`
`{:(ul(60I_(1)-65I_(2)=120)),(281I_(1)" "= 562):}`
`I_(1)= 2A`
`5I_(2)= 34-17`
`5I_(2)= 0`
`I_(2)= 0`
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