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In the circuit shown in figure below, E(...

In the circuit shown in figure below, `E_(1)` and `E_(2)` are batteries having emfs 4.0 V and 3.5 V respectively and internal resistance `1Omega` and `2Omega` respectively. Using Kirchhoff.s laws, calculate currents : `I_(1),I_(2)` and `I_(3)`.

Text Solution

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In closed circuit PQSP,
`1I_(1)+4I_(1)+10I_(3)=4" "......(1)`
Applying Kirchhoff.s first law at point S,
`I_(3)=I_(1)+I_(2)`
Using the value of Iz in equation (1).
`I_(1)+4I_(1)+10(I_(1)+I_(2))=4`
`15I_(1)+10I_(2)=4" ".......(2)`
In closed circuit RQSR,
`2I_(2)+3I_(2)+10(I_(1)+I_(2))=3.5`
`10I_(1)+15I_(2)=3.5" "......(3)`
On solving equations (2) and (3), we get
`I_(2)=0.1A,I_(1)=0.2A`
And `I_(3)=I_(1)+I_(2)=0.3A`
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