Figure shows a circuit with three ideal batteries in it. The circuit elements h ave the following values

`DeltaV_(B_(1))=3.0V,DeltaV_(B_(2))=6.0V`
`R_(1)=2.0Omega,R_(2)=4.0Omega` The currents `i_(1) ,i_(2)` and `i_(3)` as shown in the circuit have the values.

Options

A. 0.50A,-0.25A,+0.25A
B. 0.25A,-0.50A,-0.25A
C. 0.50A,0.50A,1.0A
D. `-0.25A,0.50A,0.25A`

Answer

1

Explanation

Applying kirchhoff's first law at junction E, we g-t
`i_(3)=i_(1)+i_(2)`
Applying kirchhoff's second law for the closed loop
ABEFA, we get
`-i_(1)R_(1)+DeltaV_(B_(2))+i_(2)R_(2)-i_(1)R_(1)-DeltaV_(B_(1))=0`
`-2i_(1)R_(1)+i_(2)R_(2)+DeltaV_(B_(2))=DeltaV_(b_(1))=0`
`-2i_(1)(2)i_(2)(4)+6-3=0`
`-4i_(1)+4i_(2)+3=0`
`-4i_(1)+4i_(2)=-3`
Again, applying kirchhoff's second for the closed loop BCDEB, we get
`i_(3)R_(1)+DeltaV_(B_(2))-i_(3)R_(1)-i_(2)R_(2)-DeltaV_(B_(2))=0`
`-2i_(3)R_(1)- i_(2)R_(2)=0`
`-2i_(3)(2)-i_(2)(4)=0`
`ri_(3)+4i_(2)=i_(3)+i_(2)=0`
Solving (i),(ii) and (iii), we get
`i_(1)=0.50AA,i_(2)=-0.25A,i_(3)=0.25A`