ltBrgt A circuit is as shown in the figure. Then, the current from A to B is

Options

A. `+500Ma`
B. `+250mA`
C. `-250mA`
D. `-500mA`

Answer

B

Explanation

Applying kirchhoff's second law for a closed loop CABDC, we get

`-10l_(1)-10(l_(1)-l_(2))+10=0`
`2l_(1)-l_(1)=1` . . (i)
Again, applying kirchhoff second law for a second loop ABFEA, we get
`-10(l_(1)-l_(2))-5+15l_(2)=0`
`2l_(1)-5l_(2)=-1` . . (ii)
Solving eqs. (i) and (ii) we get
`l_(1)=(3)/(4)A,l_(2)=(1)/(2)A`
The current flows from A to B is
`=l_(1)-l_(2)=(3)/(4)A-(1)/(2)A=(1)/(4)`
`A=0.25A=250mA`