Thank you for reporting, we will resolve it shortly
Q.
A circuit is as shown in the figure. Then, the current from $A$ to $B$ is
Current Electricity
Solution:
Applying Kirchhoff's second law for a closed loop $CABDC$, we get
$-10 I_{1}-10\left(I_{1}-I_{2}\right)+10=0$
$-20 I_{1}+10 I_{2}=-10$ or $2 I_{1}-I_{2}=1$...(i)
Again, applying Kirchhoff's second law for a closed loop $ABFEA,$ we get
$-10\left(I_{1}-I_{2}\right)-5+15 I_{2}=0$
$-10 I_{1}+25 I_{2}=5$ or $2 I_{1}-5 I_{2}=-1$...(ii)
Solving (i) and (ii), we get
$I_{1}=\frac{3}{4} A , I_{2}=\frac{1}{2}\, A$
The current flows from $A$ to $B$ is
$=I_{1}-I_{2}=\frac{3}{4} A -\frac{1}{2}\, A$
$=\frac{1}{4} A =0.25\, A =250\, mA$
