Question

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

• A. +500 mA
• B. +250 mA
• C. -250 mA
• D. -500 mA

Answer 1

Answer: (B)

Applying Kirchhoff's second law for a closed loop $CABDC$, we get
$-10I_1-10\left(I_1-I_2\right)+10=0$
$-20I_1+10I_2=-10$ or $2I_1-I_2=1$...(i)
Again, applying Kirchhoff's second law for a closed loop $ABFEA,$ we get
$-10\left(I_1-I_2\right)-5+15I_2=0$
$-10I_1+25I_2=5$ or $2I_1-5I_2=-1$...(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
$=I_1-I_2=\frac{3}{4}A-\frac{1}{2}A$
$=\frac{1}{4}A=0.25A=250mA$