Question

If $R_1=2\Omega$, $R_2=4\Omega$, $R_3=6\Omega$, determine the electric current that flows in the circuit below.

Answer 1

According to the kirchhoff’s voltage law (KVL) - the sum of all voltages around a closed loop in any circuit must be equal to zero.
Let us assume the direction of current (I) in the loop is clockwise.
Using Kirchoff's law's sign conventions as shown below:
Starting from Battery $E_2$ and going clockwise to applying KVL.
$-E_2-(I\times R_1)+E_1-(I\times R_2)-(I\times R_3)=0$
Substituting the value of resistances.
$-5-(I\times 2)+10-(I\times 4)-(I\times 6)=0$
$5-\left(12I\right)=0$
$12I=5$
$⟹$$I=\frac{5}{12}$
$⟹$$I=0.416A$
As the direction current comes out to be positive. So, the direction of current will be the same as assumed direction.
Hence, 0.416 A current will flow in the circuit in clockwise direction.