What is the current through the $10\Omega$resistor in $figureP28.63$? Is the current from left to right or right to left?

We need to find the current from left to right or right to left.

We will apply the loop rule where the loop rule is the statement that the electrostatic force is conservative. Suppose we go around a loop, measuring potential differences across circuit elements as we go and the algebraic sum of these differences is zero when we return to the starting point.

$\sum _{}V=0$

Let us draw a loop clockwise and apply equation $(28.2)$for the upper loop, where the voltage across the resistor $5\Omega and10\Omega$is positive because the traveling direction is in the opposite direction of the current. The $emf\epsilon =12Vand3V$are negative because the direction of traveling is from positive to negative terminal in the battery

$$\begin{gathered} \sum _{}V=0 \\ -12V+\left(5\Omega \right)I_{up}+\left(10\Omega \right)I_{mid}-3V=0 \\ {I}_{up}+2I_{mid}=3V \end{gathered}$$

Where $I_{up}$is the current in the upper branch and $I_{mid}$is the current in the middle branch, for the bottom loop, we apply loop rule clockwise, where the voltage across the resistor $10\Omega$is negative because the travelling direction is in the opposite direction of the current and $5\Omega$is positive because the travelling direction is in the opposite direction of the current. The$emf\epsilon =3V$is positive because the direction of traveling is from negative to positive terminal in the battery and $9V$are negative because the direction of traveling is from positive to negative terminal in the battery

$$\begin{gathered} \sum _{}V=0 \\ -\left(10\Omega \right)I_{mid}+3V-9V+\left(5\Omega \right)I_{bottom} \\ \left(5\Omega \right)I_{bottom}-\left(10\Omega \right)I_{mid}=6V \end{gathered}$$

Where $I_{bottom}$is the current in the bottom branch. From junction rule, we get that the current through the bottom branch is

$I_{bottom}=I_{up}-I_{mid}$

Now, Let us plug the expression of $I_{bottom}$into equation (3) to get

$$\begin{gathered} \left(5\Omega \right)(I_{up}-I_{mid})-\left(10\Omega \right)I_{mid}=6V \\ \left(5\Omega \right)I_{up}-\left(15\Omega \right)I_{mid}=6V \\ {I}_{up}-\left(3\Omega \right)I_{mid}=102V \end{gathered}$$

Let us subtract the two equations $\left(2\right)and\left(4\right)$to get the middle current by

$$\begin{gathered} I_{up}+2I_{mid}=3V \\ \frac{I_{up}-\left(3\Omega \right)I_{mid}=1.2V}{\left(5\Omega \right)I_{mid}=1.8V} \end{gathered}$$

Therefore, we get the middle current by

$I_{mid}=\frac{1.8V}{5\Omega }=0.36A$

This is the current through the resistor $10\Omega$and its direction is from left to right.