91Ó°ÊÓ

We need to find in which current flows in the bottom wire.

We will apply the loop rule where the loop rule is a statement that the electrostatic force is conservative. Suppose we go around the 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.

$∑V=0$

Let us draw a loop clockwise and apply equation $(28.2)$for the left loop, where the voltage across the resistor $6\mathrm{Î\copyright }$and $10\mathrm{Î\copyright }$is negative because the traveling direction is in the direction of the current. The e$emf\mathrm{Î\mu }=9V$is positive because the direction of traveling is from negative to positive terminal in the battery.

$$\begin{gathered} ∑V=0 \\ 9V-\left(6\mathrm{Î\copyright }\right)I_6-\left(12\mathrm{Î\copyright }\right)I_{12}=0 \\ {I}_6=105V-2I_{12} \end{gathered}$$

Draw a loop clockwise and apply equation $(28.2)$for the middle loop, where the voltage across the resistor $12\mathrm{Î\copyright }$is positive because the traveling direction is in the opposite direction of the current and $24\mathrm{Î\copyright }$is negative because the traveling direction is in the direction of the current.

$$\begin{gathered} ∑V=0 \\ -\left(24\mathrm{Î\copyright }\right)I_{24}+\left(12\mathrm{Î\copyright }\right)I_{12}=0 \\ {I}_{24}=12I_{12}/24 \\ {I}_{24}=\frac{I_{12}}{2} \end{gathered}$$

Draw a loop clockwise and apply equation $(28.2)$for the right loop, where the voltage across the resistor $10\mathrm{Î\copyright }and24\mathrm{Î\copyright }$is negative because the traveling direction is in the direction of the current. The $emf\mathrm{Î\mu }=15V$is the positive because the direction of traveling is from negative to positive terminal in the battery.

$$\begin{gathered} ∑V=0 \\ 15V-\left(10\mathrm{Î\copyright }\right)I_{10}-\left(24\mathrm{Î\copyright }\right)I_{24}=0 \\ {I}_{10}=1.5V-2.4I_{24} \end{gathered}$$

From the junction rule, we get the current through the bottom wire $I_{wire}$at the left bottom junction by

$$\begin{gathered} I_{wire}=I_6-I_{12} \\ =(1.5V-2I_{12})-I_{12} \\ =1.5V-3I_{12} \end{gathered}$$ (*)

So, for the junction rule at the right bottom junction, we get the current in the wire by

$$\begin{gathered} I_{wire}=I_{24}-I_{10} \\ =I_{24}-(105V-2.4I_{24)} \\ =-1.5V+3.4I_{24} \\ =1.5V+3.4(\frac{I_{12}}{2}) \\ =1.5V+1.7I_{12} \end{gathered}$$ (**)

Both previous equations have the same left side, so we can get $I_{12}$value by,

$$\begin{gathered} -1.5V+1.7I_{12}=105V-3I_{12} \\ {I}_{12}=0.63A \end{gathered}$$

Now, we plug the value for $I_{12}$into equation (*) to get $I_{wire}$by,

$$\begin{gathered} I_{wire}=1.5V-3I_{12} \\ =1.5V-3(0.64A) \\ =\boxed{-0.42A} \end{gathered}$$

The negative sign indicates that the current travel to the right.