Question

In FIGURE $EX28.3$, what is the magnitude of the current in the wire to the right of the junction? Does the charge in this wire flow to the right or to the left?

Short answer

The magnitude of the current in the wire to the right of the junction is$I=1A$and the charge in the wire flow to the right.

Step-by-step solution

Given information

We need to find the magnitude of the current in the wire to the right of the junction.

Simplify

The current in each resistor using Ohm's law states that the current $I$flows through a resistance due to the potential difference $V$between the resistance$R$

$I=\frac{V}{R}$

Calculating the current through each resistor using the values for $R$and $V$by

$I_{2\Omega }=\frac{V}{R}=\frac{4V}{2\Omega }=2A$

And the second resistor has current

$I_{5\Omega }=\frac{V}{R}=\frac{15V}{5\Omega }=3A$

For a resistor, the potential decreases from a positive terminal to a negative terminal. Hence, the current through the resistor travels from the positive terminal to the negative terminal. From the given figure, we can say that the current through $2\Omega$travels to the right while the current in $5\mathrm{\Omega }$travels downward.

At the junction, $I_{2\Omega }$enters while $I_{5\Omega }$leaves the junction.

For determining the current $I$we use the junction rule. Junction law states that the charge through the circuit is conserved, so at any junction in the circuit, the current entering the junction equals the leaving current from the junction. From the given figure we get

$$\begin{gathered} I_{5\Omega }=1+I_{2\Omega } \\ I=I_{5\Omega }-I_{2\Omega } \\ I=3A-2A \\ I=1A \end{gathered}$$

The leaving current is the summation of both currents that enter the junction. So, the current $I$must enters the junction.

Hence, the direction of the current is to the left.