Question

a. What are the magnitude and direction of the current in the $18\Omega$resistor in FIGURE $EX28.5$?

b. Draw a graph of the potential as a function of the distance traveled through the circuit, traveling cw from $V=0V$at the lower left corner.

Short answer

(a) The magnitude of current is$I=0.5A$and the direction is to the right or clockwise.

(b) The graph of the potential is given below in step $4$.

Step-by-step solution

Part (a) step 1: Given information 

We have given that current in the $18\Omega$resistor in FIGURE $EX28.5$.

We need to find the magnitude and direction of the current.

Part (a) step 2: Simplify

The current loop rule is a 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.

$\sum V=0...\left(1\right)$

As shown in the figure below let us use draw a loop clockwise. Let us apply equation $(28.2)$for the loop, where across the resistor resistance$∆V_{18\Omega }=-IR$is negative because of the traveling direction of the current. The emf ${\in }_1=3V$is positive because the direction of traveling is from negative to positive terminal in the battery.

localid="1648648283305" $$\begin{gathered} V=0 \\ {\in }_1+∆V_{10\Omega }+{\in }_2=0 \\ {\in }_1-IR+{\in }_2=0 \\ I=\frac{{\in }_1-{\in }_2}{R}...\left(2\right) \end{gathered}$$

Now, we plug the values for ${\in }_1,{\in }_2$and $R$into equation $\left(2\right)$to get $I$

$$\begin{gathered} I=\frac{{\in }_1-{\in }_2}{R} \\ =\frac{3V+6V}{18\Omega } \\ =0.5A \end{gathered}$$

The positive sign means the current travels in the same direction of our estimated loop, so the magnitude of the current is $I=0.5A$ to the right. or clockwise

Part (b) step 1: Given information  

We have given that the traveling cw from $V=0V$at the lower left corner.

We need to draw a graph of potential as a function of the distance traveled through the circuit.

Part (b) step 2: Simplify 

The potential difference across $18\Omega$resister is

$∆V_{18\Omega }=IR=(0.5A)\left(18\Omega \right)=9V$

As we found in part (a), The current is clockwise, so the potential decreases in the clockwise direction. When we start from the lower-left corner, the voltage gains $3V$from battery $1$. Then decreases by $9V$that losses by the resistance $18\Omega$to reach $-6V$. Finally, it increases to reach $0V$when it gains $6V$from battery $2$. So the graph of potential clockwise is shown below.

as we found in part (a),