Question

evaluation and review questions: 1. in step 8, you simulated an open resistor by removing it from the circuit. explain how the open resistor can be found in this experiment if the only reading available is the total current. 2. if one of the resistors in this experiment were shorted, what would you expect to see happen? (do not simulate this!) 3. three resistors with resistances of 620 ω, 750 ω, and 820 ω are in parallel across a 40 v source. (a) what should the total source current be? (b) if the measured source current was actually 118 ma, what fault could account for this? 4. the known currents for a circuit junction are shown in figure 9 - 10. what is the value and direction of the unknown current, (i_4)? (i_1 = 170 ma), (i_2 = 105 ma), (i_3 = 300 ma), (i_4=?) figure 9 - 10 5. show the application of the current divider rule to figure 9 - 5 by using the total current and total resistance in table 9 - 2 and the measured resistances from table 9 - 1. find (i_1), (i_2), (i_3), and (i_4).

Explanation:

Step 1: Solve question 3(a)

First, find the equivalent resistance $R_{eq}$ of the parallel - connected resistors. The formula for the equivalent resistance of three resistors $R_1$, $R_2$, $R_3$ in parallel is $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$. Given $R_1 = 620\Omega$, $R_2=750\Omega$, $R_3 = 820\Omega$.
\[

$$\begin{align*} \frac{1}{R_{eq}}&=\frac{1}{620}+\frac{1}{750}+\frac{1}{820}\\ &=\frac{750\times820 + 620\times820+620\times750}{620\times750\times820}\\ &=\frac{615000+508400 + 465000}{620\times750\times820}\\ &=\frac{1588400}{381900000}\\ R_{eq}&=\frac{381900000}{1588400}\approx240.43\Omega \end{align*}$$

\]
Then, use Ohm's law $I=\frac{V}{R}$ to find the total source current. Given $V = 40V$, so $I=\frac{40}{240.43}\approx0.166A = 166mA$.

Step 2: Solve question 3(b)

The calculated current in part (a) is $166mA$, but the measured current is $118mA$. A lower - than - expected current could be due to an open - circuit in one of the resistors. An open - circuit in a parallel circuit increases the equivalent resistance, and according to Ohm's law $I=\frac{V}{R}$, when $V$ is constant and $R$ increases, $I$ decreases.

Step 3: Solve question 4

According to Kirchhoff's current law (KCL), which states that the algebraic sum of currents entering a junction is equal to the algebraic sum of currents leaving the junction. Let currents entering the junction be positive and currents leaving be negative. So $I_1+I_2 - I_3-I_4=0$. Given $I_1 = 170mA$, $I_2 = 105mA$, $I_3 = 300mA$. Then $170 + 105-300 - I_4=0$, $I_4=170 + 105 - 300=- 25mA$. The negative sign indicates that the direction of $I_4$ is into the junction.

Answer:

3(a): The total source current is approximately $166mA$.
3(b): An open - circuit in one of the resistors could account for the lower measured current.
4: The value of $I_4$ is $25mA$ and its direction is into the junction.