https://file.aiquickdraw.com/mathbot/file/17310122558022i7p7h9a.jpg

Let's solve each question in turn based on the information provided.

---

### Question 1
**Given Data:**
- \( R_1 = 140 \, \Omega \)
- \( R_2 = 3.10 \, \text{k}\Omega = 3100 \, \Omega \)
- \( R_3 = 3.51 \, \text{k}\Omega = 3510 \, \Omega \)
- \( I_S = 7.26 \, \text{mA} \)

We are asked to find the current through \( R_3 \) using the current divider rule.

**Solution:**
1. The equivalent resistance of the parallel combination of \( R_1 \) and \( R_2 \) is:
   \[
   R_{12} = \frac{R_1 \cdot R_2}{R_1 + R_2}
   \]
   Plugging in the values:
   \[
   R_{12} = \frac{140 \times 3100}{140 + 3100} = \frac{434000}{3240} \approx 134 \, \Omega
   \]

2. Now, we apply the current divider rule to find the current \( I_{R3} \) through \( R_3 \):
   \[
   I_{R3} = I_S \times \frac{R_{12}}{R_{12} + R_3}
   \]
   Substituting the values:
   \[
   I_{R3} = 7.26 \, \text{mA} \times \frac{134}{134 + 3510} \approx 7.26 \, \text{mA} \times \frac{134}{3644} \approx 0.267 \, \text{mA}
   \]

So, the answer for **Question 1** is:
\[
\text{Answer: } 0.267 \, \text{mA}
\]

---

### Question 2
**Given Data:**
- \( R_1 = 320 \, \Omega \)
- \( R_2 = 2.46 \, \text{k}\Omega = 2460 \, \Omega \)
- \( R_3 = 3.39 \, \text{k}\Omega = 3390 \, \Omega \)
- \( I_S = 2.08 \, \text{mA} \)

We are asked to calculate the voltage across \( R_2 \).

**Solution:**
1. First, find the equivalent resistance of the parallel combination of \( R_1 \) and \( R_3 \):
   \[
   R_{13} = \frac{R_1 \cdot R_3}{R_1 + R_3}
   \]
   Plugging in the values:
   \[
   R_{13} = \frac{320 \times 3390}{320 + 3390} = \frac{1084800}{3710} \approx 292.5 \, \Omega
   \]

2. The current \( I_{R2} \) through \( R_2 \) is the same as the source current \( I_S \) since they are in series.
   \[
   I_{R2} = I_S = 2.08 \, \text{mA}
   \]

3. Now, calculate the voltage across \( R_2 \) using Ohm's law:
   \[
   V_{R2} = I_{R2} \times R_2
   \]
   Substituting the values:
   \[
   V_{R2} = 2.08 \, \text{mA} \times 2460 \, \Omega = 5.12 \, \text{V}
   \]

So, the answer for **Question 2** is:
\[
\text{Answer: } 5.12 \, \text{V}
\]

---

Do you need more details on these calculations, or is there another aspect you'd like to explore?

Here are some follow-up questions that may interest you:

1. How does the current divider rule work in circuits with more than two branches?
2. What changes in the solution if \( R_2 \) and \( R_3 \) were in series instead of parallel?
3. How would you calculate the power dissipated by \( R_3 \) in Question 1?
4. How does source current \( I_S \) affect the total voltage across the circuit?
5. What would happen if one of the resistors in parallel had a much larger resistance compared to the others?

**Tip:** Remember, using engineering notation (like kilo-ohms or milli-amps) can simplify calculations and make values more manageable.

The measured values in the above circuit are: R1=140 ohm, R2=3.10 kilo ohm, R3=3.51 kilo ohm, and the source current IS is 7.26 mA. Use the current divider rule to calculate the current in R3. For question 2: R1=320 ohm, R2=2.46 kilo ohm, R3=3.39 kilo ohm, and the source current IS is 2.08 mA. Calculate the voltage across R2.

Learn how to calculate the current through R3 and the voltage across R2 using the current divider rule and Ohm's law in electric circuits. With detailed calculations, we analyze circuits with given resistances and source currents to find precise values.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation