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.