https://file.aiquickdraw.com/mathbot/file/1726166845921bxmfsn6u.jpg

To solve for \( I_2 \) in the circuit provided, we can follow these steps:

### 1. Identify and Label the Circuit Components:
- The voltage source is \( V = 7.5 \text{ V} \).
- The resistors are as follows:
  - \( R_1 = 3 \Omega \)
  - \( R_2 = 15 \Omega \)
  - \( R_3 = 9 \Omega \)
  - \( R_4 = 5 \Omega \)
  - \( R_5 = 2 \Omega \)
  - \( R_6 = 10 \Omega \)
  - \( R_7 = 5 \Omega \)

### 2. Determine the Configuration of the Circuit:
The circuit consists of a combination of series and parallel resistors.

### 3. Simplify the Circuit:
- Start with the resistors in the middle branch: \( R_4 = 5 \Omega \) and \( R_5 = 2 \Omega \) are in series. Their combined resistance is:
  \[
  R_{45} = R_4 + R_5 = 5 \Omega + 2 \Omega = 7 \Omega
  \]
- Next, resistors \( R_3 = 9 \Omega \) and \( R_{45} = 7 \Omega \) are in parallel. Their combined resistance is:
  \[
  \frac{1}{R_{345}} = \frac{1}{R_3} + \frac{1}{R_{45}} = \frac{1}{9 \Omega} + \frac{1}{7 \Omega}
  \]
  Let's calculate this combined resistance.

### 4. Calculate \( R_{345} \):
\[
\frac{1}{R_{345}} = \frac{1}{9} + \frac{1}{7} = \frac{7 + 9}{63} = \frac{16}{63}
\]
\[
R_{345} = \frac{63}{16} \Omega \approx 3.94 \Omega
\]

### 5. Continue Simplifying:
Now, consider the upper branch where \( R_2 = 15 \Omega \) and \( R_{345} \approx 3.94 \Omega \) are in series:
\[
R_{2345} = R_2 + R_{345} = 15 \Omega + 3.94 \Omega \approx 18.94 \Omega
\]
Next, this \( R_{2345} \) is in parallel with \( R_6 = 10 \Omega \) and \( R_7 = 5 \Omega \).

Combine \( R_6 \) and \( R_7 \) first:
\[
\frac{1}{R_{67}} = \frac{1}{10} + \frac{1}{5} = \frac{1}{10} + \frac{2}{10} = \frac{3}{10} = 3.33 \Omega
\]
Now combine \( R_{2345} \approx 18.94 \Omega \) with \( R_{67} = 3.33 \Omega \):
\[
\frac{1}{R_{\text{total}}} = \frac{1}{18.94} + \frac{1}{3.33}
\]
Let's calculate this combined resistance.

### 6. Calculate the Total Resistance \( R_{\text{total}} \):
\[
\frac{1}{R_{\text{total}}} \approx \frac{1}{18.94} + \frac{1}{3.33} = 0.0527 + 0.3003 = 0.353 \text{ }\Omega^{-1}
\]
\[
R_{\text{total}} \approx \frac{1}{0.353} \approx 2.83 \Omega
\]

### 7. Calculate \( I_T \) (Total Current):
Using Ohm's Law:
\[
I_T = \frac{V}{R_{\text{total}}} = \frac{7.5 \text{ V}}{2.83 \Omega} \approx 2.65 \text{ A}
\]

### 8. Calculate \( I_2 \):
To calculate \( I_2 \), find the voltage across \( R_2 \) and then use Ohm's Law:
\[
V_{R_2} = I_T \times R_{2345} = 2.65 \times 18.94 \approx 50.2 \text{ V}
\]
Thus:
\[
I_2 = \frac{V_{R_2}}{R_2} \approx \frac{50.2}{15} \approx 3.35 \text{ A}
\]

### Conclusion:
So, the current \( I_2 \) is approximately \( 3.35 \text{ A} \).

Would you like to go into more details or do you have any questions?

---

**Relative Questions:**
1. How would you simplify a circuit with more complex resistor networks?
2. What role does Kirchhoff's Law play in solving this kind of circuit?
3. Can you explain how to identify series and parallel components in a circuit?
4. What happens to the current if the resistance in one branch of the circuit is increased?
5. How can Thevenin's theorem be applied to solve complex circuits?

**Tip:** When analyzing circuits, always start by simplifying series and parallel resistances to reduce the circuit to a simpler equivalent.

solve for I subscript 2 in the circuit provided

Solve for current I2 in a series-parallel resistor circuit with a 7.5V power source. This problem involves the application of Ohm's Law and the combination of series and parallel resistances. The total resistance and current values are calculated step by step, followed by the final determination of I2.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation