https://file.aiquickdraw.com/mathbot/file/1728994613257ap287cwm.jpg

Let's solve the problem step by step.

### 1. **Understanding the circuit:**
   - \( V_T = 120 \, \text{V} \) is the total voltage applied across the entire circuit.
   - There are four resistors:
     - \( R_1 = 30 \, \Omega \)
     - \( R_2 = 120 \, \Omega \)
     - \( R_3 = 60 \, \Omega \)
     - \( R_4 = 40 \, \Omega \)
   
   The circuit is a combination of series and parallel connections.

### 2. **Simplify the parallel combinations:**

   - **Resistors \( R_3 \) and \( R_4 \)** are in parallel:
     \[
     \frac{1}{R_{34}} = \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{60} + \frac{1}{40}
     \]
     First, find the equivalent resistance of \( R_3 \) and \( R_4 \):
     \[
     \frac{1}{R_{34}} = \frac{1}{60} + \frac{1}{40} = \frac{2}{120} + \frac{3}{120} = \frac{5}{120}
     \]
     \[
     R_{34} = \frac{120}{5} = 24 \, \Omega
     \]

### 3. **Equivalent resistance of the entire network:**

   Now, \( R_1 \), \( R_2 \), and \( R_{34} \) are in series. The total equivalent resistance \( R_{\text{eq}} \) is:
   \[
   R_{\text{eq}} = R_1 + R_2 + R_{34} = 30 \, \Omega + 120 \, \Omega + 24 \, \Omega = 174 \, \Omega
   \]

### 4. **Find the total current:**

   Using Ohm's Law, the total current \( I_T \) is:
   \[
   I_T = \frac{V_T}{R_{\text{eq}}} = \frac{120 \, \text{V}}{174 \, \Omega} \approx 0.6897 \, \text{A}
   \]

### 5. **Voltage across each resistor:**

   - For resistors in series, the current is the same through all resistors. Now, calculate the voltage drop across each resistor.
   
   - **Voltage across \( R_1 \)**:
     \[
     V_1 = I_T \times R_1 = 0.6897 \, \text{A} \times 30 \, \Omega \approx 20.69 \, \text{V}
     \]

   - **Voltage across \( R_2 \)**:
     \[
     V_2 = I_T \times R_2 = 0.6897 \, \text{A} \times 120 \, \Omega \approx 82.76 \, \text{V}
     \]

   - **Voltage across \( R_{34} \)**:
     \[
     V_{34} = I_T \times R_{34} = 0.6897 \, \text{A} \times 24 \, \Omega \approx 16.55 \, \text{V}
     \]

### 6. **Current and voltage in \( R_3 \) and \( R_4 \):**

   Since \( R_3 \) and \( R_4 \) are in parallel, they have the same voltage, which is \( V_{34} \).

   - **Current through \( R_3 \)**:
     \[
     I_3 = \frac{V_{34}}{R_3} = \frac{16.55 \, \text{V}}{60 \, \Omega} \approx 0.276 \, \text{A}
     \]

   - **Current through \( R_4 \)**:
     \[
     I_4 = \frac{V_{34}}{R_4} = \frac{16.55 \, \text{V}}{40 \, \Omega} \approx 0.414 \, \text{A}
     \]

Thus, the solution is:

- Equivalent resistance of the circuit: \( 174 \, \Omega \)
- Total current: \( 0.6897 \, \text{A} \)
- Voltage and current across each resistor:
  - \( R_1 \): \( V_1 = 20.69 \, \text{V}, I_1 = 0.6897 \, \text{A} \)
  - \( R_2 \): \( V_2 = 82.76 \, \text{V}, I_2 = 0.6897 \, \text{A} \)
  - \( R_3 \): \( V_3 = 16.55 \, \text{V}, I_3 = 0.276 \, \text{A} \)
  - \( R_4 \): \( V_4 = 16.55 \, \text{V}, I_4 = 0.414 \, \text{A} \)

Let me know if you'd like further details or explanations!

### Follow-up questions:
1. What changes in the circuit's current if \( R_1 \) increases?
2. How would adding another resistor in parallel with \( R_3 \) affect the total resistance?
3. What happens to the total current if the total voltage is reduced to 60 V?
4. How does the voltage distribution change if \( R_2 \) is replaced with a \( 60 \, \Omega \) resistor?
5. What would be the effect on power consumption if \( R_4 \) were replaced with a \( 20 \, \Omega \) resistor?

**Tip:** Remember, in parallel circuits, the voltage is the same across all components, but the current divides according to the resistance of each branch!

Compute the equivalent resistance of the network and find the current and voltage in each resistor.

Learn how to calculate the equivalent resistance, current, and voltage for a circuit containing resistors in both series and parallel. This problem involves Ohm's law and fundamental circuit theory. Detailed step-by-step solution for Grades 10-12.

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