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The circuit provided appears to be a combination of series and parallel resistances. We have the following key information from the circuit:

- \( R_1 = 10 \, \Omega \)
- \( R_2 = 20 \, \Omega \)
- \( R_3 = 20 \, \Omega \)
- \( R_4 \) is unknown.
- \( V_3 = 20 \, V \) across \( R_3 \)
- The total voltage \( V_T = 120 \, V \)

### Steps to calculate \( R_4 \):

1. **Calculate the current through \( R_3 \)**:
   Using Ohm's law, the current through \( R_3 \) is given by:
   \[
   I_3 = \frac{V_3}{R_3} = \frac{20 \, V}{20 \, \Omega} = 1 \, A
   \]
   Since \( R_2 \) and \( R_3 \) are in parallel, the current through \( R_2 \) will also be 1 A (because they have the same resistance).

2. **Total current through the parallel combination of \( R_2 \) and \( R_3 \)**:
   \[
   I_{\text{parallel}} = I_2 + I_3 = 1 \, A + 1 \, A = 2 \, A
   \]

3. **Voltage across the parallel combination**:
   Since the total current through the parallel combination is 2 A and the equivalent resistance of the parallel branch is:
   \[
   R_{\text{parallel}} = \frac{R_2 R_3}{R_2 + R_3} = \frac{20 \times 20}{20 + 20} = 10 \, \Omega
   \]
   The voltage across the parallel combination is:
   \[
   V_{\text{parallel}} = I_{\text{parallel}} \times R_{\text{parallel}} = 2 \, A \times 10 \, \Omega = 20 \, V
   \]

4. **Voltage across \( R_1 \)**:
   The total voltage \( V_T = 120 \, V \), and the voltage across the parallel branch is 20 V. Therefore, the voltage across \( R_1 \) is:
   \[
   V_1 = V_T - V_{\text{parallel}} = 120 \, V - 20 \, V = 100 \, V
   \]

5. **Current through \( R_1 \)**:
   Using Ohm's law, the current through \( R_1 \) is:
   \[
   I_1 = \frac{V_1}{R_1} = \frac{100 \, V}{10 \, \Omega} = 10 \, A
   \]

6. **Current through \( R_4 \)**:
   The current entering the junction before \( R_2 \), \( R_3 \), and \( R_4 \) is 10 A. The current through the parallel branch is 2 A, so the current through \( R_4 \) is:
   \[
   I_4 = I_1 - I_{\text{parallel}} = 10 \, A - 2 \, A = 8 \, A
   \]

7. **Calculate \( R_4 \)**:
   Using Ohm's law for \( R_4 \), where the voltage across \( R_4 \) is \( V_{\text{parallel}} = 20 \, V \):
   \[
   R_4 = \frac{V_{\text{parallel}}}{I_4} = \frac{20 \, V}{8 \, A} = 2.5 \, \Omega
   \]

Thus, the value of \( R_4 \) is **2.5 Ω**.

Would you like further details or clarification?

Here are some related questions:
1. How would the circuit change if \( R_4 \) were increased?
2. How can we use Kirchhoff’s laws to solve this circuit?
3. What happens to the total current if we reduce \( R_2 \) and \( R_3 \)?
4. What is the impact of adding another resistor in series with \( R_1 \)?
5. How would you approach this problem using Thevenin’s theorem?

**Tip:** When dealing with parallel resistors, always calculate the equivalent resistance first to simplify the analysis.

Given the measured values of voltage in the following circuit, calculate the value of resistor R4.

Learn how to calculate the value of resistor R4 in a series-parallel circuit with a total voltage of 120V. Using Ohm's law and parallel resistance formulas, this step-by-step solution helps solve the problem effectively. Ideal for high school students.

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