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In the image, the problem is asking to determine the voltage of the battery for the circuit shown, which consists of a battery and six resistors. Here is the relevant information provided:

- \( R_1 = 7.00 \, \Omega \)
- \( R_2 = 3.00 \, \Omega \)
- \( R_3 = 2.00 \, \Omega \)
- \( R_4 = 15.00 \, \Omega \)
- \( R_5 = 4.00 \, \Omega \)
- Current through resistor \( R_1 \), \( I_1 = 2.71 \, A \)

The goal is to find the voltage of the battery \( V \).

### Solution Steps:

1. **Find the total resistance of the circuit**:
   - First, the resistors \( R_3 \), \( R_4 \), and \( R_5 \) are connected in parallel.
   - Then, the combination of these parallel resistors is in series with \( R_2 \), and this result is in parallel with \( R_1 \).
   
2. **Parallel resistors**:
   For resistors in parallel, the equivalent resistance \( R_{\text{eq}} \) is given by:
   \[
   \frac{1}{R_{\text{eq}}} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}
   \]
   Substituting values:
   \[
   \frac{1}{R_{\text{eq}}} = \frac{1}{2.00} + \frac{1}{15.00} + \frac{1}{4.00}
   \]

3. **Equivalent resistance of the circuit**:
  

Determine the voltage of the battery for the circuit shown, consisting of a battery and six resistors with the following values: R1 = 7.00 Ω, R2 = 3.00 Ω, R3 = 2.00 Ω, R4 = 15.00 Ω, R5 = 4.00 Ω. The current flowing through resistor R1 is I1 = 2.71 A.

Learn how to find the voltage of a battery in a circuit with six resistors, involving resistors in both parallel and series configurations, and using Ohm's Law to solve for the unknown. Suitable for high school physics students (Grades 9-12).

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