https://r2.erweima.ai/mathbot/image/1725766926718phzbhs42.jpg

The problem involves a circuit composed of a battery and resistors with the following resistances:

- \( R_1 = 70.0 \, \Omega \)
- \( R_2 = 80.0 \, \Omega \)
- \( R_3 = 90.0 \, \Omega \)
- \( R_4 = 100.0 \, \Omega \)
- \( R_5 = 120.0 \, \Omega \)

The current flowing through resistor \( R_3 \) is given as \( I_3 = 1.85 \, \text{A} \), and we are tasked with determining the voltage of the battery \( E \).

### Step-by-Step Solution

#### 1. Analyze the Circuit

From the image, we can observe that the resistors are arranged in a mixed series-parallel configuration.

- Resistors \( R_1 \) and \( R_2 \) are in parallel.
- This combination is in series with \( R_3 \), and that branch is in parallel with the combination of \( R_4 \) and \( R_5 \).

#### 2. Use Ohm’s Law
Ohm’s Law is the foundation of solving for voltage or current in the circuit:

\[
V = IR
\]

We are given the current through \( R_3 \) as \( I_3 = 1.85 \, \text{A} \), and we know its resistance is \( R_3 = 90.0 \, \Omega \). Therefore, the voltage across \( R_3 \), \( V_3 \), can be calculated:

\[
V_3 = I_3 \cdot R_3 = 1.85 \, \text{A} \cdot 90.0 \, \Omega = 166.5 \, \text{V}
\]

This is the voltage across the parallel combination of \( R_1 \) and \( R_2 \) and also across the combination of \( R_4 \) and \( R_5 \).

#### 3. Find the Equivalent Resistance of \( R_1 \) and \( R_2 \) in Parallel

For resistors in parallel, the equivalent resistance \( R_{12} \) is given by:

\[
\frac{1}{R_{12}} = \frac{1}{R_1} + \frac{1}{R_2}
\]

Substitute the values:

\[
\frac{1}{R_{12}} = \frac{1}{70.0 \, \Omega} + \frac{1}{80.0 \, \Omega}
\]

Calculating this:

\[
R_{12} = \frac{1}{\left(\frac{1}{70} + \frac{1}{80}\right)} \approx 37.33 \, \Omega
\]

#### 4. Find the Equivalent Resistance of \( R_4 \) and \( R_5 \) in Parallel

Similarly, for \( R_4 \) and \( R_5 \):

\[
\frac{1}{R_{45}} = \frac{1}{R_4} + \frac{1}{R_5} = \frac{1}{100.0 \, \Omega} + \frac{1}{120.0 \, \Omega}
\]

Calculating this:

\[
R_{45} = \frac{1}{\left(\frac{1}{100} + \frac{1}{120}\right)} \approx 54.55 \, \Omega
\]

#### 5. Total Equivalent Resistance

Now that we have the equivalent resistances for the two parallel branches, the total resistance in the circuit is:

\[
R_{\text{total}} = R_3 + R_{12} \parallel R_{45}
\]

Since \( R_{12} \) and \( R_{45} \) are in parallel, we calculate:

\[
\frac{1}{R_{\parallel}} = \frac{1}{R_{12}} + \frac{1}{R_{45}} = \frac{1}{37.33 \, \Omega} + \frac{1}{54.55 \, \Omega}
\]

Calculating \( R_{\parallel} \):

\[
R_{\parallel} \approx 22.2 \, \Omega
\]

The total resistance is then:

\[
R_{\text{total}} = R_3 + R_{\parallel} = 90.0 \, \Omega + 22.2 \, \Omega = 112.2 \, \Omega
\]

#### 6. Find the Voltage of the Battery

Now that we have the total resistance, the total current in the circuit can be found using the voltage across \( R_3 \) and the fact that it is the same for the entire circuit.

Using Ohm’s law again:

\[
E = I_{\text{total}} \cdot R_{\text{total}}
\]

Since we know the current through \( R_3 \) is \( 1.85 \, \text{A} \), and the same current flows through the total resistance \( R_{\text{total}} \), we can find the battery voltage:

\[
E = 1.85 \, \text{A} \times 112.2 \, \Omega = 207.57 \, \text{V}
\]

Thus, the voltage of the battery is approximately \( E = 207.57 \, \text{V} \).

---

### Do you want further details or have any questions?

Here are some related questions you might find helpful:

1. What is the importance of parallel and series resistances in circuits?
2. How can we calculate current in a circuit with mixed series-parallel combinations?
3. What is the difference between total and equivalent resistance?
4. How does Ohm's law help in solving complex circuits?
5. What happens to the total current when adding more resistors in parallel?

**Tip:** When analyzing circuits, always break them into smaller parts, solving for equivalent resistances step by step.

In the figure, a circuit consists of a battery and five resistors with values R1 = 70.0 Ω, R2 = 80.0 Ω, R3 = 90.0 Ω, R4 = 100.0 Ω, and R5 = 120.0 Ω. The current flowing through R3 is 1.85 A. Determine the voltage of the battery.

Learn how to calculate the voltage of a battery in a circuit composed of both series and parallel resistors. Given resistances and current, the solution uses Ohm's Law and equivalent resistance formulas. This problem is suitable for college-level students and advanced high school learners.

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