Three resistors, R1 = 2.55 Ω, R2 = 4.77 Ω, and R3 = 6.55 Ω are connected by ideal metal wires, as shown in the figure. If the voltage dropping through R1 is 5.51 V, what is the current flowing through R2 (in A)?
Three resistors, R1 = 2.55 Ω, R2 = 4.77 Ω, and R3 = 6.55 Ω are connected by ideal metal wires, as shown in the figure. If the voltage dropping through R1 is 5.51 V, what is the current flowing through R2 (in A)?
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Three resistors, R1 = 2.55 Ω, R2 = 4.77 Ω, and R3 = 6.55 Ω are connected by ideal metal wires, as shown in the figure. If the voltage dropping through R1 is 5.51 V, what is the current flowing through R2 (in A)?
![**Series-Parallel Resistor Circuit**
In the diagram above, we see a combination of resistors arranged in both series and parallel configurations. Specifically, this circuit contains three resistors: \(R_1\), \(R_2\), and \(R_3\).
### Explanation of the Diagram:
- **Point 'a' and Point 'b':** These are the two terminals between which the entire circuit is connected.
- **Resistor \(R_1\:** This resistor is connected in series with the parallel combination of \(R_2\) and \(R_3\).
- **Resistors \(R_2\) and \(R_3\):** These two resistors are connected in parallel with each other.
### Step-by-Step Analysis:
1. **Resistor \(R_1\):**
- This resistor can be found in the path from point ‘a’ to the parallel combination of \(R_2\) and \(R_3\). The current that flows through \(R_1\) is the same as the current entering the parallel network.
2. **Parallel Resistors \(R_2\) and \(R_3\):**
- These resistors are connected in such a way that the voltage across them is identical. The total current that enters the parallel combination splits into two parts: one part goes through \(R_2\) and the other part goes through \(R_3\).
### Calculating Equivalent Resistance:
To find the equivalent resistance of the entire circuit:
1. First, calculate the equivalent resistance of \(R_2\) and \(R_3\) in parallel:
\[
R_{parallel} = \left( \frac{1}{R_2} + \frac{1}{R_3} \right)^{-1}
\]
2. Next, add the resistance \(R_1\) in series to the equivalent resistance found in step 1:
\[
R_{total} = R_1 + R_{parallel}
\]
This basic analysis helps in understanding how resistors work in both series and parallel configurations and how to simplify complex resistor networks into simpler equivalent resistances, an essential concept in circuit analysis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb234898a-ca00-4740-ae3d-79576f672843%2F12d7a37d-632d-4f0d-b31e-0aa3f407bd84%2Fhheunh_processed.png&w=3840&q=75)
Transcribed Image Text:**Series-Parallel Resistor Circuit**
In the diagram above, we see a combination of resistors arranged in both series and parallel configurations. Specifically, this circuit contains three resistors: \(R_1\), \(R_2\), and \(R_3\).
### Explanation of the Diagram:
- **Point 'a' and Point 'b':** These are the two terminals between which the entire circuit is connected.
- **Resistor \(R_1\:** This resistor is connected in series with the parallel combination of \(R_2\) and \(R_3\).
- **Resistors \(R_2\) and \(R_3\):** These two resistors are connected in parallel with each other.
### Step-by-Step Analysis:
1. **Resistor \(R_1\):**
- This resistor can be found in the path from point ‘a’ to the parallel combination of \(R_2\) and \(R_3\). The current that flows through \(R_1\) is the same as the current entering the parallel network.
2. **Parallel Resistors \(R_2\) and \(R_3\):**
- These resistors are connected in such a way that the voltage across them is identical. The total current that enters the parallel combination splits into two parts: one part goes through \(R_2\) and the other part goes through \(R_3\).
### Calculating Equivalent Resistance:
To find the equivalent resistance of the entire circuit:
1. First, calculate the equivalent resistance of \(R_2\) and \(R_3\) in parallel:
\[
R_{parallel} = \left( \frac{1}{R_2} + \frac{1}{R_3} \right)^{-1}
\]
2. Next, add the resistance \(R_1\) in series to the equivalent resistance found in step 1:
\[
R_{total} = R_1 + R_{parallel}
\]
This basic analysis helps in understanding how resistors work in both series and parallel configurations and how to simplify complex resistor networks into simpler equivalent resistances, an essential concept in circuit analysis.
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