Consider the circuit given in the drawing. Use Kirchhoff's Rules to determine the current labeled through the Ry=1.00 Q resistor 7V '5V R₁-10 R₂-30 www R₁-202

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### Example Circuit Analysis Using Kirchhoff's Rules

**Problem Statement:**
Consider the circuit given in the drawing. Use Kirchhoff's Rules to determine the current labeled through the \( R_3 = 1.00 \, \Omega \) resistor.

**Circuit Diagram:**
The diagram consists of the following components:
- A \( 7 \, V \) battery
- A \( R_1 = 2 \, \Omega \) resistor
- A \( 5 \, V \) battery
- A \( R_2 = 3 \, \Omega \) resistor
- A \( R_3 = 1 \, \Omega \) resistor

The circuit has two distinct loops:
- Loop 1: Includes the 7V battery, \( R_1 \), \( R_3 \), and the 5V battery.
- Loop 2: Includes \( R_3 \), the 5V battery, and \( R_2 \).

**Question:**
Choose the correct current value through the \( R_3 \) resistor:
- (a) 1.00 A
- (b) 5.00 A
- (c) 12.0 A
- (d) 3.00 A
- (e) 7.00 A

**Explanation:**

To analyze the circuit and find the current through \( R_3 \), we use Kirchhoff's rules:

- **Kirchhoff's Voltage Law (KVL):** The sum of all electrical potential differences around any closed network is zero.
- **Kirchhoff's Current Law (KCL):** The sum of currents entering a junction must equal the sum of currents leaving the junction.

Let's denote the current through \( R_1 \) as \( I_1 \), through \( R_2 \) as \( I_2 \), and through \( R_3 \) as \( I_3 \).

Note that the sum of currents entering and leaving any junction must be zero according to KCL.

By applying KVL in both loops:

1. For Loop 1:
\[ 7V - I_1 \times 2\Omega - I_3 \times 1\Omega - 5V = 0 \]
Simplifying:
\[ 2V = 2I_1 + I_3 \]

2. For Loop 2:
\[ 5V -
Transcribed Image Text:### Example Circuit Analysis Using Kirchhoff's Rules **Problem Statement:** Consider the circuit given in the drawing. Use Kirchhoff's Rules to determine the current labeled through the \( R_3 = 1.00 \, \Omega \) resistor. **Circuit Diagram:** The diagram consists of the following components: - A \( 7 \, V \) battery - A \( R_1 = 2 \, \Omega \) resistor - A \( 5 \, V \) battery - A \( R_2 = 3 \, \Omega \) resistor - A \( R_3 = 1 \, \Omega \) resistor The circuit has two distinct loops: - Loop 1: Includes the 7V battery, \( R_1 \), \( R_3 \), and the 5V battery. - Loop 2: Includes \( R_3 \), the 5V battery, and \( R_2 \). **Question:** Choose the correct current value through the \( R_3 \) resistor: - (a) 1.00 A - (b) 5.00 A - (c) 12.0 A - (d) 3.00 A - (e) 7.00 A **Explanation:** To analyze the circuit and find the current through \( R_3 \), we use Kirchhoff's rules: - **Kirchhoff's Voltage Law (KVL):** The sum of all electrical potential differences around any closed network is zero. - **Kirchhoff's Current Law (KCL):** The sum of currents entering a junction must equal the sum of currents leaving the junction. Let's denote the current through \( R_1 \) as \( I_1 \), through \( R_2 \) as \( I_2 \), and through \( R_3 \) as \( I_3 \). Note that the sum of currents entering and leaving any junction must be zero according to KCL. By applying KVL in both loops: 1. For Loop 1: \[ 7V - I_1 \times 2\Omega - I_3 \times 1\Omega - 5V = 0 \] Simplifying: \[ 2V = 2I_1 + I_3 \] 2. For Loop 2: \[ 5V -
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