For the circuit below. Hint, current in inversely proportional to resistance. Ix/IT=RT/Rx, see "Current Divider Rule" in your notes. For this problem. Determine the equivalent resistance in each set of branches, for the I1 branch it would be 1/RE=1/8 +1/8. Split the total (6A) current between I1 and I2 using the current divider rule. Then apply the current divider rule one more time to split I1 between I3 and the other 8 Ω resistor. Determine I1. Determine I2. Determine I3. Determine I4.
For the circuit below. Hint, current in inversely proportional to resistance. Ix/IT=RT/Rx, see "Current Divider Rule" in your notes. For this problem. Determine the equivalent resistance in each set of branches, for the I1 branch it would be 1/RE=1/8 +1/8. Split the total (6A) current between I1 and I2 using the current divider rule. Then apply the current divider rule one more time to split I1 between I3 and the other 8 Ω resistor. Determine I1. Determine I2. Determine I3. Determine I4.
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Chapter1: Units, Trigonometry. And Vectors
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For the circuit below. Hint, current in inversely proportional to resistance. Ix/IT=RT/Rx, see "Current Divider Rule" in your notes. For this problem. Determine the equivalent resistance in each set of branches, for the I1 branch it would be 1/RE=1/8 +1/8. Split the total (6A) current between I1 and I2 using the current divider rule. Then apply the current divider rule one more time to split I1 between I3 and the other 8 Ω resistor.
Determine I1.
Determine I2.
Determine I3.
Determine I4.
![**Parallel Resistor Circuit Analysis**
The given circuit diagram is a parallel resistor circuit with three distinct branches. The total current entering the parallel branches is 6 A. The detailed description of each component and the analysis is provided below:
**Branch 1:**
- **Resistors:** Two (2) resistors of 8 Ω each are connected in parallel.
- **Current Distribution:** Let the current through this branch be \(I_1\).
**Branch 2:**
- **Resistors:** Three (3) resistors of 6 Ω each are connected in parallel.
- **Current Distribution:** Let the current through this branch be \(I_2\).
**Branch 3:**
- **Resistors:** Single resistor of 6 Ω.
- **Current Distribution:** Let the current through this branch be \(I_3\).
**Current Breakdown:**
- The total current (I) entering the parallel branches is 6 A.
**Detailed Diagrams:**
1. **Current Flow:**
- The current splits into \(I_1\), \(I_2\), and \(I_3\) at the junction entering the parallel resistors.
**Calculations:**
To find the individual currents \(I_1\), \(I_2\), and \(I_3\):
1. **Effective resistance of two 8 Ω resistors in parallel (Branch 1):**
\[ \frac{1}{R_{eq1}} = \frac{1}{8} + \frac{1}{8} = \frac{1}{4} \]
\[ R_{eq1} = 4 \, \Omega \]
2. **Effective resistance of three 6 Ω resistors in parallel (Branch 2):**
\[ \frac{1}{R_{eq2}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2} \]
\[ R_{eq2} = 2 \, \Omega \]
3. **Single resistor in Branch 3:**
\[ R_{eq3} = 6 \, \Omega \]
The total effective resistance \(R_t\) of the circuit is given by:
\[ \frac{1}{R_t} = \frac{1}{R_{eq1}} + \frac{1}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a82e125-75a3-4aec-82a1-d3c8331aea9d%2F33a9350e-1d8e-4029-8d68-8d0f710217f6%2Ftp1xdkn_processed.png&w=3840&q=75)
Transcribed Image Text:**Parallel Resistor Circuit Analysis**
The given circuit diagram is a parallel resistor circuit with three distinct branches. The total current entering the parallel branches is 6 A. The detailed description of each component and the analysis is provided below:
**Branch 1:**
- **Resistors:** Two (2) resistors of 8 Ω each are connected in parallel.
- **Current Distribution:** Let the current through this branch be \(I_1\).
**Branch 2:**
- **Resistors:** Three (3) resistors of 6 Ω each are connected in parallel.
- **Current Distribution:** Let the current through this branch be \(I_2\).
**Branch 3:**
- **Resistors:** Single resistor of 6 Ω.
- **Current Distribution:** Let the current through this branch be \(I_3\).
**Current Breakdown:**
- The total current (I) entering the parallel branches is 6 A.
**Detailed Diagrams:**
1. **Current Flow:**
- The current splits into \(I_1\), \(I_2\), and \(I_3\) at the junction entering the parallel resistors.
**Calculations:**
To find the individual currents \(I_1\), \(I_2\), and \(I_3\):
1. **Effective resistance of two 8 Ω resistors in parallel (Branch 1):**
\[ \frac{1}{R_{eq1}} = \frac{1}{8} + \frac{1}{8} = \frac{1}{4} \]
\[ R_{eq1} = 4 \, \Omega \]
2. **Effective resistance of three 6 Ω resistors in parallel (Branch 2):**
\[ \frac{1}{R_{eq2}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2} \]
\[ R_{eq2} = 2 \, \Omega \]
3. **Single resistor in Branch 3:**
\[ R_{eq3} = 6 \, \Omega \]
The total effective resistance \(R_t\) of the circuit is given by:
\[ \frac{1}{R_t} = \frac{1}{R_{eq1}} + \frac{1}{
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