Answer :
Certainly! Let's break down the problem step-by-step, using the given information and intermediate results:
### Step 1: Understanding the Given Resistances
We have two resistances in parallel:
1. Resistance [tex]\( R_1 = \frac{1}{6} \)[/tex] ohms.
2. Resistance [tex]\( R_2 = \frac{1}{3\pi} \)[/tex] ohms.
### Step 2: Calculating the Total Equivalent Resistance in Parallel
For resistances in parallel, the reciprocal of the total resistance [tex]\( R_T \)[/tex] can be calculated as the sum of the reciprocals of the individual resistances:
[tex]\[ \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} \][/tex]
Substituting [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex] into the equation:
[tex]\[ \frac{1}{R_T} = \frac{1}{\frac{1}{6}} + \frac{1}{\frac{1}{3\pi}} = 6 + 3\pi \][/tex]
### Step 3: Simplifying the Calculation
The reciprocal of the total resistance is calculated by adding:
[tex]\[ \frac{1}{R_T} = 0.16666666666666666 + 0.1061032953945969 = 0.2727699620612636 \][/tex]
Thus, the total resistance [tex]\( R_T \)[/tex] is:
[tex]\[ R_T = \frac{1}{0.2727699620612636} = 3.6660928221099436 \, \text{ohms} \][/tex]
### Step 4: Intermediate Calculation Steps
Let's go through some intermediate calculations mentioned in the text:
- Simplified calculation for checking consistency:
[tex]\[ \frac{3+6}{9} = \frac{9}{9} = 1 \][/tex]
- Another simplification:
[tex]\[ 1 + \frac{2}{6} = 1 + \frac{1}{3} = 1.3333333333333333 \][/tex]
- Another intermediate result given:
[tex]\[ 0.5 \][/tex]
- Reciprocal check:
[tex]\[ \frac{3}{6} = 0.5 \][/tex]
- Calculating [tex]\( R_i \)[/tex]:
[tex]\[ \frac{6}{3} = 2 \][/tex]
### Summary of Results
Thus, the detailed results for the resistances and intermediate calculations are:
1. [tex]\( \text{Resistance 1 (R1)} = 0.16666666666666666 \, \text{ohms} \)[/tex]
2. [tex]\( \text{Resistance 2 (R2)} = 0.1061032953945969 \, \text{ohms} \)[/tex]
3. [tex]\( \text{Reciprocal of Total Resistance (1/RT)} = 0.2727699620612636 \)[/tex]
4. [tex]\( \text{Total Resistance (RT)} = 3.6660928221099436 \, \text{ohms} \)[/tex]
5. [tex]\( \frac{3 + 6}{9} = 1 \)[/tex]
6. [tex]\( 1 + \frac{2}{6} = 1.3333333333333333 \)[/tex]
7. [tex]\( 0.5 \)[/tex]
8. [tex]\( \frac{3}{6} = 0.5 \)[/tex]
9. [tex]\( \frac{6}{3} = 2 \)[/tex]
These steps illustrate a structured approach to breaking down the problem and confirming the given solution through intermediate calculations.
### Step 1: Understanding the Given Resistances
We have two resistances in parallel:
1. Resistance [tex]\( R_1 = \frac{1}{6} \)[/tex] ohms.
2. Resistance [tex]\( R_2 = \frac{1}{3\pi} \)[/tex] ohms.
### Step 2: Calculating the Total Equivalent Resistance in Parallel
For resistances in parallel, the reciprocal of the total resistance [tex]\( R_T \)[/tex] can be calculated as the sum of the reciprocals of the individual resistances:
[tex]\[ \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} \][/tex]
Substituting [tex]\( R_1 \)[/tex] and [tex]\( R_2 \)[/tex] into the equation:
[tex]\[ \frac{1}{R_T} = \frac{1}{\frac{1}{6}} + \frac{1}{\frac{1}{3\pi}} = 6 + 3\pi \][/tex]
### Step 3: Simplifying the Calculation
The reciprocal of the total resistance is calculated by adding:
[tex]\[ \frac{1}{R_T} = 0.16666666666666666 + 0.1061032953945969 = 0.2727699620612636 \][/tex]
Thus, the total resistance [tex]\( R_T \)[/tex] is:
[tex]\[ R_T = \frac{1}{0.2727699620612636} = 3.6660928221099436 \, \text{ohms} \][/tex]
### Step 4: Intermediate Calculation Steps
Let's go through some intermediate calculations mentioned in the text:
- Simplified calculation for checking consistency:
[tex]\[ \frac{3+6}{9} = \frac{9}{9} = 1 \][/tex]
- Another simplification:
[tex]\[ 1 + \frac{2}{6} = 1 + \frac{1}{3} = 1.3333333333333333 \][/tex]
- Another intermediate result given:
[tex]\[ 0.5 \][/tex]
- Reciprocal check:
[tex]\[ \frac{3}{6} = 0.5 \][/tex]
- Calculating [tex]\( R_i \)[/tex]:
[tex]\[ \frac{6}{3} = 2 \][/tex]
### Summary of Results
Thus, the detailed results for the resistances and intermediate calculations are:
1. [tex]\( \text{Resistance 1 (R1)} = 0.16666666666666666 \, \text{ohms} \)[/tex]
2. [tex]\( \text{Resistance 2 (R2)} = 0.1061032953945969 \, \text{ohms} \)[/tex]
3. [tex]\( \text{Reciprocal of Total Resistance (1/RT)} = 0.2727699620612636 \)[/tex]
4. [tex]\( \text{Total Resistance (RT)} = 3.6660928221099436 \, \text{ohms} \)[/tex]
5. [tex]\( \frac{3 + 6}{9} = 1 \)[/tex]
6. [tex]\( 1 + \frac{2}{6} = 1.3333333333333333 \)[/tex]
7. [tex]\( 0.5 \)[/tex]
8. [tex]\( \frac{3}{6} = 0.5 \)[/tex]
9. [tex]\( \frac{6}{3} = 2 \)[/tex]
These steps illustrate a structured approach to breaking down the problem and confirming the given solution through intermediate calculations.