Math Problem Statement

R1=1. R2=4/∏. Us the law (R1*R2)/(R1+R2). With steps

Solution

We are given the values R1=1R_1 = 1 and R2=4πR_2 = \frac{4}{\pi}. The formula we need to use is:

Req=R1R2R1+R2R_{\text{eq}} = \frac{R_1 \cdot R_2}{R_1 + R_2}

Let's calculate this step by step.

Step 1: Multiply R1R_1 and R2R_2

R1R2=14π=4πR_1 \cdot R_2 = 1 \cdot \frac{4}{\pi} = \frac{4}{\pi}

Step 2: Add R1R_1 and R2R_2

R1+R2=1+4πR_1 + R_2 = 1 + \frac{4}{\pi}

Step 3: Find the equivalent resistance

Now we substitute these into the formula:

Req=4π1+4πR_{\text{eq}} = \frac{\frac{4}{\pi}}{1 + \frac{4}{\pi}}

Step 4: Simplify the denominator

To simplify 1+4π1 + \frac{4}{\pi}, express it as a single fraction:

1+4π=ππ+4π=π+4π1 + \frac{4}{\pi} = \frac{\pi}{\pi} + \frac{4}{\pi} = \frac{\pi + 4}{\pi}

Step 5: Final Simplification

Now substitute this into the equation for ReqR_{\text{eq}}:

Req=4ππ+4π=4π+4R_{\text{eq}} = \frac{\frac{4}{\pi}}{\frac{\pi + 4}{\pi}} = \frac{4}{\pi + 4}

Thus, the equivalent resistance is:

Req=4π+4R_{\text{eq}} = \frac{4}{\pi + 4}

This is the simplified form of the result.

Would you like me to provide more details or clarify any step?

Related Questions:

  1. What happens if R1=R2R_1 = R_2? How does it simplify?
  2. How would the result change if R1R_1 and R2R_2 are in series instead of parallel?
  3. Can this method be extended for three resistances?
  4. What if one of the resistances approaches zero?
  5. How does the formula change for resistances in complex circuits?

Tip:

In circuits, parallel resistances always result in a smaller overall resistance than the smallest individual resistor, while series resistances add up directly.

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Math Problem Analysis

Mathematical Concepts

Algebra
Electrical Circuits
Parallel Resistance

Formulas

Equivalent resistance in parallel: (R1 * R2) / (R1 + R2)

Theorems

Law of Parallel Resistance

Suitable Grade Level

Grades 9-12