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

We are given the values \( R_1 = 1 \) and \( R_2 = \frac{4}{\pi} \). The formula we need to use is:

\[
R_{\text{eq}} = \frac{R_1 \cdot R_2}{R_1 + R_2}
\]

Let's calculate this step by step.

### Step 1: Multiply \( R_1 \) and \( R_2 \)

\[
R_1 \cdot R_2 = 1 \cdot \frac{4}{\pi} = \frac{4}{\pi}
\]

### Step 2: Add \( R_1 \) and \( R_2 \)

\[
R_1 + R_2 = 1 + \frac{4}{\pi}
\]

### Step 3: Find the equivalent resistance

Now we substitute these into the formula:

\[
R_{\text{eq}} = \frac{\frac{4}{\pi}}{1 + \frac{4}{\pi}}
\]

### Step 4: Simplify the denominator

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

\[
1 + \frac{4}{\pi} = \frac{\pi}{\pi} + \frac{4}{\pi} = \frac{\pi + 4}{\pi}
\]

### Step 5: Final Simplification

Now substitute this into the equation for \( R_{\text{eq}} \):

\[
R_{\text{eq}} = \frac{\frac{4}{\pi}}{\frac{\pi + 4}{\pi}} = \frac{4}{\pi + 4}
\]

Thus, the equivalent resistance is:

\[
R_{\text{eq}} = \frac{4}{\pi + 4}
\]

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 \( R_1 = R_2 \)? How does it simplify?
2. How would the result change if \( R_1 \) and \( R_2 \) 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.

Learn how to calculate the equivalent resistance for two resistors in parallel, where R1 = 1 and R2 = 4/π, using the law of parallel resistance. Follow a step-by-step solution that simplifies the expression to a final answer. Suitable for Grades 9-12 students studying algebra and electrical circuits.

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