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Calculate the Rx ﻿value necessary to dissipated 1 ﻿Watt power, where R1= 6 ohms

To find the value of \( R_x \) necessary to dissipate 1 Watt of power, we can start by analyzing the given circuit. The circuit includes two resistors, \( R_1 \) and \( R_x \), connected in series with a 12 V voltage source. Here, \( R_1 \) is given as 6 ohms.

### Step-by-step Solution:

1. **Total Voltage and Resistance Relation**:
   The total voltage across the series combination is 12 V.

2. **Power Dissipated by \( R_x \)**:
   The power dissipated by \( R_x \) is given as \( P = 1 \) Watt.

   \[
   P = \frac{V_x^2}{R_x}
   \]
   Here, \( V_x \) is the voltage across \( R_x \).

3. **Finding \( V_x \)**:
   Using the voltage divider rule, the voltage across \( R_x \) can be expressed as:
   
   \[
   V_x = V \frac{R_x}{R_1 + R_x}
   \]
   where \( V = 12 \) V.

4. **Substitute and Solve for \( R_x \)**:
   We substitute the expression for \( V_x \) into the power formula and set it equal to 1 Watt:
   
   \[
   1 = \frac{\left(12 \frac{R_x}{R_1 + R_x}\right)^2}{R_x}
   \]

   Simplifying and solving this equation will yield \( R_x \). First, substitute \( R_1 = 6 \) ohms into the equation and then solve:

   \[
   1 = \frac{(12 \frac{R_x}{6 + R_x})^2}{R_x}
   \]

   \[
   1 = \frac{144 R_x^2}{(6 + R_x)^2 R_x}
   \]

   \[
   (6 + R_x)^2 = 144 R_x
   \]

   \[
   36 + 12R_x + R_x^2 = 144 R_x
   \]

   \[
   R_x^2 - 132R_x + 36 = 0
   \]

   Using the quadratic formula to solve this:

   \[
   R_x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]

   Where \( a = 1, b = -132, c = 36 \).

   \[
   R_x = \frac{132 \pm \sqrt{132^2 - 4 \times 1 \times 36}}{2}
   \]

Let's calculate the exact values for \( R_x \) from the quadratic formula.The values for \( R_x \) that will allow 1 Watt of power to be dissipated are approximately \( 131.73 \) ohms and \( 0.27 \) ohms.

Do you want details on why there are two valid solutions or have any other questions about the process?

**Related Questions:**
1. What would be the total resistance of the circuit for each \( R_x \) value found?
2. How would the current flowing through the circuit change if \( R_x \) is set to \( 131.73 \) ohms versus \( 0.27 \) ohms?
3. What would be the power dissipation in \( R_1 \) for each case?
4. How would the voltage across each resistor change if the supply voltage were increased or decreased?
5. What safety precautions should be considered when designing circuits to handle specific power dissipations?

**Tip:** Always double-check the units in your calculations to ensure consistency across all variables.

Calculate the Rx value necessary to dissipate 1 Watt power, where R1 = 6 ohms.

This problem involves calculating the value of Rx in a simple series circuit with a 12V voltage source and R1 = 6 ohms, such that Rx dissipates 1 Watt of power. The solution uses the power formula, voltage divider rule, and involves solving a quadratic equation. Suitable for Grades 9-12.

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