Resistor Combination Resistance Calculation

This problem involves finding the individual resistances of two resistors, let's call them \(R_A\) and \(R_B\), given their equivalent resistances in parallel and series combinations.

Parallel and Series Resistance Formulas

The formulas for equivalent resistance (\(R_{eq}\)) are:

• Series Combination: \(R_{eq} = R_A + R_B\)
• Parallel Combination: \(R_{eq} = \frac{R_A \times R_B}{R_A + R_B}\)

Setting Up the Equations

From the problem statement, we have:

• For series combination: \(R_A + R_B = 12\,\Omega\) (Equation 1)
• For parallel combination: \(\frac{R_A \times R_B}{R_A + R_B} = 3\,\Omega\) (Equation 2)

Solving for Individual Resistances

We can substitute Equation 1 into Equation 2:

\(\frac{R_A \times R_B}{12\,\Omega} = 3\,\Omega\)

Multiply both sides by \(12\,\Omega\) to find the product of the resistances:

\(R_A \times R_B = 3\,\Omega \times 12\,\Omega = 36\,\Omega^2\)

Now we have a system of two equations:

1. \(R_A + R_B = 12\,\Omega\)
2. \(R_A \times R_B = 36\,\Omega^2\)

Consider a quadratic equation whose roots are \(R_A\) and \(R_B\). The general form is \(x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0\). Using our values:

\(x^2 - (12\,\Omega)x + 36\,\Omega^2 = 0\)

This equation is a perfect square:

\((x - 6\,\Omega)^2 = 0\)

Solving for \(x\) gives \(x = 6\,\Omega\). This means both roots are \(6\,\Omega\). Therefore:

\(R_A = 6\,\Omega \quad \text{and} \quad R_B = 6\,\Omega\)

Verification

Let's check if these values satisfy the initial conditions:

• Series: \(R_A + R_B = 6\,\Omega + 6\,\Omega = 12\,\Omega\) (Correct)
• Parallel: \(\frac{R_A \times R_B}{R_A + R_B} = \frac{6\,\Omega \times 6\,\Omega}{6\,\Omega + 6\,\Omega} = \frac{36\,\Omega^2}{12\,\Omega} = 3\,\Omega\) (Correct)

The resistances of the two resistors A and B are \(6\,\Omega\) and \(6\,\Omega\), respectively.