Problem 73 - College Algebra and Calculus: An Applied Approach

Question

When two resistors of resistances \(R_{1}\) and \(R_{2}\) are connected in parallel (see figure), the total resistance \(R\) satisfies the equation \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\) Find \(R_{1}\) for a parallel circuit in which \(R_{2}=2\) ohms and \(R\) must be at least 1 ohm.

Step-by-Step Solution

Answer

The resistance \(R_{1}\) for a parallel circuit given the condition that \(R_{2}=2\) ohms and \(R\) is at least 1 ohm is 2 ohms.

Step 1: Rearranging the equation to solve for \(R_{1}\)

Start by rearranging the given equation. The aim is to make \(R_{1}\) the subject by getting it on one side of the equation and having all other variables on the other side of the equation. \n\nLet's begin:\n\n - Start from the equation \(\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}\)\n\n - Subtract \(\frac{1}{R_{2}}\) from both sides to isolate \(\frac{1}{R_{1}}\) on one side, yielding: \(\frac{1}{R_{1}} = \frac{1}{R} - \frac{1}{R_{2}}\)\n\n - Lastly, taking reciprocal of each side will lead to \(R_{1} = \frac{1}{\frac{1}{R} - \frac{1}{R_{2}}}\).

Step 2: Substituting given values

Now that we have rearranged the equation in terms of \(R_{1}\), we can substitute the provided values into our equation. We use \(R=1\) and \(R_{2}=2\) ohm. After inserting these values, the equation for \(R_{1}\) becomes \n\n \(R_{1} = \frac{1}{\frac{1}{1} - \frac{1}{2}}\)

Step 3: Solve for \(R_{1}\)

Perform mathematical operations on the right-hand side of the equation to solve for \(R_{1}\).\n\nThus,\n\n\(R_{1} = \frac{1}{1 - 0.5} = 2\) ohms.