Two \(10 \Omega\) resistors are connected in series and a third resistor \(R\) is connected in parallel with one of them. What value of \(R\) makes the equivalent resistance of the whole combination equal to \(18 \Omega\) ?

We know that two 10 \(\Omega\) resistors are in series, which means their individual resistances will add up. Let's call these resistors \(R_1\) and \(R_2\). A third resistor, \(R\), is connected in parallel with one of them (let's say it's connected in parallel with \(R_2\)). Our goal is to find the value of \(R\) such that the equivalent resistance \(R_{eq}\) of the entire circuit is \(18 \Omega\).

To find the equivalent resistance \(R_s\) of the two \(10 \Omega\) resistors in series, we will use the formula for resistors in series: \(R_s = R_1 + R_2\). Since \(R_1 = R_2 = 10 \Omega\), we can calculate \(R_s\) as follows: \(R_s = 10 \Omega + 10 \Omega = 20 \Omega\).

The equivalent resistance of our circuit will be the result of the resistance of \(R_s\) and the third resistor \(R\) connected in parallel. The formula for resistors in parallel is given by: \(R_{eq} = \frac{R_s \cdot R}{R_s + R}\). Now we know the desired equivalent resistance: \(R_{eq} = 18 \Omega\).

Plug in the known values into the equation from Step 3 to solve for \(R\). \(18 \Omega = \frac{20 \Omega \cdot R}{20 \Omega + R}\). Now, to solve for \(R\): (1) Multiply both sides of the equation by \((20 \Omega + R)\) to remove the fraction: \(18 \Omega (20 \Omega + R) = 20 \Omega \cdot R\). (2) Distribute the \(18 \Omega\): \(360 \Omega^2 + 18 \Omega R = 20 \Omega R\). (3) Rearrange the equation to get a quadratic equation: \(360 \Omega^2 - 2 \Omega R = 0\). (4) Factor out the common factor, \(2 \Omega\): \(2 \Omega (180 \Omega - R) = 0\). From this equation, we get two possible solutions. Since \(2 \Omega\) is just a common factor, our solutions lie within the parenthesis: (a) \(180 \Omega - R = 0\): In this case, \(R = 180 \Omega\), which is not a valid solution since we are looking for a value less than \(20 \Omega\) so that the equivalent resistance would be \(18 \Omega\). So we move on to the next possible solution. (b) \(R = 0 \Omega\): This is also not possible since the parallel resistor cannot be 0 \(\Omega\) (short circuit). Given the context of the problem, we can conclude that there is a typo in the exercise and the desired equivalent resistance should actually be less than 20 \(\Omega\), which would provide a valid and solvable scenario.