Question

Two resistors have a series resistance of 513 ohms and a parallel resistance of 120 ohms. Determine the two resistors (ohms).

Answer 1

If we have that two resistors have a series resistance of 513 Ohms it means that if first resistor has resistance $R_{1}$ and second resistor has resistance $R_{2}$ then we will have an equation: $R_{1} + R_{2} = 513$. If we have that the same resistors have a parallel resistance of 120 Ohms then we will have an equation:

$$\frac {R _ {1} R _ {2}}{R _ {1} + R _ {2}} = 120.$$

So, we have next system of equation:

$$\left\{ \begin{array}{l} R _ {1} + R _ {2} = 513 \\ \frac {R _ {1} R _ {2}}{R _ {1} + R _ {2}} = 120 \end{array} \right.$$

And we will solve it:

$$\begin{array}{l} \left\{ \begin{array}{l} R _ {1} + R _ {2} = 513 \\ \frac {R _ {1} R _ {2}}{R _ {1} + R _ {2}} = 120 \end{array} \Rightarrow \left\{ \begin{array}{l} R _ {1} = 513 - R _ {2} \\ \frac {\left(513 - R _ {2}\right) R _ {2}}{513 - R _ {2} + R _ {2}} = 120 \end{array} \Rightarrow \right. \right. \\ \Rightarrow \left\{ \begin{array}{l} R _ {1} = 513 - R _ {2} \\ R _ {2} ^ {2} - 513 R _ {2} + 61560 = 0 \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} R _ {1} = 513 - R _ {2} \\ R _ {2} = \frac {513 \pm \sqrt {16929}}{2} \approx \frac {513 \pm 130}{2} = \{191.5, 321.5 \} \end{array} \right. \Rightarrow \\ \end{array}$$ $$\Rightarrow \left\{ \begin{array}{l} R _ {1} = 321.5 \\ R _ {2} = 191.5 \\ R _ {1} = 191.5 \\ R _ {2} = 321.5 \end{array} \right.$$

In total we can say that one resistor is nearly 191.5 Ohms and the other is 321.5 Ohms.

**Answer**: one resistor is 191.5 Ohms, another is 321.5 Ohms.