Two resistance $R_1$ and $R_2$ give combined resistance of 4.5 ohms when in series and 1 ohm when in parallel. The resistances are
- A. 3 ohms and 6 ohms
- B. 3 ohms and 9 ohms
- C. 1.5 ohms and 3 ohms
- D. 1.5 ohms and 0.5 ohms
Answer: C. 1.5 ohms and 3 ohms
Explanation: $R_1+R_2=4.5$ and $\frac{R_1 R_2}{R_1+R_2}=1 \Rightarrow R_1 R_2=4.5$. Using the identity $(R_1-R_2)^2=(R_1+R_2)^2-4R_1 R_2 = (4.5)^2-4(4.5) = 20.25-18 = 2.25$. So, $R_1-R_2=1.5$. Solving $R_1+R_2=4.5$ and $R_1-R_2=1.5$ gives $R_1=3~\Omega$ and $R_2=1.5~\Omega$.