Question

Two resistance R₁ and R₂ provides series to parallel equivalents as $\frac{n}{1}$ then the correct relationship is

• A. $\left( \frac{R_{1}}{R_{2}} \right)^{2} + \left( \frac{R_{2}}{R_{1}} \right)^{2} = n^{2}$
• B. $\left( \frac{R_{1}}{R_{2}} \right)^{3/2} + \left( \frac{R_{2}}{R_{1}} \right)^{3/2} = n^{3/2}$
• C. $\left( \frac{R_{1}}{R_{2}} \right) + \left( \frac{R_{2}}{R_{1}} \right) = n$
• D. $\left( \frac{R_{1}}{R_{2}} \right)^{1/2} + \left( \frac{R_{2}}{R_{1}} \right)^{1/2} = n^{1/2}$

Answer 1

Answer: (D)

$\left( \frac{R_{1}}{R_{2}} \right)^{1/2} + \left( \frac{R_{2}}{R_{1}} \right)^{1/2} = n^{1/2}$

Answer 2

Series resistance $R_{S} = R_{1} + R_{2}$ and parallel resistance $R_{P} = \frac{R_{1}R_{2}}{R_{1} + R_{2}}$ ⇒ $\frac{R_{S}}{R_{P}} = \frac{(R_{1} + R_{2})^{2}}{R_{1}R_{2}} = n$

⇒ $\frac{R_{1} + R_{2}}{\sqrt{R_{1}R_{2}}} = \sqrt{n}$

⇒ $\frac{\sqrt{R_{1}^{2}}}{\sqrt{R_{1}R_{2}}} + \frac{\sqrt{R_{2}^{2}}}{\sqrt{R_{1}R_{2}}} = \sqrt{n}$

⇒ $\sqrt{\frac{R_{1}}{R_{2}}} + \sqrt{\frac{R_{2}}{R_{1}}} = \sqrt{n}$