Question

Given AC circuit can be purely resistive if

• A. Frequency w = 0
• B. Frequency $w=\sqrt{\frac{1}{LC}-\frac{R^2}{L^2}}$
• C. If $R>\frac{L}{C}$
• D. Given circuit can never be purely resistive.

Answer 1

Answer: (A), (B)

The correct options are
A Frequency w = 0
B Frequency $w=\sqrt{\frac{1}{LC}-\frac{R^2}{L^2}}$

$\frac{1}{z}=\frac{1}{z_1}+\frac{1}{z_2}=\frac{1}{wLj+R}+\frac{1}{\frac{1}{jwc}}\Rightarrow$
$\frac{1}{z}=\frac{R-wLj}{R^2+w^2{L}^2}+jwc$
For pure resistive circuit $wc=\frac{wL}{R^2+w^2{L}^2}$
So $w^2=\frac{1}{LC}-\frac{R^2}{L^2}\Rightarrow w^2=\frac{1}{L^2}(\frac{L}{C}-R^2)$

So, there exists a real value of $w$ which satisfies this equation when $\frac{L}{C}>R^2$.