Question

A parallel RLC circuit with R1 = 20, L1 = \(\frac{1}{100}\) and C1 = \(\frac{1}{200}\) is scaled giving R2 = 10^4, L2 = 10^-4 and C2, the value of C2 is ___________

(a) 0.10 nF

(b) 0.3 nF

(c) 0.2 nF

(d) 0.4 nF

This question was addressed to me in an interview for internship.

My question comes from Series-Parallel Interconnection of Two Port Network in section Two-Port Networks of Network Theory

Answer 1

The correct answer is (c) 0.2 nF

For explanation I would say: K1 = \(\frac{R_2}{R_1}\)

= \(\frac{10^4}{20}\) = 5 X 10^2 = 500 Ω

And \(\frac{L_1}{L_2}  = \frac{kω}{k_1}\)

Or, \(\frac{kω}{k_1} =  \frac{L_1}{L_2}  X k_1\)

Or, \(\frac{10^{-2}}{10^{-4} X 5 X 10^2}\) = 5 X 10^4

Or, C2 = \(\frac{C_1}{kω.k_1} = \frac{0.5 X 10^{-2}}{5 X 10^4 X 500}\)

= 0.02 X 10^-8 = 0.2 nF.