The bode magnitude plot for the transfer function  $\frac{V_0(s)}{V_i(s)}$ of the circuit is as shown. The value of R is ______ Ω. (Round off to 2 decimal places.)

The correct answer is 0.1.

1. Analysis of the Circuit:
The given circuit is a series RLC low-pass filter with:
Inductance, L = 1 mH = 1 × 10^-3 H
Capacitance, C = 250 µF = 250 × 10^-6 F
The output voltage V₀(s) is measured across the capacitor C.

The transfer function of the series RLC circuit is given by:
$H(s)=\frac{V_0(s)}{V_i(s)}=\frac{\frac{1}{Cs}}{R+Ls+\frac{1}{Cs}}=\frac{1}{LC{s}^2+RCs+1}$

Rewriting this in standard second-order system form:
$H(s)=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$
where:
${\omega }_n=\frac{1}{\sqrt{LC}}$
and the damping ratio ζ is:
$\zeta =\frac{R}{2}\sqrt{\frac{C}{L}}$

2. Finding the Damping Ratio (ζ) from the Bode Plot:
From the given Bode magnitude plot, we observe that the resonant peak is at ω = 2000 rad/s, and the peak magnitude height is 26 dB.

The resonant peak value M_r (in linear scale) is related to the peak magnitude in dB by:
$20{\log }_{10}(M_r)=26$
$M_r={10}^{26/20}={10}^{1.3}\approx 19.95$

The standard formula for the resonant peak magnitude M_r of a second-order system is:
$M_r=\frac{1}{2\zeta \sqrt{1-{\zeta }^2}}$
Substituting M_r = 19.95:
$2\zeta \sqrt{1-{\zeta }^2}=\frac{1}{19.95}\approx 0.050125$
Since ζ is very small, we can approximate 1 - ζ² ≈ 1, which gives:
$2\zeta \approx 0.05\Rightarrow \zeta \approx 0.025$

3. Calculating the Resistance R:
Using the damping ratio formula for the series RLC circuit:
$\zeta =\frac{R}{2}$
Substituting the values of ζ, C, and L:
$0.025=\frac{R}{2}\sqrt{\frac{250\times {10}^{-6}}{1\times {10}^{-3}}}$
$0.025=\frac{R}{2}\sqrt{0.25}$
$0.025=\frac{R}{2}\times 0.5$
$0.025=0.25R$
$R=\frac{0.025}{0.25}=0.1\text{Ω}$