Question

In series LCR circuit, the plot of $I_{max}$ vs is shown in Figure. Find the bandwidth and mark in the figure.

Answer 1

Formula used:

$\Delta \omega ={\omega }_2-{\omega }_1$

$I_{rms}=\frac{I_0}{2}$

Step 1: Draw a rough diagram.


Step 2: Find maximum current.

At maximum current, we have to observe corresponding bandwidth,

so, calculate current,

$\left(I_{rms}\right)=\frac{I_0}{\sqrt{2}}$

$\left(I_{rms}\right)=\frac{1}{\sqrt{2}}=0.7A$

Step 3: Find corresponding bandwidth.

The bandwidth is from ${\omega }_{1}to{\omega }_2$ where 1 and 2 are the lowest and highest angular frequencies corresponding to $I_{rms}$ respectively.
Frequencies are,

${\omega }_1=0.8\frac{rad}{s}$

${\omega }_2=1.2\frac{rad}{s}$

Thus, bandwidth, $\Delta \omega ={\omega }_2-{\omega }_1=1.2-0.8=0.4rad/s$

Final Answer: $0.4rad/s$