draw a plot showing the variation of current i as a function of angular frequency w of the applied ac source for the two  cases of series combination of1.inductance l1 ,capacitance 1 and resistance r12.inductance l2 capacitance c2 and resistance r2 where r2>r1
​write the relation between l1,c1 and l2 c2 at resonance which one of them will be better suited for fine tunning in the recieverset 

$\begin{array}{l}\text{The relation between current and emf in a series LCR ckt. is given by}\\ i=\frac{V_0}{\sqrt{R^2+{(L\omega -\frac{1}{C\omega })}^2}}sin(\omega t+\varphi )\\ \text{So when the peak current condition is chosen }sin(\omega t+\varphi )=1\\ \text{In the first case we choose L}=l_1\text{, C}=c_1,R=r_1\\ i_o=\frac{V_0}{\sqrt{r_1^2+{(l_1\omega -\frac{1}{c_1\omega })}^2}}\\ \text{In the first case we choose L}=l_2\text{, C}=c_2,R=r_2\text{ such that }{r}_2>r_1\\ i_o=\frac{V_0}{\sqrt{r_2^2+{(l_2\omega -\frac{1}{c_2\omega })}^2}}\\ {\text{To plot the graph we choose V}}_o=311\text{ volt, }{l}_1\text{=0.1}H\text{, }{l}_2\text{=0.2}H\text{, }{c}_1={10}^{-6}F\text{, }{c}_2=2\times {10}^{-6}F,\\ r_1=50\Omega ,r_2=200\Omega \\ \end{array}$
For case 1.

For case 2.

$\begin{array}{l}\text{At resonance }\\ X_L=X_C\\ \text{Let }{\omega }_1\text{ and }{\omega }_2\text{ be angular frequencies in first and 2nd case respectively}\\ \text{So,}\\ l_1{\omega }_1=\frac{1}{c_1{\omega }_1}\\ \Rightarrow l_1{c}_1{\omega }_1^2=1\\ \text{Similarly,}\\ l_2{\omega }_2=\frac{1}{c_2{\omega }_2}\\ \Rightarrow l_2{c}_2{\omega }_2^2=1\\ \text{From both the relations }\\ l_1{c}_1{\omega }_1^2=l_2{c}_2{\omega }_2^2\end{array}$
Since from the graph it is clear that in first case the curve at resonance is sharper than that in case 2. So case 1 is better suited for tuning the receiver set.