#### Question

A series *LCR* circuit is connected across an a.c. source of variable angular frequency '*ω*'. Plot a graph showing variation of current 'i' as a function of 'ω' for two resistances R_{1} and R_{2} (R_{1} > R_{2}).

Answer the following questions using this graph :

(a) In which case is the resonance sharper and why?

(b) In which case in the power dissipation more and why?

#### Solution

he variation of current with angular frequency for the two resistances *R*_{1} and *R*_{2} is shown in the graph below.

Here,*i* = Virtual current through the circuit*ω =* Angular frequency of the source*ω _{r}* = Resonance frequency

From the graph, we can see that resonance for the resistance

*R*

_{2}is sharper than for

*R*

_{1}because resistance

*R*

_{2}is less than resistance

*R*

_{1}. Therefore, at resonance, the value of peak current will rise more abruptly for a lower value of resistance.

b) Power associated with the resistance is given by

P=Ev Iv

where*E*_{v} = Virtual voltage *I*_{v} = Virtual current

From the graph, we can say that the virtual current in case of *R*_{2 }is more than the virtual current in case of *R*_{1}. Hence, the power dissipation in case of the circuit with *R*_{2} is more than that with *R*_{1}.