Consider the L-C-R circuit shown in the fiure. Find the net current I and the phase of i. Show that `i=V/Z`. Find the impedance Z of this circuit.

Answer

A

Explanation

In the given figure `i` is the total current from the source. It is divided into two parts, `i_(1)` through R and `i_(2)` throgh series combination of C and L.
So, we can write `i=i_(1)+i_(2)`
`V_(m)sinomegat=Ri_(1)` (from the circuit diagram)
`i_(1)=(V_(m)sinomegat)/(R)`.............(i)
If `q_(2)` is charge on the capacitor at any time t, then for series combination of C and L.
Applying KVL inthe Lower circuit as shown.

`q_(2)/C + (Ldi_(2))/(dt) - V_(m)sinomegat=0`
`rArr q_(2)/C + (Ld_(2)q_(2))/(dt)^(2) = V_(m)sinomegat` `[therefore i_(2)=(dq_(2))/(dt)]`......(ii)
Let `q_(2) = q_(m) sin (omegat+phi)`..............(iii)
`(dq_(2))/(dt) = q_(m)cos(omegat+phi)`
`rArr (d^(2)q_(2))/(dt^(2))=-q_(m)omega^(2)sin(omegat+phi)`
Now putting the values in Eq. (ii), we get
`q_(m)[1/C+l(-omega^(2))]sin(omegat+phi)=V_(m)sinomegat`
If `phi = 0` and `(1/C-Lomega6(2)) gt 0,`
then `q_(m)=v_(m)/(1/C-Lomega^(2))`..............(iv)
From Eq. (iii), `i_(2) = (dq_(2))/(dt) = omegaq_(m)cos(omegat+phi)`
Using Eq. (iv), `i_(2) = (omegaV_(m)cos(omegat+phi))/(1/C-Lomega^(2))`
Taking `phi =0, i_(2)=(V_(m)cos(omegat))/(1/(omegaC)-Lomega)`
From Eqs. (i) and (v), we find that `i_(1)` and `i_(2)` are out of phase by `phi/2`
Now, `i_(1) + i_(2) = (V_(m)sinomegat)/R + (V_(m)cosomegat)/(1/(omegaC)-Lomega)`
Put `V_(m)/R = A = C cosphi` and `V_(m)/(1/(omegaC)-Lomega)=B = C sinphi`
`therefore i_(1) + i_(2) = C cos phisinomegat+ C sinphi cos omegat`
`= C =sqrt(A^(2) + B^(2)`
and `phi=tan^(-1)B/AC = [V_(m)^(2)/R^(2) + (V_(m)^(2))/(1/(omegaC)-Lomega)]^(1//2)`
and `phi = tan^(-1) =R/(1/(omegaC)-Lomega)`
Hence, `i=i_(1)+i_(2)= [V_(m)^(2)/R^(2) + V_(m)^(2)/(1/(omegaC)-Lomega)^(2)]^(1//2)sin(omegat+phi)`
or `i/V_(m)=1/Z =[1/R^(2)+1/(1/(omegaC)-Lomega)^(2)]^(1//2)`
This is the expression for impedance Z of the circuit.