Consider the L-C-R circuit shown in the figure. Find the net current i and the phase of i. Show that i=VZ. Find the impedance Z for this circuit.

                                       

Hint: The net current will be equal to the sum of currents flowing in the two branches.
Step 1: In the given figure i is the total current from the source. It is divided into two parts i2 through R and i1 through series combination of C and L.
So, we can write; i=i1+i2
As,Vmsinωt=Ri1fromthecircuitdiagram
i1=VmsinωtR...i
Step 2: If q2 is the charge on the capacitor at any time t, then for a series combination of C and L, applying KVL in the lower Circuit as shown;
q2C+Ldi2dt-Vmsinωt=0
q2C+Ld2q2dt2=Vmsinωti2=dq2dt...ii
Let,q2=qmsinωt+ϕ
dq2dt=qmωcosωt+ϕ
d2q2dt2=-qmω2sinωt+ϕ
Step 3: Now putting these values in Eq. (ii), we get;
qm1C+L-ω2sinωt+ϕ=Vmsinωt
Ifϕ=0and1C-Lω2>0,
thenqm=Vm1C-Lω2...iv
FromEq.(iii),i2=dq2dt=ωqmcosωt+ϕ
UsingEq.iv,i2=ωVmcosωt+ϕ1C-Lω2
Takingϕ=0;i2=Vmcosωt1ωC-Lω...v
Step 4: From Eqs. (i) and (v), we find that i1andi2 are out of phase by π2.
Now,i1+i2=VmsinωtR+Vmcosωt1ωC-Lω
PutVmR=A=CcosϕandVm1ωC-Lω=B=Csinϕ
i1+i2=Ccosϕsinωt+Csinϕcosωt
=Csinωt+ϕ
whereC=A2+B2
andϕ=tan-1BA
andϕ=tan-1R1ωC-Lω
Hence,i=i2+i2=Vm2R2+Vm21ωC-Lω21/2sinωt+ϕ
oriVm=1Z=Vm2R2+Vm21ωC-Lω21/2sinωt+ϕ
This is the expression for impedance Z of the circuit.