For an L-C-R circuit driven at a frequency ωω, the equation reads

                       Ldidt+Ri+qC=Vi=Vm sin ωtLdidt+Ri+qC=Vi=Vm sin ωt

 

a. Multiply the equation by i and simplify where possible.
b. Interpret each term physically.
c. Cast the equation in the form of conservation of energy statement.
d. Integrate the equation over the one cycle to find that the phase difference between V and i must be acute.
Hindi: Use Kirchoff's law.
Step 1: Consider the L-C-R circuit. Applying KVL for the loop, we can write;
                    Ldidt+qC+iR=Vmsin ωt                        ...(i)                             Ldidt+qC+iR=Vmsin ωt                        ...(i)         
Step 2: Multiplying both sides by i, we get;
Lidtdt+qCi+i2R=(Vmi)sin ωt=Vi                            ...(ii)Lidtdt+qCi+i2R=(Vmi)sin ωt=Vi                            ...(ii)
where Lididt=ddt(12Li2)where Lididt=ddt(12Li2)=rate of change of energy stored in the inductor.
Ri2Ri2=joule heating loss
qCi=ddt(q22C)=qCi=ddt(q22C)=rate of change of energy stored in the capacitor.
Vi=rate at which driving force pours in energy. It goes into (i) ohmic loss and (ii) increase of stored energy.
Hence Eq. (ii) is in the form of conservation of energy statements.
Step 3: Integrating both sides of Eq. (ii) with respect to time over one full cycle (0T)(0T), we may write;
                        T0ddt(12Li2+q22C)dt+T0Ri2dt=Vi dt                        T0ddt(12Li2+q22C)dt+T0Ri2dt=Vi dt
                                              0+(+ve)=T0Vi dt                                              0+(+ve)=T0Vi dt
T0Vi dt>0T0Vi dt>0 if the phase difference between V and i is a constant and acute angle.