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 (0→T)(0→T), 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>0⇒∫T0Vi dt>0 if the phase difference between V and i is a constant and acute angle.