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Question
For an LCR circuit driven at frequency ω, the equation reads
`L (di)/(dt) + Ri + q/C = v_i = v_m` sin ωt
- Multiply the equation by i and simplify where possible.
- Interpret each term physically.
- Cast the equation in the form of a conservation of energy statement.
- Integrate the equation over one cycle to find that the phase difference between v and i must be acute.
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Solution
Consider L-C-R series circuit with AC supply
V = Vm sin ωt
Applying voltage Kirchhoff's law over the circuit
∴ VL + VC + VR = Vm sin ωt
L = `(di)/(dt) + Ri + q/c` = Vi = Vm sin ωt
Multiply the above equation by i on both the sides.
L = `(di)/(dt) + Ri + q/c` = Vi = Vm sin ωt
Multiply the above equation by `1/2` on both sides
`1/2 Li (di)/(dt) + i q/(2C) + (i^2R)/2 = 1/2` Vmi sin ωt `(∵ i = (dq)/(dt))`
`(d(1/2 Li^2))/(dt) + 1/(2C) (dq^2)/(dt) + (i^2R)/2 = 1/2` Vmi sin ωt .....(I)
(i) `(d(1/2 li^2))/(dt)` Represents the rate of change of potential energy in inductance L.
(ii) `d/(dt) q^2/(2C)` represents energy stored in dt time in the capacitor.
(iii) i2R represents joules heating loss.
(iv) `1/2 V_m i` sin ωt is the rate at which driving force pours in energy. It goes into ohmic loss and increase of stored energy in capacitor and inductor.
Vi = rate at which driving force pours in energy. It goes into (i) ohmic loss and (ii) increase of stored energy.
Hence equation (ii) is in the form of conservation of energy statement. Integrating both sides of equation. (ii) with respect to time over one full cycle (0 → T) we may write
`int_0^T d/(dt) (1/2 Li^2 + q^2/(2C)) dt + int_0^T Ri^2 dt = int_0^T Vi dt`
⇒ 0 + (+ ve) = `int_0^T Vi dt`
⇒ `int_0^T Vi dt > 0` if phase difference between V and i is a constant and acute angle.
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