Question

Derive expression P = Vrms Irms cosδ for an A.C. circuit

Answer 1

Proof with Explanation:

• Assume that the alternating voltage given by $V(t)=vsin(wt)$
• The resulting alternating current in the circuit be $I(t)$ with a phase change of Ф, let Ф=p and amplitude be I ;therefore $I(t)=Isin(wt+p)$

power P at any instant is $P=V(t)I(t)$

we can write $I(t)$ as $\frac{V(t)}{z}$ where z is impedance.

z is given by $|z|$∠δ in phasor, where δ is same as phase difference between current and voltage in the circuit.

The power in phasor form given by $|p|$∠δ.

where |p|= is $Vrms$$Irms$

therefore,rewriting the phasor, we can write power as $P=Vrms*Irms*cos$δ

Answer 2

Derivation of P = Vrms Irms cosδ for AC circuit:

Here, we need to find the average power in LCR circuit:

The potential difference is given by the formula:

$V=V_{0} \sin \omega t$

The current is given by the formula:

$I=I_{0} \sin \omega t-\phi$

Where,

Φ is phase angle

Total work done over complete cycle is given by the formula:

$W=\int_{0}^{T} V I d t$

On substituting potential difference and current equation, we get,

$W=\int_{0}^{T} V_{0} \sin \omega t . I_{0} \sin \omega t-\phi d t$

$W =\frac{V_{0} I_{0}}{2} \int_{0}^{T} 2 \sin \omega t . \sin \omega t-\phi d t$

The above equation is similarly to the algebraic equation given below.

$2 \sin A \cdot \sin B=\cos (A-B)-\cos (A+B)$

Now, on using the above algebraic equation, we get,

$\mathrm{W}=\frac{V_{0} I_{0}}{2} \int_{0}^{T}[\cos \phi-\cos (2 \omega t-\phi)] d t$

$\mathrm{W}=\frac{V_{0} I_{0}}{2}[T \cos \phi]$

$\frac{W}{T}=\frac{\bar{V}_{0}}{\sqrt{2}} \cdot \frac{I_{0}}{\sqrt{2}} \cdot \cos \phi$

Thus, the power is obtained as,

$\mathrm{P}=V_{r m s} \cdot I_{r m s} \cdot \cos \phi$

rms = Root Mean Square