letter of favourate of english book
Answers
Answer:
⇝ Derivation :-
❒ In LCR circuit connected to AC supply ;
》 lnstantaneous voltage is given by ,
\rm E = E_{\circ}sin \omega tE=E
∘
sinωt
》 lnstantaneous current is given by ,
\rm I = I_{\circ}sin( \omega t - \phi)I=I
∘
sin(ωt−ϕ)
Therefore,
lnstantaneous Power is ;
\rm P = EI = [E_{\circ}sin \omega t] [I_{\circ}sin( \omega t - \phi)]P=EI=[E
∘
sinωt][I
∘
sin(ωt−ϕ)]
\begin{gathered}:\longmapsto \rm P=E_{\circ}I_{\circ}sin(\omega t)sin(\omega t-\phi) \\ \end{gathered}
:⟼P=E
∘
I
∘
sin(ωt)sin(ωt−ϕ)
\rm Expanding \: Using \: \red{sin(x-y) =sinxcosy-cosxsiny}ExpandingUsingsin(x−y)=sinxcosy−cosxsiny
\begin{gathered}:\longmapsto \rm P=E_{\circ}I_{\circ}sin(\omega t) [sin(\omega t)cos\phi-cos(\omega t)sin\phi] \\ \end{gathered}
:⟼P=E
∘
I
∘
sin(ωt)[sin(ωt)cosϕ−cos(ωt)sinϕ]
\begin{gathered}:\longmapsto \rm P=E_{\circ}I_{\circ}sin {}^{2} (\omega t)cos \phi - E_{\circ}I_{\circ}sin (\omega t) cos(\omega t) \sin \phi \\ \end{gathered}
:⟼P=E
∘
I
∘
sin
2
(ωt)cosϕ−E
∘
I
∘
sin(ωt)cos(ωt)sinϕ
As We Know that,
\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{dW=Pdt }}}★
dW=Pdt
⏩ Integrating Both Side ;
\begin{gathered} :\longmapsto \rm \rm W = \int Pdt \\ \end{gathered}
:⟼W=∫Pdt
⏩ Putting Value of P ;
\begin{gathered}:\longmapsto \rm W = \int^{T} _{0} E_{\circ}I_{\circ}sin ^{2} (\omega t)cos \phi \\ \rm \: \: \: \: \: \: \: \: \: \: \: \: \: - \int^{T} _{0} E_{\circ}I_{\circ}sin (\omega t) cos(\omega t) \sin \phi \\ \end{gathered}
:⟼W=∫
0
T
E
∘
I
∘
sin
2
(ωt)cosϕ
−∫
0
T
E
∘
I
∘
sin(ωt)cos(ωt)sinϕ
\begin{gathered}:\longmapsto \rm W=E_{\circ}I_{\circ}cos \bigg(\dfrac{\phi}{2} \bigg) T - 0 \\ \end{gathered}
:⟼W=E
∘
I
∘
cos(
2
ϕ
)T−0
\begin{gathered}:\longmapsto \rm W= \dfrac{E_{\circ}I_{\circ}cos \big({\phi} \big) T}2 - 0 \\ \end{gathered}
:⟼W=
2
E
∘
I
∘
cos(ϕ)T
−0
\begin{gathered}\purple{ :\longmapsto \underline {\boxed{{\bf W= \dfrac{E_{\circ}I_{\circ}cos \big({\phi} \big) T}2 } }}} \\ \end{gathered}
:⟼
W=
2
E
∘
I
∘
cos(ϕ)T
As We Know that ;
\begin{gathered}\bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ P=\frac{Work}{Time} }}} \\ \end{gathered}
★
P=
Time
Work
Therefore,
\begin{gathered}:\longmapsto \rm P_{avg}= \dfrac{E_{\circ}I_{\circ}cos \big({\phi} \big) \cancel T}{2 \cancel T} \\ \end{gathered}
:⟼P
avg
=
2
T
E
∘
I
∘
cos(ϕ)
T
\begin{gathered}\purple{ :\longmapsto \underline {\boxed{{\bf P_{avg}= \frac{1}{2} E_{\circ}I_{\circ}cos \big({\phi} \big) }}}} \\ \end{gathered}
:⟼
P
avg
=
2
1
E
∘
I
∘
cos(ϕ)
✧ Or We can say that ;
\begin{gathered}\purple{ :\longmapsto \underline {\boxed{{\bf P_{avg}= E_{rms}I_{rms}cos \big({\phi} \big)}}}} \\ \end{gathered}
:⟼
P
avg
=E
rms
I
rms
cos(ϕ)
⇝ Power Factor :-
The ratio of the actual electrical power dissipated by an AC circuit to the product of the r.m.s. values of current and voltage.