Power is the rate of doing work. In A.C current and voltage change at every moment. Since power is the product of instantaneous voltage and current, therefore it changes at every moment.
To determine average power in a circuit containing pure resistor, determine total energy spent in one cycle and divide it by the period.
Instantaneous power  \(=EI=\left[ { E }_{ 0 }sin\omega t \right] \left[ { I }_{ 0 }sin\omega t \right] \)
Instantaneous work done in small interval dt is
\( \ dW=EI \ dt={ E }_{ 0 }{ I }_{ 0 }{ sin }^{ 2 }\omega t \ dt\)
\(W=\int _{ 0 }^{ T }{ { E }_{ 0 }{ I }_{ 0 }{ sin }^{ 2 }\omega t \ dt }\)
\(W={ E }_{ 0 }{ I }_{ 0 }\int _{ 0 }^{ T }{ { \left( \frac { 1-cos2\omega t }{ 2 } \right) }dt } \)
\(=\left[ \because \int _{ 0 }^{ T }{ dt } -\int _{ 0 }^{ T }{ cos2\omega t } \right]\)
\(or \ \ W=\frac { { E }_{ 0 }{ I }_{ 0 } }{ 2 } \left[ T-0 \right] =\frac { { E }_{ 0 }{ I }_{ 0 } }{ 2 } T\)
\( \left[ \because \int _{ 0 }^{ T }{ dt } -\int _{ 0 }^{ T }{ cos2\omega t } \right] =0\)
\(or \ \ { P }_{ av }=\frac { W }{ T } =\frac { { E }_{ 0 }{ I }_{ 0 } }{ 2 } =\frac { { E }_{ 0 } }{ \sqrt { 2 } } .\frac { { I }_{ 0 } }{ \sqrt { 2 } } \)
\(\ {P }_{ av }={ E }_{ v }{ I }_{ v }\)
\(\therefore\) Average power over a complete cycle of a.c through a resistor is the product of virtual voltage and virtual current.

Here, I = 50 sin 100\(\pi\)t.
Compare it with I = I₀ sin \(\omega\)t = I₀ sin 2\(\pi\)vt
I₀ = 50A, 2\(\pi\)v = 100, v = 50c/s
Mean value of a.c. over positive half cycle
\(=\frac { 2{ I }_{ 0 } }{ \pi } =\frac { 2\times 50 }{ 3.14 } =31.8A\)
\({ I }_{ v }=\frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { 50 }{ 1.414 } =35.35A\)
\(From \ I={ I }_{ 0 }sin \ \omega t\)
\(I=50\ sin\ 2\pi \times 50\times \frac { 1 }{ 600 } =50\times \frac { 1 }{ 2 } =25A\)

\(Here, \ R=12\Omega , \ { X }_{ C }=14ohm, \ L=0.1H\)
\({ E }_{ v }=20V, \ v=50hz, \ { I }_{ v }=?, \ \phi =?\)
\( { X }_{ L }=\omega L=2\pi vL=2\times 3\times 50\times 0.1=30 \ ohm\)
\(Z=\sqrt { { R }^{ 2 }+\left( X_{ L }-{ X }_{ C } \right) ^{ 2 } } =\sqrt { { 12 }^{ 2 }+\left( 30-14 \right) ^{ 2 } } =20ohm\)
\({ I }_{ v }=\frac { { E }_{ v } }{ Z } =\frac { 200 }{ 20 } =10A\)
\(tan\phi =\frac { { X }_{ L }-{ X }_{ C } }{ R } =\frac { 30-14 }{ 12 } =1.33\)
\( \phi ={ tan }^{ -1 }\left( 1.33 \right) =53.13^{ \circ }\)

\(Here, \ { E }_{ v }=200V,\ v=50hz,\ R=100\Omega ,\ { V }_{ R }=65V,\ { V }_{ C }=415V,\ { V }_{ L }=204V\)
(i) If Iv is current in the circuit, then
\({ V }_{ R }={ I }_{ V }\times R; \ 65={ I }_{ V }\times 100, \ { I }_{ V }=0.65A\)
\( (ii) \ { V }_{ L }={ I }_{ V }{ X }_{ L }; \ { X }_{ L }=\frac { { V }_{ L } }{ { I }_{ v } } =\frac { 204 }{ 0.65 } =313.85\Omega\)
\({ X }_{ L }=\omega L=2\pi vL=313.85\)
\(L=\frac { 313.85 }{ 2\pi v } =\frac { 313.85 }{ 3.14\times 50 } =1.0H\)
\( (iii) \ { V }_{ C }={ I }_{ v }{ X }_{ C }, \ { X }_{ C }=\frac { { V }_{ C } }{ { I }_{ v } } =\frac { 415 }{ 0.65 } =638.5\Omega \)
\({ X }_{ C }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC } ; \ C=\frac { 1 }{ 2\pi v \ { X }_{ c } } =\frac { 1 }{ 2\times 3.14\times 50\times 638.5 } =4.99\times { 10 }^{ -6 }F\)
(iv) Let C′ be the capacitance that would produce resonance with L= 1.0H, then
\( v=\frac { 1 }{ 2\pi \sqrt { LC' } } ; \ C'=\frac { 1 }{ 4{ \pi }^{ 2 }{ v }^{ 2 }L } =\frac { 1 }{ 4\times \left( 3.14 \right) ^{ 2 }\times \left( 50 \right) ^{ 2 }\times 1 } =10.1\times { }^{ }F=10.1\mu F\)

\(Here, \ { E }_{ p }=2200V, \ { E }_{ s }=220V, \ { n }_{ p }=5000 \ \eta =90%\)
\({ P }_{ 0 }=E_{ s }{ I }_{ s }=8kW=8000 \ W \ { n }_{ s }=? \ \ { P }_{ i }={ E }_{ p }{ I }_{ p }=?\)
\(As \ \ \eta =\frac { { P }_{ 0 } }{ { P }_{ i } } \ \ \ \therefore \ { P }_{ i }=\frac { P_{ 0 } }{ \eta } =\frac { 8kW }{ 90/100 } =8.88kW\)
\(\\ As \ \ \frac { { E }_{ s } }{ { E }_{ p } } =\frac { { n }_{ s } }{ { n }_{ p } } \ \therefore \ { n }_{ s }=\frac { { E }_{ s } }{ { E }_{ p } } \times { n }_{ p }=\frac { 220 }{ 2200 } \times 5000 \ =500\)

\(Given \ R=10\Omega ,\)
\(C=0.01\mu F=0.01\times { 10 }^{ -6 }F\)
\(={ 10 }^{ -7 }F \)
\( v=50 \ Hz \ \therefore \omega =2\pi v=100\pi \)
\( E=100 \ V, \ I=?\)
The impedance of a C-R circuit is given by
\(Z=\sqrt { { R }^{ 2 }+\frac { 1 }{ { \omega }^{ 2 }{ C }^{ 2 } } } \\ =\sqrt { 100+\frac { 1 }{ { 10 }^{ 4 }\times { \pi }^{ 2 }\times { 10 }^{ -14 } } } \)

\(=10\sqrt { 1+\frac { { 10 }^{ 8 } }{ { \pi }^{ 2 } } } \)
\(=\frac { 10 }{ \pi } \sqrt { { \pi }^{ 2 }+{ 10 }^{ 8 } }\)
\( I=\frac { E }{ Z } =\frac { 100\times \pi }{ 10\sqrt { { \pi }^{ 2 }+{ 10 }^{ 8 } } }\)
\(= \ 0.0031A\)
If the current leads the e.m.f by a phase \(\theta \) , then
 \(tan\theta =\frac { 1 }{ R\omega C }\)
\(=\frac { 1 }{ 10\times 100\times \pi \times { 10 }^{ -7 } } \)
\(\\ or \ tan\theta =\frac { { 10 }^{ 4 } }{ 3.142 } \)
\(\tan\theta =315.09\)
\(\theta =89°50\prime \)

For series LCR resonant circuit
\(\omega L=\frac { 1 }{ \omega C } \)
\(\\ or \ C=\frac { 1 }{ { L\omega }^{ 2 } } =\frac { 1 }{ 10.5\times { \left( 400\pi \right) }^{ 2 } }\)
\( =0.06 \ \mu F\)
At resonance,
\(I=\frac { E }{ R } =\frac { 50 }{ 8 } =6.25A\)
\(\therefore \) Pot-diff. across the inductance
\(=I\times \omega L\)
\( =6.25\times 400\times \pi \times 10.5\)
\(=82430V=82.430\times { 10 }^{ 3 }V\)
\(=82.43 \ kV\)
Pot.diff.across the capacitor
\(=I\times \frac { 1 }{ \omega C }\)
\(=6.25\times \frac { 1 }{ 400\times \pi \times 0.06\times { 10 }^{ -6 } } \)
\(=\frac { 6.25 }{ 0.24 } \times { 10 }^{ 4 }\)
\(=26.4\times { 10 }^{ 4 }=264\times { 10 }^{ 3 }V\)
\(= \ 264 \ kV\)     

\(L=3mH=3\times { 10 }^{ -3 }H\)
\(v=1000kHz={ 10 }^{ 6 }Hz\)
\(Now \ v=\frac { 1 }{ 2\pi \sqrt { LC } }\)
\(or \ \sqrt { LC } =\frac { 1 }{ 2\pi v }\)
\(or \ LC=\frac { 1 }{ 4{ \pi }^{ 2 }{ v }^{ 2 } }\)
\(or \ C=\frac { 1 }{ 4{ \pi }^{ 2 }{ v }^{ 2 }L } \)
\(or \ C=\frac { 1 }{ 4\times 9.89\times { 10 }^{ 12 }\times 3\times { 10 }^{ -3 } }\)
\( =8.44\times { 10 }^{ -12 }F\)
\(\\ or \ C=8.44\mu F\)

\(Given \ V=300 \ sin \ 314t\)
\(\therefore \ { V }_{ 0 }=300 \ volt\)
\(\therefore \ \ { V }_{ rms }=\frac { { V }_{ 0 } }{ 2\sqrt { 2 } } =0.707\times { V }_{ 0 }\)
\(=0.707\times 303=212.1 \ V\)

In LR circuit, the impedance is given by
\({ Z }_{ L }=\sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } \quad ...(i)\)
In LCR circuit, the impedance is given by
\(Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } \quad ...(ii)\)
From equations, (i) and (ii), we find that
\(Z<{ Z }_{ L }\)
\(since \ I=\frac { E }{ Z } \)
As Z decreases on connecting a  capacitor in series with LR circuit, hence the current I in the circuit increases.

\(K=\frac{n_s}{n_p}=100\)

\(K=\frac{n_s}{n_p}=\frac{e_s}{e_p}\)
\( 0.1=\frac{e_s}{220} \\ \therefore e_s=22 \text { volt } \)

No, the frequency cannot be changed by a transformer.

No; power dissipated at resonance is maximum.

This is because parallel resonance circuits reject the current corresponding to parallel resonance frequencies. These circuits are used in the transmitting circuits.

Zero, in a pure inductor and in a pure capacitor.

\(V_L=V_C ; X_L=X_C ; Z=R ; \cos \phi=\frac{R}{Z}=1\)

(i) Alternating current is that current which shows periodic variation in its value with time. Direct current is that current whose magnitude and direction do not change.
(ii) The frequency of alternating current has some finite value. The frequency of direct current is zero.

Current through the bulb
\({ I }_{ \upsilon }=\frac { { E }_{ \upsilon } }{ Z } =\frac { { E }_{ \upsilon } }{ \sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } } ,where\ { X }_{ L }=\omega L\)
On introducing an iron rod into the inductor its inductance L increases. \({ X }_{ L }={ \omega }L\) increases. Therefore, impedance Z increases. Hence brightness of bulb decreases.

For definitions, see text. As \(X={ X }_{ L }-{ X }_{ C }=\) negative, when \({ X }_{ C }{ >X }_{ L },\) i.e., reactance X can be negative.

It implies, alternating current leads the applied alternating voltage in the circuit. However, Z cannot be negative.

Yes, the instantaneous power can be negative as, \(P_{\text {instantaneous }}=I_{\text {in }} \times V_{\text {in }}=I_0 \sin \omega t \times V_0 \cos \omega t\) No, because it is average, so it will be positive.

(i) \(\because\) P = V.I or I ²R
On increasing voltage, electric current reduces, therefore less power/energy is lost (as P \(\alpha\) I²) in cables during transmission.
(ii) (a) laminated iron core
(b) A.C. generator is connected to primary coil
(c) \(N_{S}=N_{P} \times \frac{V_{S}}{V_{P}}=500 \times \frac{115000}{5000}\)
N_s = 11500 turns
(iii) Transmission of 220 V
(a) needed less insulation
(b) is safer
(c) devices are designed for 220 V.

(i) The power factor is give by
\(\cos \phi=\frac{R}{Z}\)
For a circuit containing resistor and inductor
\(Z=\sqrt{R^2+(\omega L)^2}\)
\(\therefore \quad \cos \phi=\frac{R}{\sqrt{R^2+(\omega L)^2}}\)
(ii) For purely inductive circuit, \(\phi=\frac{\pi}{2}\)
\(\therefore P_{\mathrm{av}}=E_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi=E_{\mathrm{rms}} I_{\mathrm{rms}} \cos \frac{\pi}{2}=0\)
(iii) R = 2 \(\Omega\), E_rms = 220 V, f = 150 Hz
\(\therefore\) For a circuit containing only restor
\(\begin{aligned} \phi=0^{\circ} \Rightarrow \cos \phi=0 \end{aligned}\)
\(\begin{aligned} \therefore P_{\mathrm{av}}=\frac{E_{\mathrm{rms}}^2}{R}=\frac{(220)^2}{2}=24200 \mathrm{~W} \end{aligned}\)
(iv) For maximum power factor, cos \(\phi\) = 1
\(\therefore\) \(\phi\) = 0° (for purely resistive circuit)
\(\Rightarrow\) X is resistance.