The maximum emf induced in a generator is given by \(E_{0}=N B A \omega\)
When the speed of rotation (ω) is doubled:
(a) The frequency of ac will be doubled.
(b) The emf gets doubled.
(c) The mechanical power required to rotate the generator also gets doubled.

Let an alternating current of I = I_m sin ωt be passing through a network of L, C and R creating a potential difference of V = V_m sin \((\omega t \pm \phi) \text { where } \phi\) is the phase difference. Then the power consumed is given by \(P=V I=V_{m} I_{m} \sin (\omega t \pm \phi) \sin \omega t\)
\( \therefore \ P =V_{m} I_{m}(\sin \omega t \cos \phi \pm \cos \omega t \sin \phi) \sin \omega t \)
\(P =V_{m} I_{m}\left(\sin ^{2} \omega t \cos \phi \pm \frac{1}{2} \sin 2 \omega t \sin \phi\right) \)
\(P_{a v}=\frac{\int_{0}^{T} P d t}{\int_{0}^{T} P_{a v}=\frac{V_{m} I_{m}}{T}}\left[\begin{array}{l} \int_{0}^{T} \sin ^{2} \omega t \cos \phi d t \\ \quad+\frac{1}{2} \int_{0}^{T} \sin \phi \sin 2 \omega t d t \end{array}\right]\)
\(P_{a v}=\frac{V_{m} I_{m}}{T}\left[\frac{T}{2} \cos \phi+0\right]\left[\because \int_{0}^{T} \sin ^{2} \omega t d t=\frac{T}{2} \text { and } \int_{0}^{T} \sin 2 \omega t d t=0\right]\) 
\(P_{a v}=\frac{V_{m} I_{m}}{2} \cos \phi=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi\)
(i) No power is dissipated if (a) resistance in the circuit is zero and (b) phase angle between voltage and current is π/2.
(ii) Maximum power is dissipated if (a) resistance in the circuit is maximum and (b) phase angle between voltage and current is zero.

Given, input voltage, V_p = 2.5 x 10³ V
Input current, I_p = 20 A
Also, \(\frac{N_{p}}{N_{s}}=\frac{10}{1} \Rightarrow \frac{N_{s}}{N_{p}}=\frac{1}{10}\)  ...........(i)
Percentage efficiency \(=\frac{\text { Output power }}{\text { Input power }} \times 100\)
\(\Rightarrow \ \frac{90}{100}=\frac{\text { Output power }}{V_{p} I_{p}}\)
(i) Output power \(=\frac{90}{100} \times\left(V_{p} I_{p}\right)\)
\(=\frac{90}{100} \times\left(2.5 \times 10^{3} \mathrm{~V}\right) \times(20 \mathrm{~A})\)
= 4.5 x 10⁴ W
(ii) \(\because \quad \frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}\)
\(\Rightarrow \ V_{s}=\frac{N_{s}}{N_{p}} \times V_{p}\)
Voltage, \(V_{s}=\frac{1}{10} \times 2.5 \times 10^{3} \mathrm{~V}=250 \mathrm{~V}\)
(iii) V_s I_s = 4.5 x 10⁴ W
Current, \(I_{s}=\frac{4.5 \times 10^{4}}{V_{s}}=\frac{4.5 \times 10^{4}}{250}=180 \mathrm{~A}\)

(i) The town is 10 km away, length of pair of Cu wires
used, I = 20 km = 20000 m.
Resistance of Cu wires,
\(R=\rho \frac{l}{A}=\rho \frac{l}{\pi(r)^{2}}=\frac{1.7 \times 10^{-8} \times 20000}{3.14\left(0.5 \times 10^{-2}\right)^{2}} \approx 4 \Omega\)
\(I \text { at } 220 \mathrm{~V}, V I=10^{6} \mathrm{~W} ; I=\frac{10^{6}}{220}=0.45 \times 10^{4} \mathrm{~A}\)
RI² = power loss = 4 x (0.45)² x 10⁸> 10⁶ W
Therefore, this method cannot be used for transmission.
(ii) When power, P = 10⁶ W is transmitted at 11000 V.
V \('I'\)= 10⁶ W = 11000 \(I'\)
Current drawn, \(I^{\prime}=\frac{1}{1.1} \times 10^{2}\)
Power loss \(=R I^{2}=\frac{1}{121} \times 4 \times 10^{6}=3.3 \times 10^{4} \mathrm{~W}\)
\(\therefore\)  Fraction of power loss \(=\frac{3.3 \times 10^{4}}{10^{6}}=3.3 \%\)

Given, E_p = 2200V , E_s = 220 V, N_p = 5000
Efficiency, \(\eta \) = 90%
Output power, p₀ = 8kW
Since, efficiency,
\(\eta=\frac{\text { Output power }}{\text { Input power }}=\frac{P_{o}}{P_{i}}\)
\(\Rightarrow \quad P_{i}=\frac{P_{o}}{\eta}=\frac{8}{90 / 100}=8.9 \mathrm{~kW}\)
Also, \(\frac{N_s}{N_p}=\frac{E_s}{E_p} \Rightarrow N_s=\frac{E_s}{E_p} N_p=\frac{220}{2200} \times 5000\)
\(\Rightarrow\) N_s = 500

(i) Magnetic flux versus current

(ii) 

Alternatively 

When Iis increasing at constant rate
(iii) Magnetic energy stored

(i) Here, \(N=50, \ A=2.5{ m }^{ 2 }, \ \omega =60 \ rad \ { s }^{ -1 }\)
\(B=0.3T, \ R=500\Omega ,\ { I }_{ 0 }=?\)
\({ I }_{ 0 }=\frac { { e }_{ 0 } }{ R } =\frac { NAB\omega }{ R } =\frac { 50\times 2.5\times 0.3\times 60 }{ 500 } \)
\(=4.5A\)
(ii) Magnetic flux will be max. if the coil is in the vertical position. On the other hand, magnetic flux will be zero if the coil is in the horizontal position.
(iii) Yes, the generator will work if the coil is stationary and instead, the pole pieces are rotated together.

Here, \(N=100,\ A=3{ m }^{ 2 },\ \omega =60/sec.\)
\(B=0.04T,\ R=500\Omega ,\ { I }_{ 0 }=?,\ { P }_{ 0 }=?\)
\({ I }_{ 0 }=\frac { { E }_{ 0 } }{ R } =\frac { NAB\omega }{ R } =\frac { 100\times 3\times 0.04\times 60 }{ 500 } \)
\(=1.44A\)
Max. power dissipation \(={ E }_{ \upsilon }{ I }_{ \upsilon }\)
\(=\frac { { E }_{ 0 } }{ \sqrt { 2 } } \frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { { I }_{ 0 }^{ 2 }R }{ 2 } =\frac { { \left( 1.44 \right) }^{ 2 }\times 500 }{ 2 } \)
\(=518.4 \ W\)

(d) From Lenz's law we find that when A is moved towards B, current starts flowing in B and when A stops, the current also stops it means there is a constant current in anticlockwise direction in A.

(a) I_o = 3.23A
(b) 1.55ms

\(Here, \ R=40\Omega , \ { E }_{ v }=220V, \ v=50Hz,\)
\((i) \ { I }_{ v }=\frac { { E }_{ v } }{ R } =\frac { 220 }{ 40 } =5.5 \ A\)
\({ I }_{ 0 }=\sqrt { 2 } { I }_{ v }=1.414\times 5.5=7.8 \ A\)
If alternating current is given by
\(I={ I }_{ 0 } \ sin \ \omega t, \ then\)
\({ I }_{ 0 }={ I }_{ 0 } \ sin{ \omega t }; \ sin \ { \omega t }_{ 1 }=1 \ or \ { \omega t }_{ 1 }=\frac { \pi }{ 2 }\)
\(and \ { I }_{ v }=\frac { { { I } }_{ 0 } }{ \sqrt { 2 } } ={ I }_{ 0 }sin{ \ \omega t }_{ 2 }, \ which \ implies\)
\({ \omega t }_{ 2 }=\frac { \pi }{ 2 } +\frac { \pi }{ 4 } \ \ \therefore \omega \left( { t }_{ 2 }-{ t }_{ 1 } \right) =\frac { \pi }{ 2 } +\frac { \pi }{ 4 } -\frac { \pi }{ 2 } =\frac { \pi }{ 4 }\)
\({ t }_{ 2 }-{ t }_{ 1 }=\frac { \pi }{ 4\omega } =\frac { \pi }{ 4\times 2\pi v } =\frac { 1 }{ 8v } =\frac { 1 }{ 8\times 50 } =2.5\times { 10 }^{ -3 }s=2.5 \ ms\)

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
\(=\frac { 2{ I }_{ 0 } }{ \pi } =\frac { 2\times 50 }{ 3.14 } =31.8A\)
\({ I }_{ v }=\frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { 50 }{ 1.414 } =35.35A\)
\(From \ I={ I }_{ 0 }sin \ \omega t\)
\(I=50\ sin\ 2\pi \times 50\times \frac { 1 }{ 600 } =50\times \frac { 1 }{ 2 } =25A\)

\(Here,\ I=5sin\ 100\pi t\)
\( Compare\ with\ I={ I }_{ 0 }sin\ \omega t\)
\({ I }_{ 0 }=5A,\ \omega =2\pi v=100\pi\)
\(v=\frac { 100\pi }{ 2\pi } =50 \ Hz\)
\(\ { I }_{ v }=\frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { 5 }{ \sqrt { 2 } } =\frac { 5\sqrt { 2 } }{ 2 } =3.54A\)

Here, I_v = ? I₀ = 5.0 A
As \({ I }_{ v }=\frac { { I }_{ 0 } }{ \sqrt { 2 } } \)
\(\therefore \ \ { I }_{ v }=\frac { 5 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 5\times 1.414 }{ 2 } =3.54A\)
(i) An a.c. ammeter will read 3.54 ampere
(ii) An ordinary moving coil ammeter will read zero because it records average value of current over a complete cycle, which is zero in case of alternating current.

\(Here, \ C=1.0\mu F={ 10 }^{ -6 }F\)
\(R={ 10 }^{ 4 }ohm, \ { E }_{ 0 }=2.0volt\)
\(During \ charging \ of \ the \ condenser,\)
\(q={ q }_{ 0 }\left( 1-{ e }^{ -t/RC } \right) \ I=-{ I }_{ 0 }{ e }^{ -t/RC }, \ where\)
\({ I }_{ 0 }=\frac { E }{ R } =\frac { 2.0 }{ { 10 }^{ 4 } } =2.0\times { 10 }^{ -4 }amp\)
\(At \ t=0.02s, \ I={ I }_{ 0 }\left( { e }^{ -0.02/{ 10 }^{ 4 }\times { 10 }^{ -6 } } \right) =2\times { 10 }^{ -4 }{ e }^{ -2 }\)
\(=\frac { 2\times { 10 }^{ -4 } }{ { e }^{ 2 } } =\frac { 2\times { 10 }^{ -4 } }{ \left( 2.718 \right) ^{ 2 } } =0.27\times { 10 }^{ -4 }A\)
\(I=27\times { 10 }^{ -6 } \ A=27\mu A\)

(a) 1.0 J, yes
(b) \(\omega \) = 10³ rad s^-1, v = 159 Hz
(c) 1.0 J

\(Here,I=10 \ sin \ 314 \ t;\)
\(Compare \ it \ with \ I={ I }_{ 0 }sin \ \omega \ t\)
\(\ { I }_{ 0 }=10amp,{ I }_{ eff }=\frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { 10 }{ \sqrt { 2 } } =7.07A\)
\( \omega =314=3.14 \times 100=100\pi \)
\( v=\frac { \omega }{ 2\pi } =\frac { 100\pi }{ 2\pi } =50Hz\)

Yes, when the a.c. is a square wave, r.m.s. value, average value and peak value of a.c. are equal. Choice(d) is correct.

As heat produced is the same, rms voltage of a.c. source = d.c. voltage = 110 V

For transfer of power \(\left( =V\times I \right) \) at higher voltage (220 V instead of 110 V), current carried by wires is just half. Therefore, such wires need not be very thick, saving lot of transmission material and reducing the cost of transmission. This is one advantage of 220 V supply. But to design a device of particular wattage, \(P=\frac { { V }^{ 2 } }{ R } \) , as \({ V }^{ 2 }\) is 4 times, R must be four times. If not, the dissipation of power in the form of heat will be larger on 220 V supply. This is one disadvantage of this supply.

The slip ring arrangement in an a.c. generator is replaced by split ring arrangement or commutator arrangement in d.c. generator.