Light and e.m.waves are of transverse nature and they travel with the same velocity in vacuum. This led maxwell to conclude that light itself is e.m. wave

\(Given \ C=60\mu F=60\times { 10 }^{ -6 }F\)
\( { \ E }_{ v }=110V,v=60Hz\)
\(Since \ { I }_{ v }=\frac { { E }_{ v } }{ { X }_{ C } }\)
\(\therefore \ { I }_{ v }=\omega C{ E }_{ v }=2\pi v \ C \ { E }_{ v }\)
\( =2\times 3.142\times 60\times 60\times 60\times { 10 }^{ -6 }\times 110\)
\( =2.49A\)
\(or \ { I }_{ v }=2.49A\)

Self-inductance is the phenomenon of production of opposing emf in a coil. When the current through the coil changes. The self-inductance depends upon
(i) size of coil (area of cross-section and number of turns/length).
(ii) Relative magnetic permeability of the core of the coil on which winding is done.

Wattless Current The current in an A.C circuit when average power consumption in AC circuit is zero is referred as wattless current. If \(\phi\) is the phase difference between voltage and current, then power associated with I sin ϕ component of current is termed as wattless current.

E₀ = peak value of emf in a complete cycle,
(i) rms value \(\left(E_{\mathrm{rms}}\right)=\frac{E_0}{\sqrt{2}}\)
(ii) average value over a complete cycle (E) = zero

Impedance of series LCR circuit is given by \(Z=\sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 } } \)
or For Z to be minimum, \({ X }_{ L }={ X }_{ e }(or\omega =\frac { 1 }{ \sqrt { LC } } )\)
For wattless current to flow, circuit should not have any ohmic resitance R = 0
Alternatively : Power = \({ V }_{ rms }{ I }_{ rms }cos\phi \)
\(\phi ={ 90 }^{ 0 }=\frac { \pi }{ 2 } \)
Power = 0

Given, primary voltage, V_p = 2300 V
Secondary voltage, V_s = 230V
Primary turns, N_p = 4000 turns
Her, we assume that the transformer is ideal. No power loss in the form of heat, etc.,
Using the formula, \(\frac { { V }_{ s } }{ { V }_{ p } } =\frac { N_{ s } }{ N_{ p } } or\frac { 230 }{ 2300 } =\frac { { N }_{ s } }{ 4000 } \)
\({ N }_{ s }=400\)

The magnitude of emf is given by \(\left| e \right| \ or\ e=L\cfrac { { dl }_{ 0 } }{ dt } \)
​​Multiplying by I₀ to both sides, we get
\(\ e{ I }_{ 0 } \ dt={ LI }_{ 0 } \ { dI }_{ 0 }\)
\(But \ { I }_{ 0 }=\cfrac { dq }{ dt } \ or \ { I }_{ 0 }dt=dq\)
Also, work done = voltage x charge
or  \(dW=e\times dq=e{ I }_{ 0 }\ dt\)
Substituting the value from Eq. (i) into Eq. (ii), we get
\(dW={ LI }_{ 0 }{ dI }_{ 0 }\)
total work done in increasing the current from zero to I, we have by integrating both sides of Eq. we get
\(\int _{ 0 }^{ w }{ dW } =\int _{ 0 }^{ I }{ L{ I }_{ 0 } } d{ I }_{ 0 }\Rightarrow W=\frac { 1 }{ 2 } { LI }^{ 2 }\)
This work done in increasing the current flowing through the inductor is stored as the potential energy U in the magnetic field of inductor,
\(U=\cfrac { 1 }{ 2 } { LI }^{ 2 }\)

When the frequency of the supply equals the natural frequency of the circuit, resonance occurs.
\({ Z }_{ r }=R=20\Omega \)
\({ I }_{ rms }=\frac { { V }_{ rms } }{ { Z }_{ r } } =\frac { 200 }{ 20\quad } =10A\)
Average power transferred in one cycle,
\({ P }_{ av }={ V }_{ rms }{ I }_{ rms }\cos { \phi } \)
= 200 x 10 x cos 0⁰    [\(\phi ={ 0 }^{ 0 }\)]
= 2000W = 2kW

According to Lenz's law, the direction of induced current will oppose the cause of its production. So, the current in the loop will induce in such a way that it will support the current flowing in the wire, i.e in the same direction. So, the direction of current in the loop will be clockwise.

\(X_L=\omega L=2 \pi v L,\)

This is because d.c. voltage cannot produce a changing magnetic flux required in the working of a transformer.

\(V=Vo \ sin \ \omega t \ V=\frac { Q }{ C }\)
\(I=\frac { dQ }{ dt } \)
\(Io=\frac { Vo }{ (\frac { 1 }{ \omega C } ) } \)

\({ x }_{ c }=\frac { 1 }{ 2\pi fc } =212.3\Omega \)
\(Z=\sqrt { { R }^{ 2 } } +X{ c }^{ 2 }=291.5\Omega\)
\({ I }_{ rms }=\frac { { V }_{ rms } }{ Z } =\frac { 220 }{ 291.5 } =0.755A\)
\({ V }_{ R }(rms)=151 \ V\)
\(\ V_{ c }(rms)=2160.3 \ V\)

It works on the principle of electromagnetic induction, i.e. when a coil continuously rotates in a magnetic field, the magnetic flux associated with it keeps on changing; thus an emf is induced in it.
(b) When the coil rotates in a magnetic field, its effective area i.e. A cos \(\theta \) , (i.e. area normal to the magnetic field) keeps on changing. Hence magnetic flux  \(\phi \) = NBAcos S keeps on changing.
(c) Let the coil be rotating with angular velocity 'm',
at any instant 't' when the normal to the plane of the coil makes an angle \(\theta \) with the magnetic field. Hence magnetic flux, 12   \(\phi \) = NBAcos mt, Therefore induced emf (e)
\(\epsilon =-\frac { d\phi }{ dt } \)
\(\Rightarrow \epsilon =NBA\omega sin \ \omega t\)
Induced emf will be maximum when mt = 90° 12 Hence, £max = NBA\(\omega \)
Direction of induced emf can be determined using Flemming's Right hand rule. Alternatively: Statement of the above rule.

(a)

(b) Principle of a transformer: When alternating current flows through the primary coil, an emf is induced in the neighbouring (secondary) coil
Let \(\frac { d\phi }{ dt } \) be the rate of change of flux through each turn of the primary and the secondary coil
\(\frac { { \varepsilon }_{ 1 } }{ { \varepsilon }_{ 2 } } =-{ N }_{ 1 }\frac { d\phi }{ dt } /-{ N }_{ 2 }\frac { d\phi }{ dt } =\frac { { N }_{ 1 } }{ { N }_{ 2 } } \\ \frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { { N }_{ 1 } }{ { N }_{ 2 } } \)
But for an ideal transformer,
\(\frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { I_{ 2 } }{ I_{ 1 } } \)
From equation (1) and (2)
\(\frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { N_{ 1 } }{ N_{ 2 } } =\frac { I_{ 2 } }{ I_{ 1 } } \)
(c) Main assumptions
(i) The primary resistance and current are small
(ii) The flux linked with the primary and secondary coils is same / there is no leakage of flux from the core.
(iii) Secondary current is small.
(d) Reason due to which energy loses may occur Flux leakage/Resistance of the coils/Eddy currents/Hysteresis.