16.1 V, 11.4 V

\(R=12\Omega ; \ { X }_{ C }=12\Omega ; \ { X }_{ L }=30\Omega ;\)
\(E=230 \ volt; \ v=50Hz\)
\((i) \ Current=\frac { E }{ \sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } } \)
\(=\frac { 230 }{ \sqrt { { 12 }^{ 2 }+\left( 30-14 \right) ^{ 2 } } }\)
\( I=\frac { 230 }{ \sqrt { 144+256 } } =\frac { 230 }{ 20 }\)
\(I=11.5A\\ (ii) \ tan\phi =\frac { { X }_{ L }-{ X }_{ C } }{ R } =\frac { 30-40 }{ 12 } =\frac { 16 }{ 12 } \)
\(tan\phi =1.3333=tan\left( 53°,8\prime \right)\)
\(\phi =53°,8\prime\)
\( \\ Voltage \ leads \ current \ by \ 53°,8\prime \)
\((iii) \ Power \ factors \ cos\theta =\frac { R }{ Z }\)
\(=\frac { R }{ \sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } } \)
\(or \ cos\theta =\frac { 12 }{ \sqrt { { \left( 12 \right) }^{ 2 } } +{ \left( 30-14 \right) }^{ 2 } }\)
\(=\frac { 12 }{ 20 } =0.6 \ or \ Power \ factor=0.6\)

Self-Inductance When the current in a coil is changed, a back emf is induced in the same coil. This phenomenon is called self-inductance. If Lis self-inductance of coil, then
\(N\phi \propto I\Rightarrow N\phi =LI\Rightarrow L=\frac { N\phi }{ I } \)
The SI unit of self-inductance is Henry (H).
(ii) Mutual inductance of solenoid coil system
\(M=\frac { { \mu }_{ 0 }{ N }_{ 1 }{ N }_{ 2 }{ A }_{ 2 } }{ l } \)
Here, N₁ = 15, N₂ = 1, l = 1cm = 10^-2m,
A = 2.0cm² = 20 x 10^-4m²
∴ \(M=\frac { 4\pi \times { 10 }^{ -7 }\times 15\times 1\times 2.0\times { 10 }^{ -4 } }{ { 10 }^{ -2 } } \)
\(=120\pi \times { 10 }^{ -9 }H\)
Induced emf in the loop
\({ \varepsilon }_{ 2 }=M\frac { { \Delta I }_{ 1 } }{ { \Delta t } } (numerically)=20\pi \times { 10 }^{ -9 }\frac { \left( 4-2 \right) }{ 0.1 } \)
\(=120\times 3.14\times { 10 }^{ -9 }\times \frac { 2 }{ 0.1 } =7.5\times { 10 }^{ -6 }V=7.5\mu V\)

emf induced
e = Blv
= 0.1 x 10 x 10^-2 x 20 V
= 0.2 volt
(ii) Current in the loop
\(i=\frac{e}{R} \) 
\(=\frac{0.2}{2}A=0.1A\)

(i) Reactance of the capacitor will decrease, resulting in increase of the current in the circuit, Therefore the bulb will glow brighter.
(ii) The bulb will go dimmer.

(i) As the magnet falls, the magnetic flux linked with the ring increases. This induces emf in the ring which opposes the motion of the falling magnet, hence a < g.
(ii) When current is suddenly switched ON, magnetic flux linked with the solenoid and thus, with metal ring increases. Current is induced in the ring in an anti-clockwise direction (as seen from the top of the ring). Since the direction of flow of current in the ring is opposite to the current in the solenoid, therefore they will repel each other and the ring jumps up.   

A choke coil reduces the voltage across the fluorescent tube without wastage of power.

The resistance coils are double wound to avoid induction effects. Magnetic field due to current in one half of the coil is cancelled by magnetic field due to current in the other half of the coil (which is in opposite direction).

No, Lenz's law does not violate this principle.

This is because of eddy currents developed in copper sheet.

Yes, it will appear as the train is intercepting vertical component of earth's magnetic field. No, answer is not affected.

1 weber = 10⁸ Maxwell

(a) In one revolution 
Change of area, \(dA=\pi l^2\)
\(\therefore\) change of magnetic flux
\(d\phi =\vec { B } .\vec { dA } =B.dA{ \cos { 0 } }^{ \circ }\)
\(=B\pi l^{ 2 }\)
(i) Induced emf, \(\varepsilon \) \(=B\pi l^{ 2 }/T=B\pi l^{ 2 }/v\)
(ii) Induced current in the rod, \(I=\frac{\varepsilon }{R}=\frac{\pi vB l^2}{R}\)
(b) Force acting on the rod, F = IlB
\(=\frac{\pi vB^2 l^3}{R}\)
The external force required to rotate the rod opposes the Lorentz force acting on the rod/ external force acts in the direction opposite the Lorentz force
(c) Power required to rotate the rod
Power = Force X velocity
P = Fx v
\(=\frac{\pi vB^2 l^3}{R}\times v\)

Given,  
\(L=80mH=80\times { 10 }^{ -3 }H, \ R=0, \ v=50Hz\)
\(C=60\mu F=60\times { 10 }^{ -6 }F,\)
\(\omega =2\pi v=100\pi \ rad/s \)
\({ V }_{ rms }=230 \ V,\)
and \(\\ { V }_{ 0 }=\sqrt { 2{ V }_{ rms } } =\sqrt { 2 } \times 230V\)
(i) I₀ = ? and I_rms = ?
\(\begin{aligned} \Rightarrow I_0 & =\frac{V_0}{Z}=\frac{V_0}{\left|\omega L-\frac{1}{\omega C}\right|} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{230 \sqrt{2}}{\left|100 \pi \times 80 \times 10^{-3}-\frac{1}{100 \pi \times 60 \times 10^{-6}}\right|} \\ \end{aligned}\)
\(\begin{aligned} =\frac{230 \sqrt{2}}{\left|8 \pi-\frac{1000}{6 \pi}\right|}=-11.63 \mathrm{~A} \\ \end{aligned}\)
\(\begin{aligned} I_{\mathrm{rms}} & =\frac{I_0}{\sqrt{2}}=\frac{-11.63}{\sqrt{2}}=8.23 \mathrm{~A} \end{aligned}\)
(ii) For L, V_L = I_rms \(\omega \)L = 8.23 \(\times\) 100\(\pi\)\(\times\)80 \(\times\)10^-3
= 206.84 V
For C, \(V_C=I_{\mathrm{rms}} \frac{1}{\omega C}=8.23 \times \frac{1}{100 \pi \times 60 \times 10^{-6}}\)
= 436.84 V
Since, voltage across L and C are 180° out of phase, therefore they are subtracted.
Thus, applied rms voltage = 436.84 - 206.84
= 230.0 V
(iii) Average power transferred per cycle by source to inductor is always zero because of phase difference of \(\pi\)/2 between voltage and current through inductor.
(iv) Average power transferred per cycle by source to inductor is always zero because of phase difference of \(\pi\)/2 between voltage and current through inductor
(v) \(\therefore\) Total average power absorbed by the circuit is also zero.