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.

(i) Principle: When amount of magnetic flux linked with a coil changes, an e.m.f. is induced in the coil. The induced e.m.f. lasts so long as change in magnetic flux continues. The bulb fitted in the dynamo lights due to e.m.f. induced in the coil of dynamo. The bulb stops glowing when the cycle stops.
(ii) Ram showed the sprit of learning. He also thought how his classmates could benefit alongwith him. The teacher discharged his responsibility of explaning things to the entire class, encouraging them to learn new things.

(i) No, Lenz's law does not violate the principle of cnservation of energy.
(ii) Fleming's Right hand rule is yet another rule for finding the induced current.
(iii) In day to day life, Lenz's law implies that you cannot get something out of nothing. When we do mechanical work in moving the magnet towards/away from the coil, we obtain electrical energy in the form of induced current/ e.m.f. The moment we stop moving the magney, induced current/ e.m.f. disappears. If we ignor stray losses, energy obtained is always equal to energy spent or work done.

Quality factor
The quality factor of LCR-circuit is the ratio of the potential difference across inductance (or capacitance) at resonance to the applied voltage. This is also called the figure of merit or Q-value of LCR series circuit
i.e.   \(Q=\frac { Voltage \ across(L\ or\ C) }{ Voltage\ applied } \)
At resonance
Voltage across \(L=I{ X }_{ L }\)
Voltage across \(C=I{ X }_{ C }\)
Voltage applied = Voltage drop across \(R=IR\)

Consider a closed loop in which a current is flowing. The fact that a current is flowing implies of course that an electric field E exists along the loop. Small amount of work done in moving a unit charge through dl is given by \(-\overrightarrow { E } .\overrightarrow { d } l\)
When the work is done in taking unit charge along the entire closed loop is emf \(=\oint { \overrightarrow { E } .\overrightarrow { d } l } \)
\(e=\oint { \overrightarrow { E } .\overrightarrow { d } l } \)
\(\therefore \oint { \overrightarrow { E } .\overrightarrow { d } l } =-\frac { d\phi }{ dt } =e\)
\(So\oint { \overrightarrow { E } .\overrightarrow { d } l } \neq 0\)
Such fields are called non-conservative fields.
\(-\oint { \overrightarrow { E } .\overrightarrow { d } l } =-\frac { d\phi }{ dt } ,\)      ....(1)
where \(\phi \) is the flux across the loop
Consider in contrast, an electric field caused by a charge at rest. If we take a unit charge from point a to point b, such field have the property that the work done
\({ W }_{ ab }=-\int _{ a }^{ b }{ \overrightarrow { E } .\overrightarrow { d } l } \)     ....(2)
is independent of the path taken from a to b. If charge is taken along a closed path
\(\oint { \overrightarrow { E } .\overrightarrow { d } l } =0\)               ....(3)
This last equation is different from (2). The electric fields created by changing magnetic field thus have a somewhat different character compared to electric fields due to stationary charges.
The field where the closed line integral of electric field is zero \(\left( \oint { \overrightarrow { E } .\overrightarrow { d } l } =0 \right) \) is called conservative field.

Motional emf in \(AB=\left( { B }_{ 0 }\sin { \omega t } \right) d\upsilon .\left( -\hat { j } \right) \)
The direction of motional emf is along y-axis as per Fleming's right-hand rule.
At time t, the wire AB covers a distance \(x=\upsilon t\)
E.M.F. due to change in magnetic field (along OBAC)
\(=-{ B }_{ 0 }\omega \left( \cos { \omega t } \right) x.d\)
Total emf induced,  \(e=-{ B }_{ 0 }d\left[ \omega .x.\cos { \omega t+\upsilon \sin { \omega t } } \right] \)
Induced current \(I=\frac { { B }_{ 0 }d }{ R } \left[ \omega x\cos { \omega t+\upsilon \sin { \omega t } } \right] \)
Since the direction of the induced current is clockwise i.e. along OBAC hence the -ve sign is absorbed.
Force needed along \(\hat { i } ,F=BIl\)
or \(F=\frac { { B }_{ 0 }^{ 2 }{ d }^{ 2 } }{ R } \left[ \omega x\cos { \omega t+\upsilon \sin { \omega t } } \right] \sin { \omega t } \)

0.066 H, \(12\Omega \)