Phase difference between V and i = \(\frac{\pi}{2}\)red
∴ P_average = \(\frac { { V }_{ m }{ I }_{ m } }{ 2 } \). cos Φ = \(\frac { 50\times 10 }{ 2 } \) cos 0^o

P _Average = V_rms. I_rms. cost
= V_rms. I_rms. \(\frac{R}{z}\)
= 200 x 1.5 x \(\frac45\)
P_average = 240W.

(i) S₁ and S₂ are two long solenoids each length l. The solenoid S₂ is wound closely over the solenoid S_1.
(ii) N₁ and N₂ are the number of turns in the solenoids S₁ and S₂ respectively. Both the solenoids are considered to have the same area of cross-section A as they are closely wound together.

(iii) I₁ is the current flowing through the solenoid S₁ The magnetic field B₁ produced at any point inside the solenoid S₁ due to the current I₁ is
\({ B }_{ 1 }={ \mu }_{ o }\frac { { N }_{ i } }{ l } { I }_{ 1 }\) ...(1)
(iv) The magnetic flux linked with each turn of S₂ is equal to B₁A.
Total magnetic flux linked with solenoid S₂ having N₂ turns is
Φ₂ = \(\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }{ AI }_{ 1 } }{ l } \) ...(2)
But Φ₂ = MI₁ ...(3) 
where M is the coefficient of mutual induction between S₁ and S₂
From equations (2) and (3),
MI₁ = \(\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }{ AI }_{ 1 } }{ l } \); M = \(​​\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }A }{ l } \)
(v) If the core is filled with a magnetic material of permeability μ,
M = \(​​\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }A }{ l } \)

(i) During LC oscillations in LC circuits, the energy of the system oscillates between the electric field of the capacitor and the magnetic field of the inductor.
(ii) Although these two forms of energy vary with time, the total energy remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.
Total energy, U= U_E + U_B = \(\frac { { q }^{ 2 } }{ 2C } +\frac { 1 }{ 2 } { Li }^{ 2 }\)
(iii) consider 3 different stages of LC oscillations and calculate the total energy of the system.
Case (i) When the charge in the capacitor, q Q_m = and the current through the inductor, i = 0, the total energy is given by
\(U=\frac { { { Q }_{ m } }^{ 2 } }{ 2C } +0=\frac { { { Q }_{ m } }^{ 2 } }{ 2C } \)
The total energy is wholly electrical
Case (ii) When charge = 0; current = I_m, the total energy is
\(U=0+\frac { 1 }{ 2 } { { Li }^{ 2 } }_{ m }=\frac { 1 }{ 2 } { { Li }^{ 2 } }_{ m }\)
 \(=\frac { L }{ 2 } \times \left( \frac { { { Q }_{ m } }^{ 2 } }{ LC } \right) { Q }_{ m }\omega =\frac { { Q }_{ m } }{ \sqrt { LC } } \)
= \(\frac { { { Q }_{ m } }^{ 2 } }{ 2C } \)
Case (ii) When charge = 0 ; current = I_m the total energy is
\(U=0+\frac { 1 }{ 2 } { { Li }^{ 2 } }_{ m }=\frac { 1 }{ 2 } { { Li }^{ 2 } }_{ m }\)
\(=\frac { L }{ 2 } \times \left( \frac { { { Q }_{ m } }^{ 2 } }{ LC } \right) \) since I_m= \({ Q }_{ m }\omega =\frac { { Q }_{ m } }{ \sqrt { LC } } \)
= \(\frac { { { Q }_{ m } }^{ 2 } }{ 2C } \)
Case (iii) When charge = q; current = t. the total energy is
U = \(\frac { { q }^{ 2 } }{ 2C } +\frac { 1 }{ 2 } { Li }^{ 2 }\)
(iv) Since q = Qm = cos ωt, i = \(\frac { dq }{ dt } ={ Q }_{ m }=\omega \) sin ωt. The negative sign in current indicates that the charge in the capacitor decreases with time.
\(U=\frac { { { Q }_{ m } }^{ 2 }{ cos }^{ 2 }\omega t }{ 2C } +\frac { { { L{ \omega }^{ 2 }Q }_{ m } }^{ 2 }{ sin }^{ 2 }\omega t }{ 2 } \)
\(U=\frac { { { Q }_{ m } }^{ 2 }{ cos }^{ 2 }\omega t }{ 2C } +\frac { { { L{ \omega }^{ 2 }Q }_{ m } }^{ 2 }{ sin }^{ 2 }\omega t }{ 2LC } \)
since \({ \omega }^{ 2 }=\frac { 1 }{ LC } \)
= \(\frac { { { Q }_{ m } }^{ 2 } }{ 2C } { (cos }^{ 2 }\omega t+{ sin }^{ 2 }\omega t)\)
\(U=\frac { { { Q }_{ m } }^{ 2 } }{ 2C } \)
From the above three cases, it is clear that the total energy of the system remains constant.