\(v={ v }_{ o }sin \ \omega t\)
\(q=CV=c{ CV }_{ o } \ sin \ \omega t\)
\(I=\frac { dq }{ dt } =\omega C{ V }_{ o } \ cos \ \omega t\)
\(I=\omega C{ V }_{ o } \ sin(\omega t+\frac { \pi }{ 2 } )\)
so the current leads the applied voltage, in phase by\( \ \frac { \pi }{ 2 } \)

Since, average power consumption in an AC circuit is given by
P_av = V_rms X I_rms X cos
But in pure capacitive circuit, phase difference between voltage and current is given by
\(\phi={ { \pi}\over{2 } }\)
\(\therefore\)  \({P}_{av}={V}_{rms}\times{I}_{rms}\times{{\pi}\over{2}}\)
\(\Rightarrow\)  Pav = 0  \(\left( \because\ cos\ {{\pi}\over{2}}=0 \right)\)
Thus, no power is consumed in pure capacitive AC circuit.

P = 150W, V = 220V
Resistance of the bulb, R = \({ { {V}^{2} }\over{ P } }\)
R = \({ { 220\times 220 }\over{150 } }=332.7\Omega\)
As, I_ms = \({{{V}_{rms}}\over{R}}={ { 220 }\over{ 322.7 } }\)   (V_rms = V = 220V)
\(\Rightarrow\) I_rms = 0.68A

\({ X }_{ c }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC } \)
AsC decreases, Xc will increase. Hence brightness will decrease.
(ii)
\({ X }_{ c }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC } \)
As frequence (v) decreases, Xc will increase. Hence brightness will decrease

On switching on the current in a galvanometer, the coil of the galvanometer does not come to rest immediately. It oscillates about its equilibrium position but the coil of a dead beat galvanometer comes to rest immediately. It is due to the reason that the eddy currents are set up in the metallic frame, over which the coil is wound and the eddy currents oppose the oscillatory motion of the coil.

Given, L = 2.0 H, C = 32 x 10^-6 F,
\(R=10\Omega \)
\({ \omega }_{ r }\) = ? Q = ?
\({ \omega }_{ r }=\frac { 1 }{ \sqrt { LC } } =\frac { 1 }{ \sqrt { 2.0\times 32\times { 10 }^{ -6 } } } \)
\(\frac { { 10 }^{ 3 } }{ 8 } =125rad/s\)
\( Q=\frac { 1 }{ R } \sqrt { \frac { L }{ C } } =\frac { 1 }{ 10 } \sqrt { \frac { 2 }{ 32\times { 10 }^{ -6 } } } \)
\(\frac { 1 }{ 10\times 4\times { 10 }^{ -3 } } =25\)

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,
\(U=\cfrac { 1 }{ 2 } { LI }^{ 2 }\)

(i) As the electric slab is introduced between the plates of the capacitor, its capacitance. Hence, the potential drop across the capacitor will i.e \(V=\frac { Q }{ C } \)
 As a result, the potential drop across the bulb will increase as they are connected in series. Thus its brightness will increase.
(ii) As a resistance R is increased, the potential drop across the resistor will increase. As a result, the potential drop across the bulb will decrease as they are connected in series. Thus, its brightness will decrease.  

Average power/cycle in an a.c. circuit is 
\(P={ E }_{ \upsilon }{ I }_{ \upsilon }cos\phi \)
In an ideal inductor, which has no ohmic resistance, phase angle \(\phi ={ 90 }^{ \circ }\)
\(\therefore \ P={ E }_{ \upsilon }{ I }_{ \upsilon }cos{ 90 }^{ \circ }=Zero\)
Infact, power supplied during growth of current is retrieved during decay of current through the ideal inductor.

The capacitative reactance is given by 
\({ X }_{ C }=\frac { 1 }{ \omega C } =\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC }\)
\(For \ a.c.,v\neq 0.\therefore { X }_{ C }\neq \infty \)
Hence a condenser does not block a.c. It allows a.c. to pass through.

For definitions, see text. As \(X={ X }_{ L }-{ X }_{ C }=\) negative, when \({ X }_{ C }{ >X }_{ L },\) i.e., reactance X can be negative.

It implies, alternating current leads the applied alternating voltage in the circuit. However, Z cannot be negative.

(i) Alternating current is that current which shows periodic variation in its value with time. Direct current is that current whose magnitude and direction do not change.
(ii) The frequency of alternating current has some finite value. The frequency of direct current is zero.

In a polyphase generator, one end of each coil is brought to a common point through shaft of the generator. The line wire from this point is called neutral line. The line wires from other ends of different coils through their slip rings and brushes are called phase lines.

In a.d.c. circuit, reactance of inductor/solenoid is zero. Therefore, removal of iron core does not affect the brightness of bulb.