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

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

Consider the diagram.
Bandwidth = \(\omega _{ 2 }-{ \omega }_{ 1 }\),

Where , \(\omega_1\) and  \(\omega_2\)  correspond to frequencies at which magnitude of  current is \(\frac { 1 }{ \sqrt { 2 } } \) times of maximum value.
\({ I }_{ rms }=\frac { { I }_{ max } }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \approx 0.7A\)
Clearly, from the diagram, the corresponding frequencies are 0.8 rad/s and rad/s.
\(\triangle \omega \) = B and width = 1.2 - 0.8 = 0.4 rad/s

(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.