Let (I_rms)_a = rms current in circuit (a)
(I_rms)_b  = rms current in circuit (b)
\({ \left( I_{ rms } \right) }_{ a }=\frac { { V }_{ rms } }{ Z } =\frac { V }{ R } \)
\({ \left( I_{ rms } \right) }_{ b }=\frac { { V }_{ rms } }{ Z } b\)
\(=\frac { V }{ \sqrt { { R }^{ 2 }+\left( { X }_{ L }-{ X }_{ C } \right) ^{ 2 } } } \)
When \(({ I }_{ rms })_{ a } =({ I }_{ rms })_{ b }\)
\(R=\sqrt { { R }^{ 2 }+\left( { X }_{ L }-{ X }_{ C } \right) ^{ 2 } } \)
\(\Rightarrow \ { X }_{ L }={ X }_{ C }\) in resonance condition
(ii) As,  \( Z\ge R\)
\(\Rightarrow \frac { ({ I }_{ rms })_{ a } }{ ({ I }_{ rms })_{ a } } =\frac { \sqrt { R=\sqrt { { R }^{ 2 }+\left( { X }_{ L }-{ X }_{ C } \right) ^{ 2 } } } }{ R } \)
\(=\frac { Z }{ R } \ \ge 1\)
\(({ I }_{ rms })_{ a }\ge ({ I }_{ rms })_{ b }\)
No, the rms current in circuit (b) cannot be larger than that in (a)

(i) Consider the given figure,

Since, the alternating emf in the above L-C-R series circuit would be represented by
\(V={ V }_{ 0 }sin(1000t+\phi )\)
\(\Rightarrow \omega =1000Hz \)
Given, \(R=400\Omega ,C=2\mu F,L=100mH\)
Capacitive reactance, \({ X }_{ C }=\frac { 1 }{ \omega C } \)
\(\Rightarrow \ \ { X }_{ C }=\frac { 1 }{ 1000\times 2\times { 10 }^{ -6 } } \Rightarrow \quad { X }_{ C }=\frac { { 10 }^{ 3 } }{ 2 } \)
\(\Rightarrow \ \ { X }_{ L }=1000\times 100\times { 10 }^{ -3 }=100\Omega \)
So, we can see that \(\Rightarrow \ \ { X }_{ C }=\omega L\)
\(\Rightarrow \ tan\phi \) is negative.
Hence, the Voltage lags behind the current  by a phase angle \( \phi \).The  AC circuit is capacitance-dominated circuit.
Phase difference \(tan\ \phi \ =\frac { { X }_{ L }-{ X }_{ C } }{ R } \)
\(tan \ \phi \ =\frac { 100-500 }{ 400 } \Rightarrow tan \ \phi =\frac { -400 }{ 400 } \)
\(tan \ \phi \ =-1 \ \ \Rightarrow tan \ \phi =-tan\left( \frac { \pi }{ 4 } \right) \)
\(\Rightarrow \ \ \phi =-\frac { \pi }{ 4 } \)
This is the required value of the phase difference between the current and the voltage in the given series L-C-R circuit.
As, \(cos\phi <1\)
(ii) Suppose new capacitance of the circuit is C'.
\(cos\phi '=1=\frac { 1 }{ \sqrt { { R }^{ 2 }+\left( { X }_{ L }-{ X }_{ C }^{ ' } \right) ^{ 2 } } } \)
\(\Rightarrow \ { R }^{ 2 }={ R }^{ 2 }+{ R }^{ 2 }+\left( { X }_{ L }-{ X }_{ C }^{ ' } \right) ^{ 2 }\)
\(\Rightarrow \ { X }_{ L }={ X }_{ C }^{ , }=\frac { 1 }{ \omega C' } or\quad \omega L=\frac { 1 }{ \omega C' } \)
\(\Rightarrow \quad { \omega }^{ 2 }=\frac { 1 }{ LC' } \)
or    \(\Rightarrow \ (1000)^{ 2 }=\frac { 1 }{ LC' } \quad [\omega =1000]\)
\(\Rightarrow \ C^{ ' }=\frac { 1 }{ L\times { 10 }^{ 6 } } =\frac { 1 }{ 100\times { 10 }^{ -3 }\times { 10 }^{ 6 } } \)
\(\Rightarrow \ \frac { 10 }{ { 10 }^{ 6 } } =\frac { 1 }{ { 10 }^{ 5 } } ={ 10 }^{ -5 }\)
\(\Rightarrow \ C'={ 10 }^{ -5 }F\ =10\times { 10 }^{ -6 }F=10\mu F\)
As, C' > C. Hence, we have to add an additional capacitor \(10\mu F(10\mu F-2\mu F)\) in parallel with previous capacitor.

In an inductor, the current lags the voltage by 90⁰ . If the source voltage is sinusoidal, then the current is also sinusoidal but shifted in phase. The instantaneous power defined as the product of the instantaneous voltage and current can also be seen to be sinusoidal in time. However, in contrast to the resistive load, the instantaneous power in the inductor goes negative for part of the cycle of the source driving it.
As, \(V(t)={ V }_{ m }\sin { \omega t } \)
 \(I(t)=-{ I }_{ m }cos\omega t\)
instantaneous power P_i = V(t).I(t)
\(={ V }_{ m }sinn\omega t\times (-{ I }_{ m }cos\omega t)\)
 \(=-\frac { { V }_{ m }{ I }_{ m } }{ 2 } \times 2sin\omega tcos\omega t\)
 \(=-\frac { { V }_{ m }{ I }_{ m } }{ 2 } [sin\omega t+sin\ 0]\)
 \(=-\frac { { V }_{ m }{ I }_{ m } }{ 2 } sin2\omega t\)
The variation of P_i with \(\omega t\) is as given in the figure.

The instantaneous power altertes positive and negative at twice the ferquency of source supplying it.

Figure shows the variation of I_m with \(\omega \) in a L-C-R series circuit for two values of resistance R₁ and R₂(R₁>R₂).

The condition for resources in the L-C-R circuit is,
X_L = X_C
\({ \omega }_{ 0 }L\ =\ \frac { 1 }{ { \omega }_{ 0 }C } \Rightarrow { \omega }_{ 0 }^{ 2 }=\frac { I }{ LC } \Rightarrow { \omega }_{ 0 }=\frac { 1 }{ \sqrt { LC } } \)
We see that, the current amplitude is maximum at the resonance frequency \({ \omega }_{ 0 }\) .
 Since, \({ I }_{ m }=\frac { { V }_{ m } }{ R } \) ar resonance, the current amplitude for case R₂
 is sharper to that for case R₁.
Quality factor or simply the Q-factor of a resonant L-C-R circuit is defined as the ratio of voltage drop across the capacitor (or inductor) to that of applied voltage.
It is given by \(Q=\frac { 1 }{ R } \sqrt { \frac { L }{ C } } \)
The Q-factor determines the sharpness of the resonance curve and if the resonance is less sharp, not only the maximum current becomes less, but also the circuit will be closed to the resonance for a larger range \(\triangle \omega \) frequencies and the tuning of the circuit will not be good. So, less sharp the resonance, less in the selectivity of the circuit while higher is the Q, sharper is the resonance curve and lesser will be loss in energy of the circuit.

\( { I }_{ v }=\frac { { E }_{ v } }{ Z } =\frac { 130 }{ 2 } \ \ ...(i)\)
\(Since \ \ Z=\sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } =\sqrt { { \left( 12 \right) }^{ 2 }{ +\left( 2\pi vL \right) }^{ 2 } }\)
\(=\sqrt { { \left( 12 \right) }^{ 2 }+{ \left( 2\pi \times 50\times \frac { 0.05 }{ \pi } \right) }^{ 2 } } \)
\(or \ Z=\sqrt { 144+{ \left( 5 \right) }^{ 2 } } =13\Omega \)
\(So \ \ { I }_{ v } \ =\frac { 130 }{ 13 } =10 \ A\)
\(\\ And \ \ { E }_{ L }={ X }_{ L }I=5\times 10=50 \ V\)