NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current

NCERT Exemplar Class 12 Physics Solutions Chapter 7: MCQI

Question:1

If the RMS current in a 50 Hz ac circuit is 5 A, the value of the current 1/300 seconds after its value becomes zero is
a. $5\sqrt{2}A$
b.$5\sqrt{\frac{3}{2A}}$

c.$\frac{5}{6A}$

d.$\frac{5}{\sqrt{2}A}$

Answer:

The answer is the option (b)
$$\begin{gathered} I=I_0\sin \omega t \\ Itisgiventhat{I}_0=\sqrt{2}\ast 5=5\sqrt{2}A,\nu =50Hz,t=\frac{1}{300s} \\ I=5\sqrt{2}\sin 2\pi \nu t \\ =5\sqrt{2}sin2\pi \ast 50\ast \frac{1}{300} \end{gathered}$$
$$\begin{gathered} =5\sqrt{2}sin\left(\frac{\pi }{3}\right)=5\sqrt{2}\sin {60}^{\circ } \\ =5\sqrt{2}\ast \frac{\sqrt{3}}{2}=5\sqrt{\frac{3}{2}}A \end{gathered}$$

Question:2

An alternating current generator has an internal resistance R_g and an internal reactance X_g. It is used to supply power to a passive load consisting of a resistance R_g and a reactance X_L. For maximum power to be delivered from the generator to the load, the value of X_L is equal to
A. zero.
B. X_g.
C. – X_g.
D. R_g

Answer:

For maximum power to be delivered from the generator to the load
$X_L+X_g=0$(the total reactance should be equal to zero)
$\Rightarrow X_L=-X_g$

Question:3

When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220V. This means
A. input voltage cannot be AC voltage, but a DC voltage.
B. maximum input voltage is 220V.
C. the meter reads not v but <v²> and is calibrated to read $\sqrt{<v^2>}$
D. the pointer of the meter is stuck by some mechanical defect.

Answer:

The answer is the option (c)
This means that the voltage measuring device connected to AC mains is calibrated to read
the rms value $\sqrt{<v^2>}$

Question:4

To reduce the resonant frequency in an LCR series circuit with a generator
A. the generator frequency should be reduced.
B. another capacitor should be added in parallel to the first.
C. the iron core of the inductor should be removed.
D. dielectric in the capacitor should be removed.

Answer:

The answer is the option (b)
$v_0=\frac{1}{2\pi \sqrt{LC}}$
According to the formula to reduce v0 we have to increase either of the L or C.
And for increasing the capacitance or C, another capacitor should be in parallel connection to the first

Question:5

Which of the following combinations should be selected for better tuning of an LCR circuit used for communication?
A. R = 20 Ω, L = 1.5 H, C = 35μF.
B. R = 25 Ω, L = 2.5 H, C = 45μF.
C. R = 15 Ω, L = 3.5 H, C = 30μF.
D. R = 25 Ω, L = 1.5 H, C = 45μF.

Answer:

The quality factor
$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$
where R is resistance, L is the inductance and C is the capacitance of the circuit
For a higher Q, L must be larger, and R and C must be smaller in value.

Question:6

An inductor of reactance 1Ω and a resistor of 2 Ω are connected in series to the terminals of a 6 V (rms) a.c. source. The power dissipated in the circuit is
A. 8 W.
B. 12 W.
C. 14.4 W.
D. 18 W.

Answer:

$X_L=1\mathrm{\Omega },R=2\mathrm{\Omega }{E}_{rms}=6V$
The average power dissipated in the circuit
$$\begin{gathered} {P}_{av}=E_{rms}{I}_{rms}\cos \mathrm{\Phi } \\ {I}_{rms}=\frac{E_{rms}}{Z} \\ Z=\sqrt{R^2+X_L^2}=\sqrt{4+1}=\sqrt{5} \\ {I}_{rms}=\frac{6}{\sqrt{5}}A \\ \cos \mathrm{\Phi }\frac{R}{Z}=\frac{2}{\sqrt{5}} \end{gathered}$$
${P}_{av}=6\ast \frac{6}{\sqrt{5}}\ast \frac{2}{\sqrt{5}}=\frac{72}{5}=14.4W$

Question:7

The output of a step-down transformer is measured to be 24 V when connected to a 12-watt light bulb. The value of the peak current is
$$\begin{gathered} A.1/\sqrt{2}A \\ B.\sqrt{2}A \\ C.2A \\ D.2\sqrt{2}A \end{gathered}$$

Answer:

$$\begin{gathered} {V}_s=24V{P}_s=12W \\ {I}_s{V}_s=12 \\ {I}_s=\frac{12}{V_s}=\frac{12}{24}=0.5ampere \\ {I}_0=I_s\sqrt{2}=0.5\sqrt{2}=\frac{1}{\sqrt{2}}ampere \end{gathered}$$

NCERT Exemplar Class 12 Physics Solutions Chapter 7 MCQII

Question:8

As the frequency of an ac circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit?
A. Inductor and capacitor.
B. Resistor and inductor.
C. Resistor and capacitor.
D. Resistor, inductor and capacitor.

Answer:

For I to be maximum in a circuit, the reactance should be minimum. In the LCR circuit, if the frequency is increased, the inductance will increase while the capacitance decreases.
$Z=\sqrt{R^2+{(X_L-X_C)}^2}=\sqrt{R^2+{(2\pi \nu L-\frac{1}{2\pi \nu C})}^2}$
On increasing v, there will come a stage where we will reach resonance
$(X_L=X_C)$

Question:9

In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?
A. Only resistor.
B. Resistor and an inductor.
C. Resistor and a capacitor.
D. Only a capacitor.

Answer:

The correct answer is the options (c, d)
In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of the supply. When the current increases, the capacitive reactance of circuit decreases, and resistance does not depend on frequency.

Question:10

Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?
A. For a given power level, there is a lower current.
B. Lower current implies less power loss.
C. Transmission lines can be made thinner.
D. It is easy to reduce the voltage at the receiving end using step-down transformers.

Answer:

Since the power is to be transmitted over the large distances at high alternating voltages, the current flowing through the wires will below.
The power loss due to transmission lines with resistance R and current
$I_{rms}$ in the circuit is given by $I_{rms}^{2}R$
$P=E_{rms}{I}_{rms}(I_{rms}islowwhen{E}_{rms}ishigh)$
$Powerloss=I_{rms}^{2}R=low$
The voltage at the receiving end is reduced by step-down transformers.

Question:11

For an LCR circuit, the power transferred from the driving source to the driven oscillator is $P=I^{}2Z\cos \varphi$

A. Here, the power factor cos$\varphi \ge 0,P\ge 0.$
B. The driving force can give no energy to the oscillator (P = 0) in some cases.
C. The driving force cannot syphon out (P < 0) the energy out of oscillator.
D. The driving force can take away energy out of the oscillator.

Answer:

The answers are the options (a, b, c)
According to the question
$P=I^{}2Z\cos \varphi$
Power factor
$\cos \varphi =\frac{R}{Z}AsR>0andZ>0$
$$\begin{gathered} Thus\cos \varphi =\frac{R}{Z}Ispositive \\ andwhere\varphi =\frac{\pi }{2},P=0 \end{gathered}$$

Question:12

When an AC voltage of 220 V is applied to the capacitor C
A. the maximum voltage between plates is 220 V.
B. the current is in phase with the applied voltage.
C. the charge on the plates is in phase with the applied voltage.
D. power delivered to the capacitor is zero.

Answer:

The correct answer are the options (c, d) On connecting the capacitor to ac supply the plate of the capacitor connected to the +ve terminal is at higher potential compared to the plate connected to negative terminal.
The power applied to the circuit is given by
$$\begin{gathered} P_{av}=V_{rms}{I}_{rms}\cos \varphi \\ \varphi ={90}^{\circ }forpurecapacitorcircuit \\ {P}_{av}=V_{rms}{I}_{rms}\cos {90}^{\circ }=0 \end{gathered}$$

Question:13

The line that draws power supply to your house from street has
A. zero average current.
B. 220 V average voltage.
C. voltage and current out of phase by 90°.
D. voltage and current possibly differing in phase $\varphi$ such that $\left|\varphi \right|<\frac{\pi }{2}$

Answer:

The correct answer are the options (a,d)
It is a known fact that AC supply is used in houses.
The average current over a cycle of AC is zero. L and C are connected in household circuit so R and Z cannot be equal.
$SoPowerfactor=\cos \varphi =\frac{R}{Z}\ne 0=\varphi \ne \frac{\pi }{2}$
$\varphi <\frac{\pi }{2}i.e.phaseanglebetweenvoltageandcurrentliesbetween0and\frac{\pi }{2}$

NCERT Exemplar Class 12 Physics Solutions Chapter 7: Very Short Answer

Question:14

If a LC circuit is considered analogous to a harmonically oscillating spring block system, which energy of the LC circuit would be analogous to potential energy and which one analogous to kinetic energy?

Answer:

If a L-C circuit is considered analogous to a harmonically oscillating spring block system, energy due to motion of charge particle i.e., magnetic energy is analogous to kinetic energy and electrostatic energy due to charging of capacitor is analogous to potential energy.

Question:15

Draw the effective equivalent circuit of the circuit shown in Fig 7.1, at extremely high frequencies and find the effective impedance.

Answer:

Inductive reactance
$X_L=\omega L=2\pi fL$
At very high frequencies, $X_L$ will be high or L can be considered an open circuit for the high frequency of ac circuit.
Capacitive reactance
$X_C=\frac{1}{2\pi \nu C}$
At high frequency, $X_C$will below.
Therefore reactance of capacitance can be considered negligible and capacitor can be considered short-circuited.

Question:16

Study the circuits (a) and (b) shown in Fig 7.2 and answer the following questions.

(a) Under which conditions would the rms currents in the two circuits be the same?
(b) Can the rms current in circuit (b) be larger than that in (a)?

Answer:

$I_{rms}in(a)=\frac{V_{rms}}{R}$
$$\begin{gathered} {I}_{rms}in(b)=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C)}} \\ (a)Now(I_{rms}{)}_a=(I_{rms}{)}_b \\ \frac{V_{rms}}{R}=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C{)}^2}} \\ R=\sqrt{R^2+(X_L-X_C)} \end{gathered}$$
Squaring both sides
$$\begin{gathered} {R}^2=R^2+{(X_L-X_C)}^2 \\ Or{(X_L-X_C)}^2=0 \\ {X}_L=X_C \end{gathered}$$
So, $I_{rms}$ in circuits a and b will be equal if ${X}_L=X_C$
$(b)For(I_{rms}{)}_b>(I_{rms}{)}_a$

$$\begin{gathered} \frac{V_{rms}}{\sqrt{R^2+(X_L-X_C)}}>\frac{V_{rms}}{R} \\ As{V}_{rms=V} \\ So\sqrt{R^2+(X_L-X_C)}<R \\ Squaringbothsides{R}^2+(X_L-X_C{)}^2<R^2 \end{gathered}$$
$(X_L-X_C{)}^2<0$
Square of a number can'tbe negative.
Therefore, the rms current in circuit (b) has to be larger than that in (a).

Question:17

Can the instantaneous power output of an ac source ever be negative? Can the average power output be negative?

Answer:

The instantaneous power output of an AC source can be negative because there are chances for power to get absorbed at some point of time. However, the average output over a cycle cannot be negative at any instant. It can only be either positive or zero as in any LC circuit.
$$\begin{gathered} Lettheappliede.m.f=E=E_0\sin \omega t \\ I=I_0\sin (\omega t-\varphi ) \\ InstantaneouspoweroutputofacsourceP=EI \\ {E}_0\sin \omega t{I}_0\sin (\omega t-\varphi )=E_0{I}_0\sin \omega t[\sin \omega t\cos \varphi -\cos \omega t\sin \varphi ] \\ =E_0{I}_0[\frac{1-\cos 2\omega t}{2}\cos \varphi -\frac{1}{2}\sin 2\omega t\sin \varphi ] \end{gathered}$$
$$\begin{gathered} =\frac{E_0{I}_0}{2}[\cos \varphi -\cos 2\omega t\cos \varphi -\sin 2\omega t\sin \varphi ] \\ =\frac{E_0{I}_0}{2}[\cos \varphi -(\cos 2\omega t\cos \varphi +\sin 2\omega t\sin \varphi )] \\ P=\frac{E_0{I}_0}{2}[\cos \varphi -\cos (2\omega t-\varphi )] \end{gathered}$$
The phase angle taken φ, ±ve

Question:18

In series LCR circuit, the plot of $I_{max}vs\omega$ is shown in Fig 7.3. Find the bandwidth and mark in the figure.

Answer:

$I=\frac{E_0}{\sqrt{2}}=\frac{1}{\sqrt{2}}=0.707Amp$
From graph we get
${\omega }_1=0.8\frac{rad}{s}and{\omega }_2=1.2\frac{rad}{s}$
$Bandwidth=1.2-0.8=0.4rad/sec$

Question:19

The alternating current in a circuit is described by the graph shown in Fig 7.4. Show RMS current in this graph.

Answer:

$I_{rms}=\sqrt{\frac{(1^2+2^2)}{2}}=\sqrt{\frac{5}{2}}\cong 1.6A$

Question:20

How does the sign of the phase angle $\theta$, by which the supply voltage leads the current in an LCR series circuit, change as the supply frequency is gradually increased from very low to very high values.

Answer:

Phase angle $\theta$ by which Voltage leads current is given by,

$\tan \theta =\frac{X_L-X_C}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}$
For low frequencies, $X_L<X_C\Rightarrow \tan \theta <0$
For resonance

$X_L<X_C\Rightarrow \tan \theta =0(\nu ={\nu }_0=\frac{1}{2\pi \sqrt{2C}})$

NCERT Exemplar Class 12 Physics Solutions Chapter 7: Short Answer

Question:21

A device ‘X’ is connected to an a .c source. The variation of voltage, current and power in one complete cycle is shown in Fig 7.5.

(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device ‘X’.

Answer:

(a) $\text{Power (P)}=\text{VI}$
Power consumption will have the maximum amplitude between the three curves.
(b) Average Power consumption is zero covering a positive and a similar negative area in a complete cycle.

(c) For the given circuit, the phase difference between V and I is $\frac{\pi }{2}$. The device ‘X’ can be an inductor or capacitor or a series combination of inductors and capacitors.

Question:22

Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current?

Answer:

In a DC circuit, 1 Ampere is defined as 1 Coulomb per second, but for AC circuit, the direction of current keeps changing. Instead, Joule’s heating is used to define the rms value of AC current. This is done because Joule’s heating is independent of the direction of the current. RMS value of AC is equal to the value of DC required to generate the same amount of heat through a given resistor in a given time.

Question:23

A coil of 0.01 henry inductance and 1-ohm resistance is connected to 200 volts, 50 Hz ac supply. Find the impedance of the circuit and time lag between max. alternating voltage and current.

Answer:

$Inductance(L)=0.01H$
$Resistance(R)=1\mathrm{\Omega }$
$Voltage(V)=200V$
$Frequency(f)=50Hz$
$X_L=2\pi fL=3.14\mathrm{\Omega }$
$z=\sqrt{R^2+X_L^2}=\sqrt{10.86}=3.3\mathrm{\Omega }$
$\tan \varphi =\frac{\omega L}{R}=\frac{2\pi fL}{R}=3.14$
$\text{Phase different}(\varphi )={\tan }^{-1}3.14={72}^o=\frac{72}{180}\ast \pi$
$\text{Time lag}(\mathrm{\Delta }t)=\frac{\varphi }{\omega }=\frac{\frac{72\pi }{180}}{2\pi \ast 50}=0.004s$

Question:24

A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of transformer being used.

Answer:

$P_s=60W$
$I_s=0.54A$
$V_s=\frac{P_s}{I_s}=\frac{60}{0.54}=110V$
$V_P=220V$
As, $V_P>V_s,$ it is a step down transformer
$r=\frac{V_s}{V_p}=\frac{I_P}{I_S}$
$I_P=0.54\ast \frac{110}{220}=0.27A$

Question:25

Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.

Answer:

Under the effect of AC current, capacitor plates charge and discharge alternately. Current is a direct result of this charging and discharging. On increasing the frequency of current, the capacitor will pass more current through it.

Question:26

Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.

Answer:

Under the flow of current, Inductor develops a back emf. On decreasing the current, the induced emf will be such that it tries to maintain the existing flow of current. Induced emf is directly proportional to the rate of change of current.

NCERT Exemplar Class 12 Physics Solutions Chapter 7: Long Answer

Question:27

An electrical device draws 2kW power from AC mains $(Voltage223V(rms)=\sqrt{50,000}v)$ The current differs (lags) in phase by $\varphi (\tan \varphi =\frac{-3}{4})$ as compared to voltage. Find (i) R, (ii) XC – XL, and (iii) I_M.

Answer:

Impedance,
$Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}$
$P=2000W$
$\tan \varphi =-\frac{3}{4}$
$V_{rms}=223V$
$P=\frac{V^2}{Z}$
$Z=25\mathrm{\Omega }$
$\frac{X_c-X_L}{R}=-\frac{3}{4}$
$X_L-X_C=-\frac{3}{4}R$
$R^2+(X_L-X_C{)}^2=Z^2$
$R^2+\frac{9}{16}{R}^2={25}^2$
$R^2=625\ast \frac{16}{25}=400$
$R=20\mathrm{\Omega }$
$X_C-X_L=-\frac{3}{4}\ast 20=-15\mathrm{\Omega }$
$I_M=\sqrt{2}I=\sqrt{2}\frac{V}{Z}=12.6A$

Question:28

1MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if

(i) power is transmitted at 220V. Comment on the feasibility of doing this.
(ii) a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transformer is used to bring voltage to 220 V.
$({\rho }_{cu}=1.7\times {10}^{-8}SIunit)$

Answer:

$L=20km=20000m$
$R={\rho }_{cu}\frac{L}{A}=\frac{1.7\ast {10}^{-8}\ast 2\ast {10}^4}{\pi (0.5\ast {10}^{-2}{)}^2}=4\mathrm{\Omega }$
$I=\frac{{10}^6}{220}=0.45\ast {10}^{4}A$
Power loss = $I^{2}R=4\ast {0.45}^2\ast {10}^8>{10}^{6}W$
ii) $P={10}^6$
$V^{\prime }=11000V$
$I^{\prime }=\frac{P}{V^{\prime }}=\frac{1}{1.1}\ast {10}^2$
Power Loss =
$I^{\prime 2}R=\frac{1}{1.21}\ast 4\ast {10}^4=3.3\ast {10}^{4}W$
Fraction of power loss =
$\frac{3.3\ast {10}^4}{{10}^6}=3.3{}^o{/}_o$

Question:29

Consider the LCR circuit shown in Fig 7.6. Find the net current i and the phase of i. Show that $i=\frac{v}{Z}$. Find the impedance Z for this circuit.

Answer:

Taking Resistor andAlternating EMF circuit, $i=i_1+i_2$
$i_{2}R=V_m\sin \omega t\Rightarrow i_2=\frac{V_m\sin \omega t}{R}$
Taking Capacitor-Inductor and Alternating EMF circuit,
$V_m\sin \omega t=\frac{q_1}{C}+L\frac{d^{2}q}{d{t}^2}$
Assuming $q_1=q_m\sin (\omega t+\varphi )$
$\frac{d{q}_1}{dt}=\omega q_m\cos (\omega t+\varphi )$
$\frac{d^2{q}_1}{d{t}^2}=-{\omega }^2{q}_m\sin (\omega +\varphi )$
$V_m\sin \omega t=\frac{q_m\sin (\omega t+\varphi )}{C}+L(-{\omega }^2{q}_m\sin (\omega t+\varphi ))$
$V_m\sin \omega t=q_m(\frac{1}{C}-{\omega }^{2}L)\sin (\omega t+\varphi )$
$q_m=\frac{V_m}{\frac{1}{C}-{\omega }^{2}L}$
$\sin \omega t=\sin (\omega t+\varphi )$
$\varphi =0$
$i_1=\frac{d{q}_1}{dt}=\omega \left(\frac{V_m}{\frac{1}{C}-{\omega }^{2}L}\right)\cos (\omega t)$
$i=i_1+i_2=\left(\frac{V_m}{\frac{1}{\omega C}-\omega L}\right)\cos (\omega t)+\frac{V_m(\sin \omega t)}{R}$
$=A\sin (\omega t+\varphi )$
$A=\sqrt{({\left(\frac{V_m}{(\frac{1}{\omega C}-\omega L)}\right)}^2+{\left(\frac{V_m}{R}\right)}^2)}=V_m\sqrt{({\left(\frac{V_m}{(\frac{1}{\omega C}-\omega L)}\right)}^2+{\left(\frac{1}{R}\right)}^2)}$
$\tan \varphi =\frac{R}{\frac{1}{\omega C}-\omega L}$
$\varphi ={\tan }^{-1}\left(\frac{R}{\frac{1}{\omega C}-\omega L}\right)$

Question:30

For an LCR circuit driven at frequency $\omega$, the equation reads
$L\frac{di}{dt}+Ri+\frac{q}{C}=v_1=v_m\sin \omega t$
(i) Multiply the equation by i and simplify where possible.
(ii) Interpret each term physically.
(iii) Cast the equation in the form of a conservation of energy statement.
(iv) Integrate the equation over one cycle to find that the phase difference between v and i must be acute.

Answer:

$V=V_m\sin \omega t$
$L\frac{di}{dt}+iR+\frac{q}{C}=V_m\sin \omega t$
$Li\frac{di}{dt}+i^{2}R+\frac{q}{C}i=i{V}_m\sin \omega t$
$Li\frac{di}{dt}+i^{2}R+\frac{q}{C}i=iV$
Power loss = $i^{2}R$
Rate of change of energy stored in Inductor =
$Li\frac{di}{dt}=\frac{d}{dt}\left(\frac{1}{2}L{i}^2\right)$
Rate of change of energy stored in Capacitor =
$\frac{q}{C}i=\frac{d}{dt}\left(\frac{q^2}{2C}\right)$
${\int }_0^T\frac{d}{dt}(\frac{1}{2}L{i}^2+\frac{q^2}{2C})dt+{\int }_0^T{i}^{2}Rdt={\int }_0^{T}iVdt$
$0+(+ve)={\int }_0^{T}iVdt>0$

Question:31

In the LCR circuit shown in Fig 7.7, the ac driving voltage is $v=v_m\sin \omega t$

(i) Write down the equation of motion for q (t).
(ii) At $t=t_0$, the voltage source stops and R is short circuited. Now write down how much energy is stored in each of L and C.
(iii) Describe subsequent motion of charges.

Answer:

a) $V=V_m\sin \omega t$
$L\frac{di}{dt}+iR+\frac{q}{C}=V_m\sin \omega t$
$L\frac{d^{2}q}{d{t}^2}+\frac{dq}{dt}R+\frac{q}{C}=V_m\sin \omega t$
b) Let $q=q_m\sin (\omega t+\varphi )$
$\frac{dq}{dt}=\omega q_m\cos (\omega t+\varphi )$
$i_m=\frac{V_m}{Z}=\frac{V_m}{\sqrt{(R{)}^2+(X_c-X_L{)}^2}}$
$\varphi ={\tan }^{-1}\left(\frac{X_c-X_L}{R}\right)$
At $t=t_0,$ Resistance is short - circuited and the inductor and capacitor store energy
$U_L=\frac{1}{2}L{i}^2=\frac{1}{2}L{\left[\frac{V_m}{\sqrt{(R{)}^2+(X_c-X_L{)}^2}}\right]}^2{\sin }^2(\omega t_0+\varphi )$
$U_c=\frac{q^2}{2C}=\frac{1}{2C{\omega }^2}{\left[\frac{V_m}{\sqrt{(R{)}^2+(X_C-X_L{)}^2}}\right]}^2{\cos }^2(\omega t_0+\varphi )$
(c) When R is short-circuited, the circuit becomes an L-C oscillator. When the capacitor discharges, all its energy goes from capacitor to inductor. Energy oscillates from electrostatic to magnetic and from magnetic to electrostatic.