Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 7 - Alternating Current

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Page No 40:

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{3 / 2} A (c) 5 / 6 A (d) 5 / \sqrt{2} A

Answer:

Frequency, f = 50 Hz
RMS current, I_rms = 5A
Time, t = $\frac{1}{300} s$
Instantaneous current, I = I₀sinωt
$\Rightarrow I = I_{o} \sin (ω . t) = \sqrt{2} I_{rms} \sin (2 π f . t) (I_{o} = \sqrt{2} I_{rms} and ω = 2 π f) \Rightarrow I = \sqrt{2} \times 5 \sin (2 π \times 50 \times \frac{1}{300}) \Rightarrow I = \sqrt{2} \times 5 \times \frac{\sqrt{3}}{2} = 5 \sqrt{\frac{3}{2}} A$
Hence, the correct answer is option (B).

Page No 40:

Question 2:

An alternating current generator has an internal resistance Rg and an internal reactance Xg. It is used to supply power to a passive load consisting of a resistance Rg 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.

(d) R_g.

Answer:

For maximum power to be delivered from generator to load, the total reactance must be zero.
i.e., x_L + x_g_{= 0

⇒}x_L = – x_g
Hence, the correct answer is option (C).

Page No 41:

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.

\sqrt{< v^{2} >} .

(d) the pointer of the meter is stuck by some mechanical defect.

Answer:

The voltmeter connected to A-C mains reads mean value (<v²>) and is caliberated in such a way that it gives value of <v²>, which is multiplied by form factor to give 'rms' value.

Hence, the correct answer is option (C).

Page No 41:

Question 4:

To reduce the reasonant 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.

(d) dielectric in the capacitor should be removed.

Answer:

Resonant frequency for an LCR circuit,
$f_{0} = \frac{1}{2 π \sqrt{L C}}$

To reduce the resonant frequency either inductance (L) or capacitance (C) should be increased.
In the given case, if we add another capacitor in parallel, the net capacitance will increase and hence the resonant frequency will decrease.

Hence, the correct answer is option (B).

Page No 41:

Question 5:

Which of the following combinations should be selected for better tuning of an LCR circuit used for communication?

(a) R = 20 â„¦, L = 1.5 H, C = 35μF.

(b) R = 25 â„¦, L = 2.5 H, C = 45μF.

(d) R = 25 â„¦, L = 1.5 H, C = 45μF.

Answer:

For better tuning of L-C-R circuit. The quality factor should be as high as possible.
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
where, R is resistance
L is the inductance
C is the capacitance
For 'Q' (quality factor) to be high:
'R' should be low
'L' should be high and 'C' should be low.

Hence, the correct answer is option (C).

Page No 41:

Question 6:

An inductor of reactance 1 â„¦ and a resistor of 2 â„¦ 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.

(d) 18 W.

Answer:

Given,
X_L= 1Ω
R = 2Ω
E_rms = 6 V
Here, I_rms= $\frac{E_{rms}}{Z}$
where, Z is the impedance.
$Z = \sqrt{X_{L}^{2} + R^{2}} = \sqrt{1^{2} + 2^{2}} = \sqrt{5} \Rightarrow I_{rms} = \frac{6}{\sqrt{5}} A \Rightarrow cosϕ = \frac{R}{Z} = \frac{2}{\sqrt{5}}$
Average power,
$P_{av} = I_{rms} E_{rms} cosϕ$
$P_{av} = \frac{6}{\sqrt{5}} \times 6 \times \frac{2}{\sqrt{5}} = 14.4 W$
Hence, the correct answer is option (C).

Page No 41:

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

(a) 1 / \sqrt{2} A . (b) \sqrt{2} A . (c) 2 A . (d) 2 \sqrt{2} A .

Answer:

Given,
Secondary voltage, V_s= 24 V
Power associated with secondary, P_s = 12 V
P_s= I_sV_s
$\Rightarrow I_{s} = \frac{P_{s}}{V_{s}} = \frac{12}{24} = \frac{1}{2} A$
Peak value of the current,
$I_{0} = I_{s} \sqrt{2} = \frac{1}{2} \times \sqrt{2} = \frac{1}{\sqrt{2}} A$
Hence, the correct answer is option (A).

Page No 42:

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.

(d) Resistor, inductor and capacitor.

Answer:

Impedance of a series LCR circuit containing resistor, inductor and capacitor is given by,
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$\Rightarrow Z = \sqrt{R^{2} + {(2 π f L - \frac{1}{2 π f C})}^{2}}$
At resonance frequency, Z is minimum and current is maximum.
For any other value of frequency, the value of current will decrease (when compared to current at resonance).
Clearly the circuit must have capacitor and inductor. However, it may be possible that the circuit also has a resistor besides inductor & capacitor.

Hence, the correct options are (a) and (d).

Page No 42:

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.

(d) Only a capacitor.

Answer:

According to question, the value of current is increasing with the increase in frequency. That is possible for a capacitive circuit.
$X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C} \Rightarrow So, current I = \frac{E}{X_{C}}$

Here, $f ↑, X_{C} ↓$ and $I ↑$
This is also true for an R-C circuit.

$Z = \sqrt{R^{2} + X_{C}^{2}} So, I = \frac{E}{Z}$
(Here, all symbols have their usual meaning.)

Hence, the correct options are (C) and (D).

Page No 42:

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.

(d) It is easy to reduce the voltage at the receiving end using step-down transformers.

Answer:

Power, P = E_rmsI_rms.
For a given power level, if 'E_rms' is high then 'I_rms' will be low.
Also, power loss = $I_{rms}^{2} R$ ...(1)
If I_rmsis low then power loss is also low. (from equation 1)
The high alternating voltage can be easily reduced by using step down transformer at the receiving end.

Hence, the correct options are (A), (B) and (D).

Page No 42:

Question 11:

For an LCR circuit, the power transferred from the driving source to the driven oscillator is P = I²Z cos Ï•.

(a) Here, the power factor cos Ï• ≥ 0, P ≥ 0.

(b) The driving force can give no energy to the oscillator (P = 0) in some cases.

(d) The driving force can take away energy out of the oscillator.

Answer:

According to the question,
P = I²Z cos Ï•
where, I = Current
Z = Impedance
cos Ï• = Power factor
Power factor, $\cos ϕ = \frac{R}{Z}$
where R > 0 and Z > 0
⇒ cos Ï• > 0
⇒ P > 0
Hence, the correct options are (A), (B) and (C).

Page No 42:

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.

(d) power delivered to the capacitor is zero.

Answer:

The maximum value of voltage,
$V_{peak} = \sqrt{2} V_{rms} = \sqrt{2} \times 220 V = 311.08 V$
V_rmsI_rms As the circuit is pure capacitive, so the current developed leads the applied voltage by a phase angle of 90°.
Thus, the power delivered,
P_av = V_rmsI_rmscos90° = 0
When an AC voltage of 220 V is applied to a capacitor C, the charge on the plates is in phase with the applied voltage.

Hence, the correct options are (C) and (D).

Page No 43:

Question 13:

The line that draws power supply to your house from street has

(a) zero average current.

(b) 220 V average voltage.

(d) voltage and current possibly differing in phase

ϕ

such that

|ϕ| < \frac{π}{2} .

Answer:

For household supplies, AC currents are used which are having zero average value over a cycle.
The line is having some resistance so power factor, $\cos ϕ = \frac{R}{Z} \neq 0$
So, $ϕ \neq \frac{π}{2} \Rightarrow |ϕ| < \frac{π}{2}$
i.e., phase lies between $0 & \frac{π}{2}$ .
Hence, the correct options are (A) and (D).

Page No 43:

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:

The electrostatic energy $\frac{1}{2} C V^{2}$ is analogous to potential energy.
Energy associated with moving charges (current) that is magnetic $(\frac{1}{2} L I^{2})$ is analogous to kinetic energy.

Page No 43:

Question 15:

Draw the effective equivalent circuit of the circuit shown in the given figure, at very high frequencies and find the effective impedance.

Answer:

Inductive reactance, $X_{L} = 2 π f L$
Capacitive reactance, $X_{C} = \frac{1}{2 π f C}$
For very high frequencies
$\Rightarrow X_{L} \to \infty$ and $X_{C} \to 0$
For infinite resistance, the circuit can be considered as open circuit.
If the reactance of circuit is zero, it will be considered as short circuited.

Effective impedance, Z = R₁ + R₃

Page No 43:

Question 16:

Study the circuits (a) and (b) shown in the given figure 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:

Let the supply voltage be V.
For pure resistive circuit:
$I = \frac{V}{R}$
For series LCR circuit:
$I' = \frac{V}{Z} = \frac{V}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} (For, LCR circuit, Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}})$
(a).
If capacitive reactance and inductive reactance in circuit (b) becomes equal.
i.e., X_L = X_C
Then, $I' = \frac{V}{Z} = \frac{V}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} = \frac{V}{R} = I$
Then, same current will from in both circuit (a) and (b)

(b).
No, the rms current in circuit (b) can not be larger than circuit (a).

Page No 43:

Question 17:

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

Answer:

Let the applied e.m.f., E = E₀â€‹sin(ωt)
I = I₀â€‹sin(ωt−Ï•)
Then, instantaneous power, $P = E I = \frac{E_{0} I_{0}}{2} [cosϕ - \cos (2 ω t + ϕ)]$
where Ï• is the phase difference.
when, cos Ï• < cos (2ωt + Ï•)
Then, P < 0
So, the instantaneous power output source can be negative.

Average power, $P_{av} = \frac{V_{0}}{\sqrt{2}} \frac{I_{0}}{\sqrt{2}} cosϕ$
$\Rightarrow P_{av} = V_{rms} I_{rms} cosϕ$

But, we know that,
$cosϕ = \frac{R}{Z} > 0$
$\Rightarrow P_{av} > 0$
Hence, the average power output of an AC source cannot be negative.

Page No 43:

Question 18:

In series LCR circuit, the plot of I_max vs ω is shown in the given figure. Find the bandwidth and mark in the figure.

Answer:

Bandwidth, $∆ ω = ω_{2} - ω_{1}$

where, $ω_{1} & ω_{2}$ corresponds to the frequencies at which magnitude of current $\frac{1}{\sqrt{2}}$ times of maximum value.
$I_{rms} = \frac{I_{0}}{\sqrt{2}} = \frac{1}{\sqrt{2}} = 0.7 A (maximum current, I_{0} = 1 A)$

From the given diagram,
$∆ ω = ω_{2} - ω_{1} = 1.2 - 0.8$
$∆ ω = 0.4 rad s^{- 1}$

Page No 44:

Question 19:

The alternating current in a circuit is described by the graph shown in the given figure . Show rms current in this graph.

Answer:

Square root of mean of squared value of current is called rms current.

$I_{rms} = \sqrt{\frac{1^{2} + 2^{2}}{2}} \approx 1.6 A$
Desired graph is as given below (horizontal line parallel to time-axis):

Page No 44:

Question 20:

How does the sign of the phase angle Ï•, 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:

For an L-C-R circuit,
Let the phase angle be Ï• and Z be impedence.
And the voltage leads the current in LCR series circuit.
$tanϕ = \frac{X_{L} - X_{C}}{Z}$
$\Rightarrow tanϕ = \frac{2 π f L - \frac{1}{2 π f C}}{Z}$
tan Ï• < 0 (for f < f₀)
tan Ï• > 0 (for f > f₀)
tan Ï• < 0 (for f = f₀)

Page No 44:

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 the given figure.

(a) Which curve shows power consumption over a full cycle?

(b) What is the average power consumption over a cycle?

Answer:

(a) Power is given by product of voltage and current. So it will have maximum amplitude. Thus, the curve A represent the power consumption over a full cycle.
(b) Average power consumption over a complete cycle is zero.
(c) The device X may be L or C or LC.

Page No 44:

Question 22:

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

Answer:

In DC, the current is defined as the ratio between the charge flowing through a circuit in a particular time period.
An AC current changes direction with the source frequency and the attractive force would average to zero. Thus, the a.c ampere must be defined in terms of some property that is independent of the direction of current. Joule’s heating effect is such property and hence it is used to define rms value of a.c

Page No 44:

Question 23:

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

Answer:

Given, L = 10^–2 H
R = 1 Ω
v = 200 V
f = 50 Hz
Impedance, $Z = \sqrt{R^{2} + x_{L}^{2}} = \sqrt{R^{2} + {(2 π f L)}^{2}}$
⇒ Z = 3.3 Ω
$\tan ϕ = \frac{ω L}{R} = 3.14$
⇒ Ï• â‰ƒ 72°
Phase difference, $ϕ = \frac{72 \times π}{180} rad$
Time lag, $∆ t = \frac{ϕ}{ω} = \frac{72 π}{180 \times 2 π \times 50} = \frac{1}{250}$
âˆ† t = 4 × 10^–3 s

Page No 44:

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:

Given, P_s = 60 ω
I_s = 0.54 A
⇒ P_s = V_sI_s
$\Rightarrow V_{S} = \frac{P_{s}}{I_{s}} = \frac{60}{0.54} = 11 V$
Voltage in secondary is less than voltage in primary.
â€‹i.e., Vs < Vp
The transformer is step down transformer.
⇒ Power impact = Power output
$\Rightarrow V_{p} I_{p} = V_{s} I_{s} \Rightarrow I_{p} = \frac{V_{s} I_{s}}{V_{p}} \Rightarrow I_{p} = \frac{111 \times 0.54}{220} = 0.27 A$
Hence, the current drawn in primary coil is 0.27 A

Page No 45:

Question 25:

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

Answer:

Reactance provided by a capacitor is given by, $x_{c} = \frac{1}{ω c} = \frac{1}{2 π f C}$
For a given value of capacitance (C).
We know that, a capacitor does not allow flow of direct current through it as the resistance across the gap is infinite. If an alternating voltage is applied across the capacitor plates, then the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge). Hence, a capacitor will pass more current through it if the voltage is changing at a faster rate, i.e. if the frequency of supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency.

Page No 45:

Question 26:

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

Answer:

Inductive reactance, X_L= ωL
⇒X_L = 2πfL
We know that, an inductor opposes flow of current through it by developing a back emf according to Lenz’s law.
The induced voltage has a polarity so as to maintain the current at its present value. Now if the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice versa. But the induced emf is proportional to the rate of change of current, so it will provide greater reactance to the flow of current if the rate of change is faster, i.e., when frequency increases.
Hence, the reactance of an inductor, is proportional to the frequency, being given by ωL.

Page No 45:

Question 27:

An electrical device draws 2kW power from AC mains

(voltage 223 V (rms) = \sqrt{50, 000} V) .

The current differs (lags) in phase by

ϕ (\tan ϕ = \frac{- 3}{4})

as compared to voltage. Find (i) R, (ii) X_C – X_L, and (iii) I_M. Another device has twice the values for R, X_C and X_L. How are the answers affected?

Answer:

P = 2 kω = 2000 ω
tanÏ• = $- \frac{3}{4}$
V_rms = V = 223 v
Power, $P = \frac{V^{2}}{Z}$
$Z = \frac{V^{2}}{P} = 25 Ω$
Impedance, $\Rightarrow 625 = R^{2} + {(X_{L} - X_{C})}^{2} . . . . . (1)$
But, $tanϕ = \frac{X_{L} - X_{C}}{R} = \frac{3}{4}$
$\Rightarrow X_{L} - X_{C} = \frac{3 R}{4} . . . . . (2)$
Putting values in equation (1)
$625 = R^{2} + {(\frac{3 R}{4})}^{2}$
(a) Resistance, R = $\sqrt{25 \times 16} = 20 Ω$
(b) $X_{L} - X_{C} = \frac{3 R}{4} = 15 Ω$
(c) I_rms= $\sqrt{2} I$ = 12.6 A
If R, X_C, X_L are all doubled, tanÏ• does not change. Z is doubled, current is solved. So power is also solved.

Page No 45:

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 transfomer is used to bring voltage to 220 V.
(ρ_Cu = 1.7 × 10^–8 SI unit)

Answer:

(i) The distance between power station and tower is 10 km. Length of copper wire to used, L = 20 km = 20,000 m.
Resistance of copper wire, $R = ρ \frac{l}{A}$
$\Rightarrow R = \frac{1.7 \times 10^{- 8} . 20, 000}{3.14 \times {(0.5 \times 10^{- 2})}^{2}} = 4 Ω$
p = 10⁶ ω
If, V = 220 V
then, I = $\frac{P}{V} = \frac{10^{6}}{220}$ = 0.45 × 10⁴ A
Power loss = I²R = 0.4 × (0.45 × 10⁴)²> 10⁶ ω
Thus, this method cannot be used for transmission.
(ii) When power is transmitted at voltage, V = 1100 V
Current drawn, I¹ = 91 A
Power loss = (I')²R = (91)² × 4 = 3.3 × 10⁴ ω
Fraction of power loss $= \frac{3.3 \times 10^{4}}{10^{6}} = 3.3 %$

Page No 45:

Question 29:

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

Answer:

In the given figure 'L' & 'C' are connected in series and resistor 'R' is connected in parallel to both.
From the circuit, i = i₁ + i₂
where 'i' is the total current flowing through the circuit.

Applying KVL in lower circuit.
$\frac{q_{2}}{C} + L d i_{2}$
⇒ V_C + V_L = V_m sinωt
$\Rightarrow \frac{q_{1}}{C} + \frac{L d i_{1}}{d t} = V_{m} sinω t \Rightarrow \frac{q_{1}}{C} + \frac{L d^{2} q_{1}}{d t^{2}} = V_{m} sinω t . . . . . (1)$
Let q₁ = q_m sin (ωt + Ï•) .....(2)
i₁= $\frac{d q_{1}}{d t}$ = q_mω²sin(ωt + Ï•) .....(3)
$\frac{d^{2} q_{1}}{d t^{2}}$ = – qmω²sin(ωt + Ï•) .....(4)
Substitute the values of equation (2) & (4) in equation (1)
$\frac{q_{m} \sin (ω t + ϕ)}{c}$ – Lq_mω²sin(ωt + Ï•) = v_msinωt
q_msin(ωt + Ï•) $[\frac{1}{C} - ω^{2} L]$ = v_msinωt
at Ï• = 0
$q_{m} = \frac{V_{m}}{ω [\frac{1}{ω C - ω L}]} . . . . . (5)$
Applying KCL at junction, i = i₁ + i₂
i = $\frac{v_{m} sinω t}{R}$ + q_mωcos (ωt + Ï•)
Using relation (5) for q_mat Ï• = 0
$i = \frac{v_{m} sinω t}{R} + \frac{v_{m} ωcosω t}{ω (\frac{1}{ω c} - ω L)}$
$i = \frac{v_{m}}{R} sinω t + \frac{v_{m}}{(\frac{1}{ω c} - ω L)} cosω t$
$Let, A = \frac{v_{m}}{R} = C cosϕ . . . . . (6) B = \frac{v_{m}}{\frac{1}{ω C} - ω L} = C sinϕ . . . . . (7)$
i = C sin(ωt + Ï•)
Squaring and adding (6), (7)
A² + B² = C²cos²Ï• + C²sin²Ï•
$\Rightarrow C = \sqrt{A^{2} + B^{2}} ϕ = \tan^{- 1} (\frac{B}{A}) = \tan^{- 1} (\frac{v_{m}}{\frac{\frac{1}{ω C - ω L}}{\frac{V_{m}}{R}}}) ∴ tanϕ = \frac{R}{\frac{1}{ω C} - ω L} \Rightarrow C = \sqrt{\frac{V_{m}^{2}}{R^{2}} + \frac{V_{m}^{2}}{{(\frac{1}{ω C_{2}} - ω L)}^{2}}} ∴ i = [\frac{V_{m}^{2}}{R} + \frac{V_{m}}{{(\frac{1}{ω C} - ω L)}^{2}}] \sin (ω t + ϕ) . . . . . (8) And ϕ = \tan^{- 1} \frac{R}{(\frac{1}{ω C} - ω L)} ∴ i = \frac{V}{Z}$
For AC
$i = \frac{V}{Z} \sin (ω t + ϕ) . . . . . (9)$
Comparing (8) & (9)
So, $\frac{1}{Z} = \sqrt{\frac{1}{R^{2}} + \frac{1}{{(\frac{1}{ω C} - ω L)}^{2}}}$

Page No 45:

Question 30:

For an LCR circuit driven at frequency ω, the equation reads

L \frac{d i}{d t} + R i + \frac{q}{C} = v_{i} = v_{m} \sin ω 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) Intergrate the equation over one cycle to find that the phase difference between v and i must be acute.

Answer:

$L \frac{d i}{d t} + R_{i} + \frac{q}{c} = V_{i} = V_{m} sinω t . . . . . (1)$
Multiplying both the sides by i, we get
$L i \frac{d i}{d t} + \frac{q}{c} i + i^{2} R = (V_{m} i) sinω t = V_{i} . . . . . (2)$
where, $L_{i} \frac{d i}{d t} = \frac{d}{d t} (\frac{1}{2} L i^{2})$ = rate of change of energy stored in an inductor
$i^{2} R$ = Joule heating loss,
$\frac{q}{C} i = \frac{d}{d t} (\frac{q^{2}}{2 C})$ = rate of change of energy stored in the capacitor.
V_i= Rate at which driving force pours in energy.
It goes into (i) ohmic loss and (ii) increase of stored energy.
Hence, equation (2) in the form of conservation of energy statement. Integrating both the sides of equation (2) with respect to time over one full cycle (0 → T) we may write.
$\int_{0}^{T} \frac{d}{d t} (\frac{1}{2} L i^{2} + \frac{q^{2}}{2 C}) d t + \int_{0}^{T} R i^{2} d t = \int_{0}^{T} v i d t \Rightarrow 0 + (+ v e) = \int_{0}^{T} v i d t \Rightarrow \int_{0}^{T} v i d t > 0$
If phase difference between v & i is a constant and acute angle.

Page No 46:

Question 31:

In the LCR circuit shown in the given figure, the ac driving voltage is v = v_m sin ωt.

(i) Write down the equation of motion for q (t).

(ii) At t = t₀, 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:

(i) The ac driving voltage in the given LCR circuit,
$v = v_{m} \sin ω t$
At a time t,
$Sum of potential drop across inductor, resistor and the capacitor, L \frac{d^{2} q}{d t^{2}} + R \frac{d q}{d t} + \frac{q}{C} = v_{m} \sin ω t L e t, q = q_{m} \cos (ω t + ϕ) (let, ϕ = phase) Hence, the current at time t, |i| = |\frac{d q}{d t}| = | \frac{d}{d t} [q_{m} \cos (ω t + ϕ)] | = q_{m} ω \sin (ω t + ϕ)$
$i = i_{m} \sin (ω t + ϕ) (Let, i_{m} = q_{m} ω)$
Now for given LCR circuit,
$i_{m} = \frac{v_{m}}{Z} = \frac{V_{m}}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} Hence, \tan ϕ = \frac{(X_{L} - X_{C})}{R}$
(ii) At, t = t_o, resistance R is short circuited.
Due to which energy stored in inductor L:
$U_{L} = \frac{1}{2} L i^{2} = \frac{1}{2} L {[\frac{v_{m}}{Z}]}^{2} = \frac{1}{2} L {[\frac{v_{m}}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}]}^{2} \sin^{2} (ω t_{ο} + ϕ) Energy stored in capacitor : U_{L} = \frac{1}{2} \frac{q^{2}}{C} = \frac{1}{2 C} {[\frac{v_{m}}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}]}^{2} [\frac{1}{ω^{2}} \cos^{2} (ω t_{ο} + ϕ)] (As, ω q_{m} = i_{m})$

(iii) Now further the circuit behaves as an LC oscillator. The capacitor will go on discharge further and all energy will go to inductor L and back and forth continuously.

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