Solutions-Class-12-Physics-Chapter-13-AC Circuits-MSBSHSE

(B) 5\(\sqrt{\frac{3}{2}}\) A

(A) \(\frac{1}{\sqrt{2}}\)

(B) \(\frac{1}{2πf(2πfL-R)}\)

(C) 5625W

(B) 0.707

Capacitive reactance: Xc = \(\frac{1}{2πfC}\)

The capacitive reactance is inversely proportional to the capacitance. The capacitive reactance decreases as the capacitance increases. The current flow increases as capacitive reactance decreases. The brightness is proportional to the magnitude of the current, therefore the brightness of the bulb increases as the current increases. Therefore it can be concluded that the brightness of the bulb increases as the capacitance increases.

For an LR circuit, the impedance, Z_LR = \(\sqrt{R^2+X_L^2}\), where X_L is the reactance of the inductor.

When a capacitor of capacitance C is added in series with L and R, the impedance,

Z_LCR = \(\sqrt{R^2+(X_L-X_C)^2}\) because in the case of an inductor the current lags behind the voltage by a phase angle of π/2 rad while in the case of a capacitor the current leads the voltage by a phase angle of π/2 rad. The decrease in net reactance

decreases the total impedance (Z_LCR < Z_LR).

The reactance of a capacitor is Xc = \(\frac{1}{2πfC}\), where f is the frequency of the AC supply and C is the capacitance of the capacitor. For very high frequency, f, Xc is very small. Hence, for very high frequency AC supply, a capacitor behaves like a pure conductor.

Wattless current : The current that does not lead to energy consumption, hence zero power consumption, is called wattless current.

In the case of a purely inductive circuit or a purely capacitive circuit, average power consumed over a complete cycle is zero and hence the corresponding alternating current in the circuit is called wattless current.

The natural frequency of LC circuit is \(\frac{1}{2π\sqrt{LC}}\)

where L is the inductance and C is the capacitance.

The reactance of this circuit at this frequency is \(\frac{1}{2πfC-\frac{1}{2πfL}}\) = \(\frac{1}{\frac{2πC}{2π\sqrt{LC}}-\frac{1}{\frac{2πL}{2π\sqrt{LC}}}}\) = \(\frac{1}{\sqrt{\frac{C}{L}}-\sqrt{\frac{C}{L}}}\) = \(\frac{1}{0}\) = ∞

For a series LR circuit, power factor,

cos Φ = \(\frac{R}{\sqrt{R^2+X_L^2}}\)

If X_L = R, power factor P₁ = \(\frac{R}{\sqrt{2R^2}}\) = \(\frac{1}{\sqrt{2}}\)

For a series LCR circuit, power factor,

cos Φ = \(\frac{R}{\sqrt{R^2+(X_L-X_C)^2}}\)

If X_L = Xc, power factor P₂ = \(\frac{R}{R}\) = 1

∴ P₁/P₂ = \(\frac{1}{\sqrt{2}}\)

In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of π/2 rad. Here, for e = e₀ sin ωt, we have,

i  = i₀ sin(ωt − π/2)

Instantaneous power,

P = ei = (e₀ sin ωt)[i₀(sin ωt cos π/2 − cos ωt sin π/2)]

          = − e₀i₀ sin ωt cos ωt  (… as cos π/2 =0 and sin π/2 =1)

Average power over one cycle,

P_av = \(\frac{\text{work done in one cycle}}{\text{time for one cycle}}\)

     = \(\frac{\int_{0}^{T}P\,dt}{T}\) = \(\frac{-\int_{0}^{T}e_0i_0\,sin\,ωt\,cos\,ωt\,dt}{T}\)

     = \(-\frac{e_0i_0}{T}\int_{0}^{T}sin\,ωt\,cos\,ωt\,dt\)

Now, \(\int_{0}^{T}sin\,ωt\,cos\,ωt\,dt\) = \(\frac{1}{2}\int_{0}^{T}sin\,2ωt\,dt\)

                                                              = \(\frac{1}{2}[\frac{-cos\,2ωt}{2ω}]_0^T\) 

                                                              = \(-\frac{1}{4ω}[cos\,2\frac{2π}{T}-cos\,0]\) 

                                                              = \(-\frac{1}{4ω}[1-1]=0\) 

 ∴ P_av = 0 

In an AC circuit containing only an ideal capacitor, the current i leads the emf e by a phase angle of π/2 rad.

Here, for e = e₀ sin ωt, we have, i  = i₀ sin(ωt + π/2)

Instantaneous power,

P = ei = (e₀ sin ωt)[i₀(sin ωt cos π/2 + cos ωt sin π/2)]

=  e₀i₀ sin ωt cos ωt  (… as cos π/2 =0 and sin π/2 =1)

Average power over one cycle,

P_av = \(\frac{\text{work done in one cycle}}{\text{time for one cycle}}\)

     = \(\frac{\int_{0}^{T}P\,dt}{T}\) = \(\frac{\int_{0}^{T}e_0i_0\,sin\,ωt\,cos\,ωt\,dt}{T}\)

     = \(\frac{e_0i_0}{T}\int_{0}^{T}sin\,ωt\,cos\,ωt\,dt\)

Now, \(\int_{0}^{T}sin\,ωt\,cos\,ωt\,dt\) = \(\frac{1}{2}\int_{0}^{T}sin\,2ωt\,dt\)

                                                              = \(\frac{1}{2}[\frac{-cos\,2ωt}{2ω}]_0^T\) 

                                                              = \(-\frac{1}{4ω}[cos\,2\frac{2π}{T}-cos\,0]\) 

                                                              = \(\frac{1}{4ω}[1-1]=0\) 

  ∴ P_av = 0, i.e., the circuit does not dissipate power.

(a) Let e = e₀ sin ωt be the alternating emf applied across the series combination of pure inductor, capacitor and resistor as shown in Fig.

There is a phase difference φ between the applied emf and current given by

i = i₀ sin (ωt ± Φ )

Instantaneous power is given by

P = ei = (e₀ sin ωt)[ i₀ sin (ωt ± Φ )]

= (e₀ sin ωt)[i₀(sin ωt cos Φ ± cos ωt sin Φ)]

= e₀i₀ sin ωt (sin ωt cos Φ ± cos ωt sin Φ)

= e₀i₀ cos Φ sin²ωt ± e₀i₀ sin Φ sin ωt cos ωt

Average power over one cycle,

P_av = \(\frac{\text{work done in one cycle}}{\text{time for one cycle}}\)

     = \(\frac{\int_{0}^{T}P\,dt}{T}\)

     =   \(\frac{\int_{0}^{T}[e_0i_0cos\,Φ\,sin^2 ωt±e_0i_0sin\,Φ\,sin\,ωt\,cos\,ωt]dt}{T}\)

= \(\frac{e_0i_0}{T}[cos\,Φ\int_{0}^{T}sin^2 ωt±sin\,Φ\int_{0}^{T}sin\,ωt\,cos\,ωt]dt]\)

Now, \(\int_{0}^{T}i^2\,sin^2ωt\,dt\) = \(\int_{0}^{T}\frac{1-cos2ωt}{2}dt\)

= \(\int_{0}^{T}\frac{1}{2}dt-\int_{0}^{T}\frac{cos2ωt}{2}dt\)

= \(\frac{T}{2}-\frac{1}{2}(\frac{sin2ωt}{2ω})_{0}^{T}\)

= \(\frac{T}{2}-\frac{1}{4ω}(sin2ωT-sin0)\)

= \(\frac{T}{2}-\frac{1}{4ω}[sin2(\frac{2π}{T})T-0]\)

= \(\frac{T}{2}-\frac{1}{4ω}[0-0]\) = \(\frac{T}{2}\)

Also

\(\int_{0}^{T}sin\,ωt\,cos\,ωt\,dt\) = \(\frac{1}{2}\int_{0}^{T}sin\,2ωt\,dt\)

                                                              = \(\frac{1}{2}[\frac{-cos\,2ωt}{2ω}]_0^T\) 

                                                              = \(-\frac{1}{4ω}[cos\,2\frac{2π}{T}-cos\,0]\) 

                                                              = \(\frac{1}{4ω}[1-1]=0\) 

Hence  P_av = \(\frac{e_0i_0}{T}cos\,Φ×\frac{T}{2}\) = \(\frac{e_0i_0}{2}cos\,Φ\)                      

                   = \(\frac{e_0}{\sqrt{2}}.\frac{i_0}{\sqrt{2}}cos\,Φ\)

                   = \(e_{rms}.i_{rms}cos\,Φ\) = \(e_{rms}.i_{rms}\frac{R}{Z}\)

where the impedance Z = \(\sqrt{R^2+(X_L-X_C)^2}\)

(b) P_av = \(e_{rms}.i_{rms}cos\,Φ\)

• The factor cos Φ is called as power factor. For circuits used for transporting electric power, a low power factor means the power available on transportation is much less than e_rms.i_rms. It means there is significant loss of power during transportation.

(a) The device Y is a capacitor. Its reactance is X_C = \(\frac{1}{ωC}\)

where ω is the angular frequency of the applied emf and C is the capacitance of the capacitor.

(b) Graphs showing variation of emf and current with time over one cycle of AC for Y.

(c) X_C = \(\frac{1}{ωC}\)= \(\frac{1}{2πfC}\), Thus X_C ∝ \(\frac{1}{f}\)   where f is the frequency of AC.

Suppose C = \(\frac{1000}{2π}\) pF, ∴  Xc =  M Ω

For f = 100 Hz, Xc = 1 x 10⁷ Ω = 10 M Ω;

for f = 200 Hz, Xc = 5 M Ω;

for f = 300 Hz, Xc = 10/3 M Ω;

for f = 400 Hz, Xc = 2.5 M Ω

for f = 500 Hz, Xc = 2 M Ω and so on  (1 ΜΩ = 10⁶Ω).

(d) The phasor representing the peak emf (e₀) makes an angle ot in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (i₀) is turned 90° anticlockwise with respect to the phasor representing emf e₀. The projections of these phasors on the vertical axis give instantaneous values of e and i.

Now let us consider the total opposition offered by a resistor, pure inductor and capacitor connected in series with the alternating source of emf as shown in Fig.

Let a pure resistor R, a pure inductance L and an ideal capacitor of capacitance C be connected in series to a source of alternative emf. As R, L and C are in series, the current at any instant through the three elements has the same amplitude and phase. Let it be represented by i = i₀ sin ωt.

The voltage across the resistor, e_R = Ri, is in phase with the current. The voltage across the inductor, e_L = X_Li, leads the current by π/2 rad and that across the capacitor, e_c = X_Ci, lags behind the current by π/2 rad. This is shown in the phasor Diagram (Fig.).

From this figure, e₀² = e_R² + (e_L − e_C)²

= R²i₀² + (X_Li₀ − X_Ci₀)² = i₀² [R² + (X_L − X_C)²]

∴ e₀ = \(i_0\sqrt{R^2+(X_L-X_C)^2}\) = i₀Z, Where

Z = \(\frac{e_0}{i_0}\) = \(\sqrt{R^2+(X_L-X_C)^2}\)  is the effective resistance of the circuit. It is called the impedance.

Comparison between resistance and reactance :

Let us connect the source of alternating emf to a circuit containing pure inductor (L) only as shown in Fig.

Let us assume that the inductor has negligible resistance. The circuit is therefore a purely inductive circuit.

Suppose the alternating emf supplied is represented by

e = e₀ sin ωt

On closing the key K, an emf is induced in the inductor as the magnetic flux linked with it changes with time. This emf opposes the applied emf and according to the laws of electromagnetic induction by Faraday and Lenz, we have,

e' = \(L\frac{di}{dt}\) .....(1)

where e’ is the induced emf and i is the current through the inductor. To maintain the current, e and e’ must be equal in magnitude and opposite in direction.

According to Kirchh0ff’s voltage law, as the resistance of the inductor is assumed to be zero, we have,

e = −e’ = \(L\frac{di}{dt}\)    …….(2)

∴ \(L\frac{di}{dt}\) = e/L = (e₀ sin ωt)/L

∴ ∫ di = \(\int \frac{e_0\,sin\,ωt}{L}dt\)

∴ i = \(-\frac{e_0}{ωL}cos\,ωt+C\)

where C is the constant of integration. C must be time independent and have the dimension of current. As e oscillates about zero, i also oscillates about zero and hence there cannot be any time independent component of current.

∴ C = 0.  ∴ i = \(-\frac{e_0}{ωL}cos\,ωt\)  = \(-\frac{e_0}{ωL}sin(\frac{π}{2}-ωt)\) …….(3)

∴ i  = \(\frac{e_0}{ωL}sin(ωt-\frac{π}{2})\)   … [ as sin(-θ) = − sin θ]

From Eq. (3), i_peak = i₀ = \(\frac{e_0}{ωL}\)

∴ i  = \(i_0sin(ωt-\frac{π}{2})\)

Comparison of this equation with e = e₀ sin ωt shows that e leads i by π/2 rad, i.e., the voltage is ahead of current by π/2 rad in phase.

Let us consider a capacitor with capacitance C connected to an AC source with an emf having instantaneous value e = e₀ sin ωt. This is shown in Fig.

The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle, the capacitor is alternately charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is V = q/C, ∴ q = CV.

q and V are functions of time, with V = e = e₀ sin ωt. The instantaneous current in the circuit is

i = dq/dt = \(\frac{d}{dt}(CV)\) = \(C\frac{dV}{dt}\)  = \(C\frac{d}{dt}(e_0\,sinωt)\) = ωCe₀ cos ωt

∴ i = \(\frac{e_0}{(1/ωC)}sin(ωt+\frac{π}{2})\) = \(i_0\,sin(ωt+\frac{π}{2})\)

Where i₀ = \(\frac{e_0}{(1/ωC)}\) is the peak value of the current,

From Eq. e = e₀ sin ωt and Eq. \(i=i_0\,sin(ωt+\frac{π}{2})\) we find that in an AC circuit containing a capacitor only, the alternating current i leads the alternating emf e by phase angle of π/2 radian as shown in above graph.

Given : f = 50 Hz, i_rms = 5A, t = \(\frac{1}{600}\) s

The peak value of the current, i₀ = i_rms\(\sqrt{2}\) = (5)(1.414) = 7.07 A

i = i₀ sin(2πft)

= 7.07 sin [2π(5)\((\frac{1}{600})\)]

= 7.07 sin \(\frac{π}{6}\)

= (7.07)(0.5) = 3.535 A

3.535 A is the required current.

Given : Power (V_rms i_rms) = 100W, V_rms = 220 V, f = 50 Hz

The rms current through the bulb,

i_rms = \(\frac{power}{V_{rms}}\) = \(\frac{100}{220}\) = 0.4545 A

The resistance of the bulb, R = \(\frac{V_{rms}}{i_{rms}}\) = \(\frac{220}{100/220}\) = 484 Ω

Given : C = 15 μF = 15 x 10⁻⁶F, V_rms = 220V, f = 50 Hz,

The capacitive reactance = \(\frac{1}{2πfC}\) = \(\frac{1}{2×3.142×50×(15×10^{-6}}\) = 212.2 Ω

i_rms = \(\frac{V_{rms}}{capacitive\,reactance}\) = \(\frac{220}{212.2}\) = 1.037 A

i_peak = i_rms\(\sqrt{2}\) = (1.037)(1.414) = 1.466 A

If the frequency is doubled, the capacitive reactance will be halved and the current will be doubled.

Given : L = 2H, i₀ = 0.25 A, f = 60 Hz, π = 3.142,

ωL = 2πfL = 2(3.142)(60)(2) = 754.192

The effective potential difference across the inductor = ωLi_rms = ωL\(\frac{i_0}{\sqrt{2}}\) = \(\frac{(754.1)(0.25)}{1.414}\) = 133.3 V

Given : e = 220sin 100 πt, L = (1/π)H

Comparing e = 220sin 100 πt with e = e₀ sin ωt, we get

ω = 100π,  ∴ ωL = 100π × \(\frac{1}{π}\) = 100 Ω

∴ The instantaneous current through the circuit = i = \(\frac{e_0}{ωL}\) sin(100 πt – π/2)

=  sin(100 πt – π/2) = 2.2sin(100 πt – π/2) A

i_rms = \(\frac{i_0}{\sqrt{2}}\) = \(\frac{2.2}{1.414}\) = 1.556 A is the reading of the AC galvanometer connected in the circuit.

Data: C = 25 μF = 25 x 10⁻⁶ F, L = 0.10 H, R = 25 Ω, e = 310 sin (314t) [volt]

Comparing e = 310 sin (314 t) with e = e₀ sin (2πft), we get, the frequency of the alternating emf as f = \(\frac{314}{2π}\) = \(\frac{314}{2×3.14}\) = 50 Hz

Reactance = \(|ωL-\frac{1}{ωC}|\) = \(|2πfL-\frac{1}{2πfC}|\) = \(|(2)(3.142)(50)(0.10)-\frac{1}{(2)(3.142)(50)(25×10^{-6}}|\) = |31.4 − 127.4| = 96 Ω

Z² = R² + \((ωL-\frac{1}{ωC})^2\) = 25² + 96² = 9841 Ω²

∴ Impedance Z = \(\sqrt{9841}\) = 99.2 Ω

Peak current i₀ = e₀/Z = 310/99.2 A

∴ i_rms = \(\frac{i_0}{\sqrt{2}}\) = (\frac{310}{(1.414)(99.2)}\) = 2.21 A

cos Φ = R/Z = 25/99.2 = 0.2520

∴ Phase angle Φ = cos⁻¹ (0.2520) = 75.40⁰ = 1.316 rad

Given : C = 100 µF = 100 x 10⁻⁶ F = 10⁻⁴ F, R = 50 Ω, L = 0.5H, f = 50 Hz,

V_rms = 110 V

∴ ωL = 2πfL = 2(3.142)(50)(0.5) = 157.152

And \(\frac{1}{ωC}\) = \(\frac{1}{2πfC}\) = \(\frac{1}{(2)(3.142)(50)(10^{-4}}\) = 31.83 Ω

Z² = R² + \((ωL-\frac{1}{ωC})^2\) = (50)² + (157.1 - 31.83)² = 2500 + 15700 = 18200 Ω²

∴ Impedance, Z = \(\sqrt{18200}\) Ω = 134.9 Ω

The rms value of the current in the circuit,

i_rms = \(\frac{V_{rms}}{Z}\) = \(\frac{110}{134.9}\) = 0.8154 A

Given : R = 10 Ω, power factor = 0.5, f = 100 Hz,

Power factor = \(\frac{1}{2πfCR}\)

0.5 = \(\frac{1}{(2)(3.142)(100)C(10)}\)

∴ C = \(\frac{1}{3.142×10^3}\) = 3.182 × 10⁻⁴ F

This is the capacity of the capacitor.

Given : f = 50 Hz, i = \(\frac{i_0}{\sqrt{2}}\)  ∴ \(\frac{i}{i_0}\) = \(\frac{1}{\sqrt{2}}\)

i = i₀ sin ωt,  ∴ sin ωt = \(\frac{i}{i_0}\) = \(\frac{1}{\sqrt{2}}\)

∴ ωt = π/4 rad

∴ 2πft = π/4

∴ t = \(\frac{1}{8f}\) = \(\frac{1}{8×50}\) = 2.5 × 10⁻³ s

2.5 × 10⁻³ s is the required time

Given: f = 10⁶ Hz, L = 101.4 x 10⁻⁶ H

f_r = \(\frac{1}{2π\sqrt{LC}}\), 

∴ f_r² = \(\frac{1}{4π^2LC}\) ,

∴ C = \(\frac{1}{4π^2f_r^2L}\) = \(\frac{1}{4(3.142)^2(10^6)^2(101.4×10^{-6}}\) = 2.497 × 10⁻¹⁰ F

= 249.7 × 10⁻¹² F = 249.7 picofarad

This is the value of the capacity.

Given : C = 10 µF = 10 x 10⁻⁶ F = 10⁻⁵ F, L = 100 mH = 100 × 10⁻³ H = 10⁻¹ H,

V = 25 V

The energy stored in the electric field in the capacitor = \(\frac{1}{2}\)Cv²

The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li²

Here, \(\frac{1}{2}\)Cv² = \(\frac{1}{2}\)Li²

∴ i² = \(\frac{C}{L}V^2\) = \(\frac{10^{-5}}{10^{-1}}(25)^2\) = 25² x 10⁻⁴

∴ i = 25 x 10⁻² A = 0.25 A

This is the maximum current in the coil.

Given : C = 100 µF = 100 x 10⁻⁶ F = 10⁻⁴ F, V = 50 V, i = 5A

The energy stored in the electric field in the capacitor = \(\frac{1}{2}\)CV²

The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li²

Here, \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)Li²

∴ L = \(C(\frac{V}{i})^2\) = \(10^{-4}(\frac{50}{5})^2\) = 10⁻² H

This is the value of the inductance.