Alternating Current

Given:

Resistance, R = 100Ω

Voltage, V_rms = 220 V

Frequency of A.C. supply, F = 50 Hz

(a) The root mean square value of the current in the circuit can be calculated as follows:

I_rms = V_rms/R

⇒ I_rms = (220V)/(100Ω)

⇒ I_rms = 2.20 A

(b) The net power consumed over a full cycle can be calculated as follows:

P = Voltage × Current

P = V × I

P = (220V) × (2.20A)

⇒ P = 484 W

Given: Peak voltage of an A.C. supply, V_peak = V₀ = 300 V

The root mean square voltage can be calculated as follows:

V_rms = V₀/√2

V_rms = (300V)/√2

⇒ V_rms = 212.1 V

Given: the root mean square value in an A.C. circuit, I_rms = 10A A

The peak current can be calculated as follows:

I_rms = I₀ /√2

⇒ I₀ = I√2 = (10 A)√2

⇒ I₀ = 14.1 A

Given: Inductance = 44 mH = 44 × 10^-3H

Voltage V = 220 V

Frequency, v = 50 Hz

The diagram is:

Angular frequency, ω = 2πv

ω = 2 × 3.14 × 50Hz

Inductive reactance is given by the formula X_L = ω × L

Substituting the value of ω from above

X_L = 2 × 3.14 × 50Hz × 44 × 10^-3H

The root mean square value of the current in the circuit is given as:

I_rms = V/X_L

Substituting the value of inductive reactance from equation (i), we get

⇒

⇒ I_rms = 15.92 A

Given: capacitance = 60 μF = 60 × 10^-6 F

Voltage V = 110 V

Frequency f = 60 Hz

The diagram is:

Capacitive reactance is given by the following relation:

X_c = 1/ωC

Substituting ω = 2πv in the above formula, we get

X_c = 1/2πvC

Substituting the values, we get

⇒

Now, the root mean square value of the current can be calculated as follows:

I_rms = V/X_c

I_rms = 110V × 2 × 3.14 × (60Hz) × 60 × 10^-6(F)

On calculating, we get

I_rms = 2.49 A

In inductive circuit

Root mean square value of current, I = 15.92 A

Root mean square value of voltage, V = 220 V

The net power absorbed by each circuit over a complete cycle is given by the solution:

Power = VI cosФ

Where Ф is the phase difference between the voltage and current

For a pure inductive circuit, the phase difference between voltage and current in 90⁰.

Substituting the above values, we get

⇒ P = 0

In capacitor circuit

Root mean square value of current, I = 2.49 A

Root mean square value of voltage, V = 110 V

The net power absorbed by each circuit over a complete cycle is given by the solution:

Power = VI cos Ф

For a pure capacitor circuit, the phase difference between voltage and current in 90⁰.

Substituting the above values, we get

⇒ P = 0

∴ Zero power is absorbed by each circuit.

Given: L = 2.0H

C = 32mF = 32 × 10^-6 F

R = 10 Ω

The diagram of the question is:

Resonant frequency ω_r is calculated as follows:

ω_r = 125 s^-1

Q- value of the LCR circuit can be calculated as follows:

On calculating, we get

Q = 25

The capacitor C = 30 μF = 30 × 10 ^-6 F

Inductor L = 27 mH = 27 × 10^-3 H

The diagram is given :

Angular frequency ω_r is given by the following:

ω_r = 1/√ LC

ω_r = 1/{√(27 × 10^-3H × 30 × 10^-6F)}

on calculating, we get

ω_r = 1.11 × 10³ rad/s

Given: Capacitance C = 30 uF = 30 × 10 ^-6F

Inductance L = 27 mH = 20 × 10 ^-6H

The charge on capacitor, Q = 6 mC = 6 × 10^-3C

The diagram is given :

The total energy stored in the circuit can be calculated as followed:

E = 1/2 Q²C

E = 1/2 × (6 × 10^-3C)² × 30 × 10^-6F

On calculating, we get

⇒ E = 0.6 J

At later time, total energy will be same i.e. E = 0.6 J as energy is shared between inductor and capacitor.

Given:

R = 20 _Ω

L = 1.5 H

C = 35 μF = 35 × 10^-6F

Voltage supplied V = 200 V

The diagram given is:

Impedance of the circuit can be calculated as follows:

At resonance condition i.e. when ωL = 1/ωC

Z = R = 20 Ω

Current is given by the relation

I = V/R

I = 200V /20Ω

⇒ I = 10A

Therefore, the average power transferred to the circuit in one complete cycle can be calculated as follows:

Power = VI

Power = 200V × 10A

Power = 2000 W.

Given: frequency range of radio waves = 800 kHz to 1200 kHz

Therefore, low tuning frequency of radio wave v₁ = 800 kHz, or 800 × 10^-3 Hz

Upper tuning frequency v₂ = 1200 kHz or 1200 × 10^-3 Hz

Inductance L = 200 μH or 200 × 10^-6H

Capacitance of the variable capacitor for low tuning frequency v₁ is given by the following relation:

C₁ = 1/ ω₁²L……………..(i)

Where ω is equal to 2πv₁

Substituting in equation (i), we get

C₁ = 1/{(2π × 800 × 10^-3Hz)² × 200 × 10^-6H}

On calculating we get

C₁ = 1.9809 × 10^-10F

Or C₁ = 198.1 pF

Capacitance of the variable capacitor for upper tuning frequency v₂ is given by the following relation:

C₂ = 1/ ω₂²L……………..(ii)

Where ω is equal to 2πv₂

Substituting in equation (i), we get

C₂ = 1/{(2π × 1200 × 10^-3Hz)² × 200 × 10^-6H}

On calculating we get

⇒ C₂ = 88.04 pF

Therefore, the range of its variable capacitor is from 88.04 pF to 198.1 pF.

(a) Given: L = 5.0 H

C = 80 Μf = 80 × 10^-6F

R = 40Ω

_{Voltage source V = 230 V}

_{The diagram is given as:}

_{Resonance angular frequency can be calculated as follows}

_ωR = 1/√LC

_ωR = 1/ √(5H × 80 × 10^-6F)

_{On calculating, we get}

_ωR = 50 rad/s

Therefore, source frequency which drives the circuit in resonance frequency ω_R = 50 rad/s.

(b) Impedance can be calculated by using the formula:

At resonance condition i.e. when

Z = R = 40 Ω

_{At resonating frequency, the amplitude of the current can be calculated using the formula:}

_{Where V0} is the peak voltage

_{Z is the impedance}

_V0 = √2V_rms

_I0 = (√2 × 230V)/40Ω

_⇒I0 = 8.13 A

The impedance of the circuit is 40 Ω and the amplitude of current at the resonating frequency is 8.13 A.

_{(c) The root mean square potential drop across the inductor is given by the following formula:}

_(VL) _rms = I × 1/ω_R L

_{Where I = I0}/√ 2

_{I = √ 2V/√2Z}

_{Substituting values, we get}

_{I = 230V/40Ω = 230/40 A}

_Now,

_(VL) _rms = I × 1/ω_R L

(V_L)_r.m.s.

(V_L)_r.m.s = 1437.5 V

Potential drop across capacitor can be calculated as follows:

_(VC) _rms = I × 1/ω _R C

(V_C)_r.m.s.

(V_C)_r.m.s = 1437.5 V

Potential drop across resistor can be calculated as follows:

_(VR) _rms = I × R

_(VR) _{r. m. s}

_(VR) _rms = 230 V

Potential drop across LC combination can be calculated as follows:

_VLC

At resonance condition i.e. when ωL = 1/ωC

Therefore, _VLC = 0

Inductor L = 20 mH

In henry L = 20 × 10^-3H

capacitor C = 50 μF = 50 × 10^-6F

Charge Q = 10 mC = 10 × 10^-3C

(a) Total energy stored in the capacitor can be calculated as follows:

E = 1/2 Q²C

Substituting values in above formula, we get

E =

On calculating, we get

E = 1 Joule

Yes, the total energy will be conserved during LC oscillations because there is no resistor connected in the circuit.

(b) Natural frequency of the circuit can be calculated using the following expression:

V = 1/(2π√LC)

V = 1/[2 × 3.14 × √ (20 × 10^-3H × 50 × 10^-6F)]

On calculating, we get

V = 159.24 Hz

Natural angular frequency is given by the following expression:

ω_r = 1/√ LC

or ω_r = 1/√ (20 × 10^-3H × 50 × 10^-6F)

ω_r = 10³ rad/s

(c) (i) Time period T = 1/v

T = 1/159.24 Hz

Or T = 6.29 ms

Therefore, for time period T = 6.29 ms, the total charge on the capacitor can be calculated by the following relation:

Q’ = Q cos 2π/T

Since energy stored is electrical, therefore Q’ = Q

Therefore, energy stored is electrical at time t = 0, T/2, T, 3T/2

(ii) Since when electrical energy is zero, the magnetic energy will be maximum

Therefore, energy stored is electrical at time t = T/4, 3T/4, 5T/4

(d) Q’ is equal to the charge on the capacitor when the total energy is shared between capacitor and inductor

When the energy is shared between the capacitor and inductor then the total energy stored in the capacitor is given as half of the maximum energy

⇒

⇒ Q’ = Q/√ 2………………(1)

But as we know, Q’ = Q cos 2π/T

Substituting eqn (1) in above

Thus, the total energy is shared between capacitor and inductor at time t = T/8, 3T/8, 5T/8

(e) If a resistor is inserted in the circuit, then total energy is eventually dissipated as heat.

Given: Inductance = 0.50 H

Resistance R = 100 Ω

Voltage V = 240 V

Frequency = 50 Hz

The diagram is:

(a) Peak voltage is given by the following formula:

V₀ = √2 V

V₀ = √2 × 240V

On calculating, we get

V₀ = 339.41 V

Angular frequency is given by the formula ω = 2πv

Substituting value of v in above formula

ω = 2 × π × 50(Hz) = 100 π rad/s.

The maximum current in the coil can be calculated as follows:

On calculating, we get

I₀ = 1.82 A

(b) The equation for voltage and current are as follows:

V = V₀ cos ωt

I = I₀ cos (ωt-Ф)

At time t = 0, V = V₀

At t = Ф/ ω, I = I₀

Therefore, the time lag between maximum current and maximum voltage is Ф/ ω. Phase angle is given by the following relation:

tan Ф = ωL/R

tan Ф = (2 π × 50(Hz) × 0.5(H))/100

On calculating, we get

tan Ф = 1.57

or Ф = 57.5⁰

in rad Ф = 57.5π/180 rad

ωt = 57.5π/180

or

On calculating, we get

t = 3.2 ms

Given: Inductor = 0.5 Hz

Resistor R = 100 Ω

Voltage V = 240 V

Frequency v = 10 kHz

In Hz, the frequency v = 10⁴ Hz

(a) Angular frequency is given by the following relation

ω = 2πv

ω = 2π × 10⁴ rad/s

Peak voltage is given by the relation: V = √2 × V

On substituting the values

V = 240√2V

The maximum current is given by the relation:

Substituting the values we get

⇒

On calculating, we get

⇒ I₀ = 1.1 × 10^-2A

(b) tan Ф = ωL/R

tan Ф =

on calculating, we get

or Ф = = 89.82⁰

in rad, Ф = 89.82 π/180 rad

ωt = 89.82π/180

or

On calculating, we get

t = 25 μs

Since I₀ is very small, therefore at higher frequency the inductor is at open circuit.

In case of a dc circuit, when steady state is obtained i.e. ω = 0. Therefore, the inductor behaves like a conducting object.

Given: capacitor C = 100 μf = 100 × 10^-6F

Resistance R = 40Ω

Voltage V = 110 V

Frequency v = 60 Hz

(a) Angular frequency is given by the relation: ω = 2πv

ω = 2π × 60 rad/s

The impedance is given by the relation

Or Z² =

Peak voltage is given by the relation V₀ = V √2

Or V₀ = 110 √ 2 V

The maximum current is given by the following relation:

I₀ = V₀/Z

I₀ =

I₀ =

On calculating, we get

I₀ = 3.24 A

(b) Since in capacitor circuit, the voltage lags behind the current by a phase angle Ф. Therefore, the phase angle is given by the expression

tanФ =

tanФ = 1/ωCR

substituting the values, we get

tanФ = 1/(120π(rad/s) × 10^-4(F) × 40 (Ω))

On calculating, we get

tanФ = 0.6635

or Ф = 33.56°

in radians, Ф = (33.56 π) /180

Time lag is given by the relation

t = Ф/ω

t =

on calculating, we get

t = 1.55 × 10^-3s

on converting in ms, we get time lag t = 1.55 ms.

Given: Capacitance C = 100 μF

In Farad, Capacitance C = 100 × 10^-6F

Resistance = 40Ω

Voltage V = 110 V

Frequency = 12 kHz

In Hz, frequency v = 12 × 10³Hz

Angular frequency ω = 2πv = 2 × π × 12 × 10³(Hz)

ω = 24 × 10³ rad/s

Peak voltage is calculated as follows:

V₀ = V/ √2

V₀ = 110√2V

The peak current is calculated as follows:

I₀ =

I₀ =

On calculating, we get

I₀ = 3.9 A

Since for RC circuit, the voltage lags behind the current by a phase angle of Ф

tanФ =

tanФ = 1/ωCR

substituting the values, we get

tanФ = 1/(24π × 10³(Hz) × 100 × 10^-6(F) × 40 (Ω))

On calculating, we get

tanФ = 1/96π

or Ф = 0.2⁰

in radians, Ф = (0.2 π) /180

At high frequency, Ф approaches to zero and capacitor acts as a conductor.

In d.c. circuit, ω is zero, and the steady state is attained and therefore, capacitor C tends to be an open circuit.

Given: L = 5.0 H

capacitor C = 80 × 10^-6F

Resistance = 40 Ω

Voltage = 230 V

Impedance is expressed as:

At resonance condition,

ω = 1/ √LC

substituting the values, we get

ω = 1/ √(5(H) × 80 × 10^-6(F))

On calculating, we get

ω = 50 rad/s

Since, the magnitude of impedance is maximum at 50 rad/s due to which the total current is minimum.

The root mean square current flowing through the inductor is given by the relation:

I_L = V/ωL

Substituting the values, we get

I_L = 230(V)/(50 (rad/s) × 5 (H))

On calculating, we get

I_L = 0.92 A

Root mean square value of the current through capacitor is given by the following relation:

I_c = ωCV

I_c = 50rad/s × 80 × 10^-6F × 230V = 0.92 A

Root mean square value of the current through resistor is given by the following relation:

I_R = V/R

I_R = 230V/40Ω

I_R = 5.75A

Given: L = 80 mH = 80 × 10^-3 H

capacitor C = 60 Μf = 60 × 10^-6 F

Voltage V = 230 V

Frequency v = 50Hz

(a) Peak voltage is given by the formula:

V₀ = V/√ 2

V₀ = 230√2 V

Substituting the values we get

On calculating, we get

I₀ = -11.63 A.

The amplitude of maximum current is given by the following relation:

I₀ = 11.63 A.

The root mean square value of the current I = I₀/√ 2

⇒ I = -11.63(A)/√ 2

⇒ I = - 8.22 A

(b) The potential difference across the inductor can be calculated as follows:

V_L = I × ω × L = 8.22(A) × 100 π(Hz) × 80 × 10^-3 (F) = 206.61 V

Potential difference across the capacitor is given by the formula as follows:

V_c = I × 1/ωC

Substituting values we get

V_c = 8.22(A) × 1/ 100 π(Hz) × 60 × 10^-6(F)

On calculating, we get

V_c = 436.3 V

(c) Since the actual voltage leads the current by π/2, so the average power consumed by the inductor is zero.

(d) Since voltage lags the current by π/2, so the average power consumed by the capacitor is zero.

(e) The total power over a complete cycle is zero.

Given: Average power transferred to the resistor = 788.44 W

Average power transferred to the capacitor = 0 W

Power absorbed by the circuit = 788.44 W

Inductance L = 80 mH

In henry, Inductance L = 80 × 10^-3H

Capacitance C = 60 μF = 60 × 10^-6F

Resistance R = 15 Ω

Voltage V = 230 V

Frequency v = 50 Hz

Angular frequency is calculated as follows:

ω = 2πv

ω = 2 × 3.14 × 50(Hz) = 100π rad/s

The impedance can be calculated as follows:

On calculating, we get

Z = 31.728 Ω

Current can be calculated as follows: I = V/Z

I = 230 (A)/31.728 (Ω) = 7.25 A

The average power transferred to resistor can be calculated as follows:

P_r = I²R

Substituting values, we get

P_r = (7.25)²(A) × 15(Ω)

P_r = 788.44 W

The average power transferred to capacitor is equal to the average power transferred to resistor = 0

Total power absorbed = P_R + P_c + P_L = 788.4W

(a) Given: L = 0.12 H

C = 480 nF

Converting into Farad, we get 480 × 10^-9F

R = 23Ω

_{Voltage = 230 V}

_{Peak voltage can be calculated as follows:}

_V0 = √2V

_{Substituting values we get}

_V0 = 230√2

_V0 = 325.22 V

_{Current flowing through the circuit is given by the following:}

I₀ =

At resonating frequency, ω_R L - 1/ω_RC = 0

ω_R = 1/√LC

= 1/ √0.12 (H) × 480 × 10^-9 (F) = 4166.67 rad/s

Resonant frequency can be calculated using the formula:

v_R = ω_R/2π

v_R = 4166.67(Hz)/2 × 3.14

v_R = 663.48 Hz

_{Maximum current is calculated as follows:}

_(I0)_max = V₀/R = 325.22(Hz)/23 = 14.14 A

_{(b) The maximum power absorbed by the circuit can be calculated as follows:}

_Pav = 1/2(I₀)²_maxR

_{Substituting the values, we get}

_Pav = 1/2 × (14.14)²(A) × 23 (Ω)

_Pav = 2299.33 W

_{Therefore, the resonating frequency is 663.48 Hz.}

_{(c) The power transferred is equal to the half of the power at the resonating frequency.}

_{Frequency at which power is half = ωR} �∆ω

_{Or = 2π (vR} �∆v)

_{Where ∆ω = R/2L}

_{Substituting the values we get}

_{∆ω = 23/2(Ω) × 0.12(H)}

_{∆ω = 95.83 rad/s}

_{Therefore, change in frequency is written as follows:}

_{∆v = 1/2π × ∆ω}

_{Substituting the values, we get}

_{∆v = 95.83(rad/s)/2π = 15.26 Hz}

_vR + ∆v is calculated as follows:

_{663.48(Hz) + 15.26(Hz) = 678.74 Hz}

_vR-∆v is calculated as follows:

_{663.48(Hz) – 15.26(Hz) = 648.22 Hz}

_{The current amplitude is calculated as follows:}

_{I’ = 1/√2 × (I0})_max

_{I’ = 14.14(A)/ √2}

_{I’ = 10 A}

_{(d) Q factor can be calculated as follows:}

_{Q = ωR}L/R

_{Q = (4166.67 rad/s) × 0.12H/23Ω}

_{⇒Q = 21.74}

Given: L = 3.0 H

C = 27 μF,

In Farad, capacitance C = 27 × 10^-6F

R = 7.4Ω

Angular frequency is given by the relation:

ω_R = 1/√LC

ω_R = 1(√27 × 10^-6F × 3 H)

On calculating, we get

ω_R = 111.11 rad/s

_{Q factor can be calculated as follows:}

_{Q = ωR}L/R

_{Q = (111.11(rad/s) × 3H)/7.4(Ω)}

_{⇒Q = 45.0446}

The sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2 by reducing R to half.

i.e. R = R/2 = 7.4(Ω)/2 = 3.7Ω

The statement is not true for root mean square voltage. In any A.C. circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit but is not true for root mean square voltage because the voltages are not in the same phase across different elements.

This statement is true that A capacitor is used in the primary circuit of an induction coil. This is due to the reason, that when a circuit is broken, capacitor is charged through high induced voltage so as to avoid the sparkling.

Dc signal will appear across capacitor. This is because impedance of an inductor is negligible whereas the impedance of the capacitor is high and almost infinite. An A.C. singal of high frequency appears across inductor because for an A.C. signal of high frequency, the impedance of inductor is high while impedance of capacitor is low.

When an iron core is connected in series with the lamp connected to the A.C. line then the choke coil and iron will increase the impedance and the lamp will glow dim.

Choke coil needed in the use of fluorescent tubes with A.C. mains because choke coil reduces the voltage across the tube and doesn’t waste the power whereas while using ordinary resistor, it waste the power in form of heat dissipation. Therefore, we can’t use an ordinary resistor instead of choke coil.

Given: Input voltage = 2300 V

No. of turns = 4000

Output voltage = 230 V

The relation between voltage and no. of turns is written as follows:

V₁/V₂ = n₁/n₂

Substituting the values

2300(V)/230(V) = 4000/ n₂

On calculating, we get

n₂ = 400

Given: height of water pressure h = 300 m

Height of water flow = 100 m³s^–1

Efficiency of turbine n = 60%

Acceleration due to gravity g = 9.8 ms^–2

Density of water = 10³Kg/m³

The electric power available from the plant is given by the relation:

h × η × р × g × V

substituting the values, we get

300m × 0.6 × 10³ × 9.8ms^–2 × 100 m³s^–1 = 176.4 MW

(a) Given: Power P = 800 kW

Voltage = 220 V

Voltage at which power is generated P’ = 440 V

Distance between town and power generating station = 15 km

The resistance of the two-wire line carrying power is 0.5 Ω per km.

Total resistance of wire = (15(km) + 15(km))0.5 (Ω per km)

∴ R_total = 15 Ω

Input voltage v₁ = 4000 V

Output voltage V₂ = 220 V

The root mean square current in the lines can be calculated as follows:

I = P/V₁

= (800(kW) × 10³)/4000(V) = 200 A

Line power loss can be calculated as follows:

P = I²R_total

= (200)²A × 15Ω

On calculating we get P = 600 Kw

(b) Let the power loss due to current leakage is negligible

Total power supplied by the power plant = 800kW + 600kW = 1400 kW

(c) Voltage drop V = IR_Total

Substituting the values, we get

200V × 15Ω = 3000V

Therefore, the total voltage supplied = 3000V + 4000V = 7000V

And since the power generated = 440 V

Therefore, the rating should be 440V - 7000 V

Given: Rating of step down transformed = 40,000-220 V

Input voltage V₁ = 40000 V

Output voltage V₂ = 220 V

Total electric power P = 800 KW

In watt, P = 800 × 10³ W

Potential V = 220 V

Voltage of power generated by the plant V’ = 440 V

Distance between town and power generating station = 15 km

The resistance of the two wire line carrying power is 0.5Ω per km.

Total resistance of wire = (15(km) + 15(km))0.5 = 15 Ω

Input voltage v₁ = 4000 V

Output voltage V₂ = 220 V

The root mean square current in the lines can be calculated as follows:

I = P/V₁

= (800 × 10³W)/40000V = 20 A

Line power loss can be calculated as follows:

P = I²R

= (20A)² × 15Ω = 6 kW

B. Let the power loss due to current leakage is negligible

Total power supplied by the power plant = 800kW + 6kW = 806 Kw

C. Voltage drop V = IR

Substituting the values, we get

20A × 15Ω = 3000V

Therefore, the total voltage supplied = 300V + 40000V = 40300V

And since the power generated = 440 V

Therefore, the rating should be 440 V- 40300 V