Alternating CurrentClass 12 Physics NCERT Solutions
8 Solutions
Generated by KedovoAI
Solution 1 of 8
Q1EXERCISES
A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a)
What is the rms value of current in the circuit?
(b)
What is the net power consumed over a full cycle?
Solution
Given:
- Resistance, R = 100 Ω
- RMS voltage, V = 220 V
- Frequency, f = 50 Hz
(a) The rms value of current in the circuit
The relation between rms voltage (V), rms current (I), and resistance (R) for a resistive circuit is given by Ohm's law, similar to a DC circuit:
V = I R
Therefore, the rms current I is:
I = V / R = 220 V / 100 Ω = 2.2 A
The rms value of the current in the circuit is 2.2 A.
(b) The net power consumed over a full cycle
The net power consumed over a full cycle in an AC circuit is given by:
P = V I cos(φ)
For a purely resistive circuit, the voltage and current are in phase, so the phase angle φ = 0. Therefore, the power factor cos(φ) = cos(0) = 1.
The power consumed is:
P = V * I = 220 V * 2.2 A = 484 W
Alternatively, using the formula P = I² R:
P = (2.2 A)² * 100 Ω = 4.84 * 100 W = 484 W
The net power consumed over a full cycle is 484 W.
Q2EXERCISES
(a) The peak voltage of an ac supply is 300 V . What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A . What is the peak current?
Solution
(a) RMS voltage from peak voltage
Given:
- Peak voltage, vₘ = 300 V
The relationship between the root mean square (rms) voltage (V) and the peak voltage (vₘ) is:
V = vₘ / √2
V = 300 V / 1.414 ≈ 212.1 V
The rms voltage is approximately 212.1 V.
(b) Peak current from rms current
Given:
- RMS current, I = 10 A
The relationship between the rms current (I) and the peak current (iₘ) is:
I = iₘ / √2
Therefore, the peak current is:
iₘ = I * √2
iₘ = 10 A * 1.414 = 14.14 A
The peak current is 14.14 A.
Q3EXERCISES
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Solution
Given:
- Inductance, L = 44 mH = 44 × 10⁻³ H
- RMS voltage, V = 220 V
- Frequency, f = 50 Hz
First, we need to calculate the inductive reactance (Xₗ) of the circuit. Inductive reactance is the opposition offered by an inductor to the flow of alternating current.
The formula for inductive reactance is:
Xₗ = ωL = 2πfL
Substituting the given values:
Xₗ = 2 × 3.14 × 50 Hz × (44 × 10⁻³ H)
Xₗ = 13.816 Ω
Now, we can find the rms value of the current (I) in the circuit using the relation similar to Ohm's law:
I = V / Xₗ
I = 220 V / 13.816 Ω ≈ 15.92 A
The rms value of the current in the circuit is approximately 15.92 A.
Q4EXERCISES
A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Solution
Given:
- Capacitance, C = 60 μF = 60 × 10⁻⁶ F
- RMS voltage, V = 110 V
- Frequency, f = 60 Hz
First, we need to calculate the capacitive reactance (X꜀) of the circuit. Capacitive reactance is the opposition offered by a capacitor to the flow of alternating current.
The formula for capacitive reactance is:
X꜀ = 1 / (ωC) = 1 / (2πfC)
Substituting the given values:
X꜀ = 1 / (2 × 3.14 × 60 Hz × 60 × 10⁻⁶ F)
X꜀ = 1 / 0.022608 Ω ≈ 44.23 Ω
Now, we can find the rms value of the current (I) in the circuit using the relation similar to Ohm's law:
I = V / X꜀
I = 110 V / 44.23 Ω ≈ 2.49 A
The rms value of the current in the circuit is approximately 2.49 A.
Q5EXERCISES
In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Solution
The net power absorbed by each circuit over a complete cycle is zero.
Explanation:
The average power (P) dissipated in an AC circuit is given by the formula:
P = V I cos(φ)
where V is the rms voltage, I is the rms current, and cos(φ) is the power factor. The angle φ is the phase difference between the voltage and the current.
-
For the purely inductive circuit (Exercise 7.3): In a circuit containing only a pure inductor, the current lags the voltage by a phase angle of π/2 (or 90°). Therefore, φ = π/2. The power factor is cos(φ) = cos(π/2) = 0. So, the net power absorbed is P = V I × 0 = 0.
-
For the purely capacitive circuit (Exercise 7.4): In a circuit containing only a pure capacitor, the current leads the voltage by a phase angle of π/2 (or 90°). Therefore, φ = -π/2. The power factor is cos(φ) = cos(-π/2) = 0. So, the net power absorbed is P = V I × 0 = 0.
In both cases, although a current flows (known as wattless current), no net energy is dissipated over a complete cycle. The energy stored in the magnetic field (for the inductor) or electric field (for the capacitor) during one-quarter cycle is returned to the source in the next quarter cycle. Thus, the average power consumption is zero.
Q6EXERCISES
A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Solution
Given:
- Capacitance, C = 30 μF = 30 × 10⁻⁶ F
- Inductance, L = 27 mH = 27 × 10⁻³ H
This setup describes an LC circuit. The angular frequency (ω₀) of free oscillations in an LC circuit is its natural resonant frequency. It is determined by the values of the inductance (L) and capacitance (C).
The formula for the angular frequency of free oscillations is:
ω₀ = 1 / √(LC)
Substituting the given values into the formula:
ω₀ = 1 / √((27 × 10⁻³ H) × (30 × 10⁻⁶ F))
ω₀ = 1 / √(810 × 10⁻⁹) s⁻¹
ω₀ = 1 / √(81 × 10⁻⁸) s⁻¹
ω₀ = 1 / (9 × 10⁻⁴) s⁻¹
ω₀ ≈ 1111.1 rad/s
The angular frequency of free oscillations of the circuit is approximately 1.11 × 10³ rad/s.
Q7EXERCISES
A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
Solution
Given:
- Resistance, R = 20 Ω
- Inductance, L = 1.5 H
- Capacitance, C = 35 μF = 35 × 10⁻⁶ F
- RMS voltage of the supply, V = 200 V
The problem states that the frequency of the supply equals the natural frequency of the circuit. This is the condition for resonance.
At resonance in a series LCR circuit:
- The inductive reactance (Xₗ) is equal to the capacitive reactance (X꜀).
- The impedance (Z) of the circuit is minimum and is equal to the resistance (R). Z = R.
- The phase difference (φ) between the voltage and current is zero, making the power factor cos(φ) = 1.
First, we find the rms current (I) in the circuit at resonance. The impedance Z = R = 20 Ω.
I = V / Z = V / R
I = 200 V / 20 Ω = 10 A
The average power (P) transferred to the circuit is given by:
P = V I cos(φ)
At resonance, cos(φ) = 1, so the power transferred is maximum.
P = V * I = 200 V * 10 A = 2000 W
Alternatively, we can use the formula P = I²R, since power is dissipated only across the resistor.
P = (10 A)² * 20 Ω = 100 * 20 W = 2000 W
The average power transferred to the circuit in one complete cycle is 2000 W or 2 kW.
Q8EXERCISES
Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Ω.
(a)
Determine the source frequency which drives the circuit in resonance.
(b)
Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c)
Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Solution
Given:
- Inductance, L = 5.0 H
- Capacitance, C = 80 μF = 80 × 10⁻⁶ F
- Resistance, R = 40 Ω
- RMS source voltage, V = 230 V
(a) Source frequency for resonance
The circuit is in resonance when the inductive reactance equals the capacitive reactance. The resonant angular frequency (ω₀) is:
ω₀ = 1 / √(LC)
ω₀ = 1 / √(5.0 H × 80 × 10⁻⁶ F) = 1 / √(400 × 10⁻⁶) = 1 / (20 × 10⁻³) = 50 rad/s
The source frequency (νᵣ) is:
νᵣ = ω₀ / (2π) = 50 / (2 × 3.14) ≈ 7.96 Hz
(b) Impedance and amplitude of current at resonance
At the resonating frequency, the impedance (Z) of the circuit is minimum and equal to the resistance.
Z = R = 40 Ω
The rms current (I) at resonance is:
I = V / Z = 230 V / 40 Ω = 5.75 A
The amplitude of the current (iₘ) is related to the rms current by iₘ = I√2.
iₘ = 5.75 A × √2 ≈ 5.75 A × 1.414 ≈ 8.13 A
(c) RMS potential drops and potential drop across LC combination
First, calculate the reactances at the resonant frequency (ω₀ = 50 rad/s):
- Inductive Reactance, Xₗ = ω₀L = 50 rad/s × 5.0 H = 250 Ω
- Capacitive Reactance, X꜀ = 1 / (ω₀C) = 1 / (50 rad/s × 80 × 10⁻⁶ F) = 1 / (4000 × 10⁻⁶) = 250 Ω
(As expected, Xₗ = X꜀ at resonance.)
Now, calculate the rms potential drops across each element using the rms current I = 5.75 A:
- Potential drop across R: Vᵣ = I R = 5.75 A × 40 Ω = 230 V
- Potential drop across L: Vₗ = I Xₗ = 5.75 A × 250 Ω = 1437.5 V
- Potential drop across C: V꜀ = I X꜀ = 5.75 A × 250 Ω = 1437.5 V
Note that the potential drops across the inductor and capacitor can be much larger than the source voltage.
Potential drop across the LC combination:
In a series LCR circuit, the voltage across the inductor (Vₗ) and the voltage across the capacitor (V꜀) are 180° out of phase. Therefore, the total potential drop across the LC combination is their algebraic difference.
Vₗ꜀ = Vₗ - V꜀
Vₗ꜀ = 1437.5 V - 1437.5 V = 0 V
This shows that at the resonating frequency, the potential drop across the LC combination is zero. The entire source voltage appears across the resistor (Vᵣ = 230 V), which is consistent with our calculation.