Unlike direct current (DC), which flows in one direction, alternating current (AC) periodically reverses its direction.
Key Concepts
AC Voltage and Current:
AC voltage and current oscillate sinusoidally.
They can be expressed as:
V = Vm sin(ωt)
I = Im sin(ωt ± φ)
where:
Vm and Im are the peak values
ω is the angular frequency
φ is the phase difference
Root Mean Square (RMS) Value:
The RMS value is the equivalent DC value of an AC quantity.
For sinusoidal AC: Vrms = Vm / √2 and Irms = Im / √2
AC Circuits with Components:
Resistive Circuit:
Inductive Circuit:
Capacitive Circuit:
LCR Circuit:
Power in AC Circuits:
Average power:
P =
Vrms * Irms * cos φ
(power factor)
Transformers:
Transformer equation: V₁ / V₂ = N₁ / N₂ = I₂ / I₁
Key Points to Remember
AC is efficient for transmission and distribution.
Power factor impacts circuit efficiency.
Transformers are essential for voltage transformations.
Understanding these concepts is essential for analyzing and solving problems related to AC circuits.
1. A 100 W 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?
Ans :(a) To determine the root-mean-square (RMS) current (Irms), we can employ Ohm’s Law:
Vrms = Irms * R
Where:
Vrms is the RMS voltage (220 V)
Irms is the RMS current (unknown)
R is the resistance (100 Ω)
Rearranging the equation:
Irms = Vrms / R = 220 V / 100 Ω = 2.2 A
Hence, RMS current in the circuit is 2.2 amperes.
(b) The net power consumption over a complete AC cycle is given by:
P = Vrms * Irms * cosφ
Where:
The phase difference between voltage and current( φ )
For a purely resistive circuit, since voltage and current are in phase, the phase difference is 0°, making cosφ equal to 1.
Therefore, the net power consumed is:
P = 220 V * 2.2 A * 1 = 484 W
Consequently, the net power consumption over a full cycle is 484 watts.
2. (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?
Ans :(a) The RMS (Root Mean Square) voltage is related to the peak voltage (V_peak) by the following formula:
V_rms = V_peak / √2
Given V_peak = 300 V, we can calculate V_rms:
V_rms = 300 V / √2 ≈ 212.13 V
Hence, the RMS voltage is 212.13 V.
(b) Similarly, the peak current (I_peak) is related to the RMS current (I_rms) by the formula:
I_peak = √2 * I_rms
Given I_rms = 10 A, we can calculate I_peak:
I_peak = √2 * 10 A ≈ 14.14 A
So, the peak current is approximately 14.14 A.
3. A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Ans :We can use the formula for the inductive reactance (XL) of an inductor:
XL = 2πfL
Where:
* frequency of the AC supply(f)
(50 Hz)
* L = inductance of the inductor
(44 mH = 0.044 H)
Calculating XL:
XL = 2π * 50 Hz * 0.044 H ≈ 13.82 Ω
to find the RMS current (Irms),we follow use Ohm’s law
Irms = Vrms / XL
Where:
* Vrms = RMS voltage
(220 V)
* XL = inductive reactance
(13.82 Ω)
Substituting the values:
Irms = 220 V / 13.82 Ω ≈ 15.9 A
Hence, the root mean square (RMS) current flowing through the circuit is nearly 15.9 amperes.
4. A 60 mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Ans :The given information is:
Capacitor capacitance
C = 60 μF
= 60 × 10⁻⁶ F
Supply voltage, V = 110 V
Frequency, f = 60 Hz
Angular frequency, ω = 2πf
The capacitive reactance (Xc) is given by:
Xc = 1 / (ωC)
= 1 / (2πfC)
1
= ————————————————-
2π(60 Hz)(60 x 10^-6 F)^-1
The root mean square (RMS) value of the current (I) is given by Ohm’s law:
I = V / Xc
110
= ——————————————–
2π(60 Hz)(60 x 10^-6 F)^-1
= 2.49 A
Therefore, the RMS value of the current in the circuit is 2.49 amperes.
5. In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Ans :For the inductive circuit:
RMS value of current,
I = 15.92 A
RMS value of voltage,
V = 220 V
The net power absorbed (P) can be calculated using the formula:
P = VI cos Φ
Where,
phase difference between voltage and current(Φ )
For a purely inductive circuit, the voltage leads the current by 90 degrees, resulting in a phase difference (Φ) of 90°.
Therefore,
P =
VI cos 90°
= 0
Hence, the net power absorbed in the inductive circuit is zero.
For the capacitive circuit:
RMS value of current,
I = 2.49 A
RMS value of voltage,
V = 110 V
The net power absorbed (P) can be calculated as:
P = VI cos Φ
For a purely capacitive circuit, the phase difference between voltage and current is also 90 degrees (Φ = 90°).
Therefore,
P
= VI cos 90°
= 0
Hence, the net power absorbed in the capacitive circuit is also zero.
6. A charged 30 mF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Ans :Given:
Capacitance,
C = 30 μF
= 30 x 10^-6 F
Inductance,
L = 27 mH
= 27 x 10^-3 H
by given formula we find, angular frequency (ω) of free oscillations in an LC circuit
ω = 1 / √(LC)
Substituting the given values:
ω = 1 / √((27 x 10^-3)(30 x 10^-6))
ω = 1 / (9 x 10^-4)
ω = 1.11 x 10^4 rad/s
Therefore, the angular frequency of free oscillations in the circuit is 1.11 x 10^4 radians per second.
7. A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 mF 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?
Ans :At resonance, the inductive reactance (XL) equals the capacitive reactance (XC):
XL = XC
2πfL = 1/(2πfC)
Solving for the resonant frequency (f):
f = 1 / (2π√(LC))
Substituting the given values:
f = 1 / (2π√(1.5 H × 35 × 10⁻³ F)) ≈ 7.75 Hz
At resonance, the impedance (Z) of the circuit is purely resistive, equaling the resistance (R).
Z = R = 20 Ω
We can determine the root-mean-square (RMS) current (Irms) by applying Ohm’s Law.
Irms = Vrms / Z = 200 V / 20 Ω = 10 A
The mean power (P) dissipated in the circuit is given by:
P = Irms² * R = (10 A)² * 20 Ω = 2000 W
Therefore, the average power transferred to the circuit in one complete cycle at resonance is 2000 W.
8. Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80mF, R = 40 W.
(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.
Ans :Given:
(a)source frequency
Inductance of the inductor,
L = 5.0 H
Capacitance of the capacitor,
C = 80 μF
= 80 x 10^-6 F
Resistance of the resistor,
R = 40 Ω
The resonance angular frequency (ωR) of an LCR circuit is given by:
ωR = 1 / √(LC)
Substituting the given values:
ωR = 1 / √(5 x 80 x 10^-6)
ωR = 10^3 / 20
ωR = 50 rad/s
Therefore, the circuit will resonate when the source frequency is 50 radians per second.
(b) Impedance and Current at Resonance
At resonance, the impedance (Z) of the circuit is purely resistive, equaling the resistance (R).
Z = R = 40 Ω
The amplitude of the current (I₀) can be calculated using Ohm’s law:
I₀ = V₀/Z
Where V₀ is the peak voltage, which is √2 times the RMS voltage (V):
V₀ = √2 × V = √2 × 230 V ≈ 325 V
So, the amplitude of the current is:
I₀ = 325 V / 40 Ω ≈ 8.13 A
(c) RMS Potential Drops Across Elements
Across the resistor (VR):
VR = IR = 8.13 A × 40 Ω ≈ 325 V (RMS)
Across the inductor (VL):
At resonance, the voltage across the inductor and capacitor are equal in magnitude but opposite in phase, so their net voltage is zero.
Across the capacitor (VC):
Similarly, at resonance, the voltage across the capacitor is also equal in magnitude but opposite in phase to the voltage across the inductor. Hence, the net voltage across the LC combination is zero.
Therefore, at the resonant frequency, the potential drop across the LC combination is zero.
