Alternating Current

Alternating Current (AC) and Direct Current (DC) are two types of electricity based on the direction of flow. AC, used in households, it reverses direction periodically, while DC used in batteries, flows in one direction. It is an electric current that periodically reverses its direction, with a voltage that varies continuously over time. AC is commonly used for household and industrial power supply, typically at a frequency of 50 Hz or 60 Hz.

Alternating Current Voltage Symbol

The alternating current voltage symbol or alternating current symbol consists of a circle with a sine wave inside it, where the sine wave within the circle represents the waveform of the alternating current (AC) being generated by the source.

AC Generators

Alternating currents can be generated using devices known as alternators. Apart from these, many circuits can produce alternating voltages and currents.

One of the most basic ways to produce an AC voltage or current is by using a single-coil AC generator. The generator consists of two-pole magnets and a single rectangular loop of wire. For supplying AC, the following three wires are used:

• Hot wire – for power transmission.
• Neutral wire – connected to the earth, provides a return path for the current in the hot wire.
• Earth wire – connected to the metallic parts to prevent electric shock hazards.

Alternating Current Waveform

The AC current forms a wave similar to the sinusoidal wave. The wave starts from zero and reaches its maximum value, then started decreasing to reach the zero value it continues to decrease till it reaches its minimum value(negative maximum value), and then it increases again to reach the zero value. The AC waveform is positive in the first half and negative in the second half.

Characteristics of Alternating Current (AC)

1. Sinusoidal Waveform

Alternating current has a smooth, periodic waveform that resembles a sine wave. It represents the instantaneous value of current or voltage with time and is therefore called a sinusoidal waveform.

2. Frequency and Amplitude

Frequencyis the number of AC cycles per second (measured in HZ), usually 50 or 60 Hz, while amplitude is the maximum value of the current or voltage.

3. Alternating Current Peak Value

The peak value of an AC waveform refers to the maximum positive or negative value reached during each cycle. It represents the highest magnitude of the waveform. Generally, this is represented as I_m or I₀.

4. Alternating Current Average Value

Unlike DC, AC does not have a constant value as it oscillates with time. The average value of AC is therefore calculated over a half cycle, since over a full cycle it becomes zero due to the cancellation of positive and negative halves. The average value for a positive half-cycle is

I_{av} = \frac{2I_m}{\pi}

Where,

• I_av is the average value of AC, and
• I_m is the peak value of AC.

5. Alternating Current Root Mean Square (RMS) Value

The RMS value of AC is the equivalent DC value that produces the same amount of heat in a conductor as the AC. Hence, it is also called the effective value of sinusoidal AC. Mathematically,

I_{rms} = \frac{I_m}{\sqrt{2}}

Where,

• I_rms is the RMS value of AC, and
• I_m is the peak value of AC.

6. Phase and Phase Difference

In AC systems, multiple wave forms can exist with a phase difference between them. Phase refers to the relative position of one waveform with respect to another and is measured in degrees or radians. The phase difference indicates the angular displacement between wave forms and shows how they align in time.

Alternating Current (AC) Vs Direct Current (DC)

AC Circuit Analysis

1. Reactance is the opposition offered by capacitive or inductive components in an AC circuit due to energy storage. It is measured in ohms, denoted by X, and depends on frequency.

2. Impedance is the total opposition to AC flow, combining resistance and reactance. It is measured in ohms and represented by Z.

Ohm's Law for AC Circuits

Ohm's Law for AC circuits relates the RMS values of voltage, current, and impedance. It is expressed as

V = I × Z

Where,

• V is the RMS value of voltage across the circuit (in volts),
• I is the RMS value of current (in amperes), and
• Z is the impedance of the circuit (in ohms).

AC Circuit with only Resistance

The source produces a sinusoidal varying potential difference across its terminals. Let us assume that this potential difference is called AC voltage. Then, this sinusoidally varying voltage can be expressed by the equation given below, 

v = v_msinωt 

Here, v_m is the amplitude of the oscillating voltage, and ω denotes its angular frequency. 

The amplitude of the current in the given circuit is given by, 

i = -\frac{v_m}{R}

The figure given below plots both of the values on the graph. Notice that both current and voltage go to a maximum and become zero at the same time. This means that they have zero phase difference. 

It can be inferred from the figure that the average current is also zero in a single cycle is zero. 

AC Circuit with only Capacitor

Capacitive Reactance (X_c): Capacitors store electrical energy in an electric field. In an AC circuit, the capacitor charges and discharges, creating a reactance that opposes the current flow. The formula for capacitive reactance is,

X_c = \frac{1}{2\pi f C}

Where,

• "f" is the frequency of the AC signal in hertz
• "C" is the capacitance in farads

AC Circuit with only an Inductor

Inductive Reactance (X_L): Inductors store energy in a magnetic field. In an AC circuit, the changing magnetic field induces a voltage that opposes the current flow. The formula for inductive reactance is,

X_L = 2πfL

Where,

• "f" is the frequency of the AC signal in hertz
• "L" is the inductance in Henries

Power Dissipation in an AC Circuit

Even though the average current through the cycle is zero, that does not mean that the average power dissipation through the cycle is also zero. Dissipation of electrical energy is there. It's known that Joule's heating is given by I²R and depends on I². This term is always positive, irrespective of the sign of "i". Thus average dissipation cannot become zero. 

P = (I_m)²sin²ωtR

This is the instantaneous power in the circuit. Average power dissipation is given by, 

P_{avg} = \frac{1}{2} I_m^2 R

P_{avg} = \frac{V_m^2}{2R}

Expressing this expression of power is similar to the usual expression. 

P_{avg} = I^2 R

P_{avg} = \frac{V^2}{R}

Where,

• I is RMS value of Current i.e., I = \frac{I_m}{\sqrt{2}} and 
• V is RMS value of Voltage i.e., V = \frac{V_m}{\sqrt{2}}

Applications of Alternating Current

There are various use cases of Alternating Current, some of these use cases are as follows:

• It is widely used in electrical networks for distributing power efficiently.
• Motors are preferred in industries due to their efficiency and speed control.
• Most modern electrical devices like LED lights, washing machines, and cooling systems operate on it.
• Signals are used in telephony and data transmission, and medical machines like X-rays and MRI scanners require them.

Related Articles:

• Power in AC Circuit
• How to Convert AC to DC?

Solved Problems

Problem 1: A light bulb is rated at 100W for a 200V supply. Find the resistance of the bulb. 

Solution:  Average power is given by, 

P = V²/R

Given,

• V = 200 
• R =? 
• P = 100 

Plugging these values into the equation, 

P = V²/R

⇒ 100 = (200)²/R 

⇒ R = 400 Ohms

Problem 2: Find the expression for the current flowing in the circuit with a resistance of 5 ohms. The voltage source works on the expression given, v = 10sin(30t) 

Solution:  For calculating the value of the current Kirchhoff's law can be used. 

∑ε(t) = 0

Applying this law to the circuit shown above, 

v + iR = 0 

10sin(30t) + i5 = 0 

i = -2sin(30t)

Problem 3: Find the expression for the current flowing in the circuit with aresistance of20 ohms. The voltage source works on the expression given, v = 5sin(20t) 

Solution:  For calculating the value of the current Kirchhoff's law can be used. 

∑ε(t) = 0

Applying this law to the circuit shown above, 

v + iR = 0 

5sin(20t) + i(20) = 0 

i = -0.25sin(20t)

Problem 4: Average power dissipation in a circuit is given as 100W. The resistance of the circuit is 10 ohms. Find the peak value of the voltage in the circuit. 

Solution:  Average power is given by, 

P = V²/R

Given: 

• V =? 
• R = 10 
• P = 100 

Plugging these values into the equation, 

P = V²/R

⇒ 100 = (V)²/10 

⇒ V = √1000

V = 10√10 V 

This is the RMS value of the voltage. 

Peak Value will be, 

V_peak = V_rms√2

⇒ V_peak = 10√20

⇒ V_peak = 20√5 V

Question 5: A sinusoidal AC voltage has a peak value of 200 V and is applied across a 50 Ω resistor. Calculate the RMS voltage, the average voltage over a half cycle, and the average power dissipated in the resistor.

Solution:  The RMS voltage of a sinusoidal AC is given by

V_{rms} = \frac{V_m}{\sqrt{2}}

Here, V_m = 200 V, so

V_{rms} = \frac{200}{\sqrt{2}} \approx 141.42\ \text{V}

The average voltage over a half cycle is given by:

V_{av} = \frac{2V_m}{\pi}

V_{av} = \frac{2 \times 200}{\pi} \approx 127.32\ \text{V}

The average power dissipated in the resistor is calculated using:

P_{avg} = \frac{V_{rms}^2}{R}

P_{avg} = \frac{(141.42)^2}{50} \approx 400\ \text{W}

RMS Voltage = 141.42 V

Average Voltage = 127.32 V

Average power = 400 W

Unsolved Problems

Question 1: A sinusoidal AC voltage has a peakof value of 250 V. Find the RMS voltage and the average voltage over a half cycle.

Question 2: An AC source of RMS voltage 220 V is connected to a 44 Ω resistor. Calculate the average power dissipated in the circuit

Question 3: A sinusoidal AC current has a peak value of 12 A. Determine its RMS value and average value over a half cycle.

Question 4: A 50 Hz AC supply is connected to a circuit containing a 30 Ω resistor and a capacitor of 50 µF in series. Find the capacitive reactance and the impedance of the circuit.

Question 5: A sinusoidal AC voltage of peak value 200 V is applied across a pure resistor of 40 Ω. Calculate the RMS current, average power, and instantaneous power when ωt = π/4.