# Values of Alternating Voltage and Current

## Values of AC voltage and current in a circuit may be given as peak value, average value, or rms value.

An AC voltage is described completely when either the peak or RMS value is given along with the frequency of the sine wave. For example, the common value of voltage used in any home in the United States is 120 VAC effective with a frequency of 60 Hz. Since an AC sine wave for voltage or current has many different and instantaneous values throughout the cycle, we need to define the relationships between the different AC values.

## Peak Value

The maximum instantaneous value of either the positive or negative sine wave alternation is the peak value of an AC current or voltage. Assuming that our AC voltage is a sine wave with the initial phase of zero, the positive peak value is above the 0 V baseline and exists at 90°, whereas its negative peak value is below the 0 V baseline and exists at 270° (see *Figure 1*).

**Figure 1.** The peak value of a sine wave with the initial phase of zero occurs at 90° and 270°. The positive peak value occurs at 90°, and the negative peak value occurs at 270°.

**Figure 1.**The peak value of a sine wave with the initial phase of zero occurs at 90° and 270°. The positive peak value occurs at 90°, and the negative peak value occurs at 270°.

The peak-to-peak value of an AC waveform is a value measured from the positive peak to the negative peak. Voltages and currents in an AC circuit are sometimes given in their peak-to-peak values. Because a pure sine wave with an average value of zero has equal values above and below the 0 V baseline, the positive and negative peaks are equal in magnitude. When either the positive or negative peak value of the voltage is known, then the peak-to-peak value of the voltage can be calculated by applying the following formula:

V_{P-P} = 2 × V_{P}

where

V_{P-P} = AC peak-to-peak value (in V)

V_{P} = AC positive peak value (in V)

**Example:** What is the peak-to-peak value of an AC voltage if the positive peak voltage value is equal to 170 V?

V_{P-P} = 2 × V_{P}

V_{P-P} = 2 × 170V = 340 V

Formulas for voltage apply equally to alternating current, with peak current I_{P} and peak-to-peak current I_{P-P} substituted for V_{P} and V_{P-P}. In certain applications, the peak value is the preferred metric when explaining the characteristics of a circuit or a signal. For example, in a bridge rectifier, the filter capacitor charges to the peak of the applied voltage. The deflection of the electron beam in a CRT computer monitor is responsive to the peak-to-peak voltages. Typically, the vertical scale on an oscilloscope is adjusted depending on the peak-to-peak or peak values.

## Average Value

The average value of an AC voltage or current in an AC circuit is the mathematical mean of all instantaneous values in a sine wave. For certain AC applications, the average value of a sine wave is used. Either voltage or current can be represented. When a signal is symmetric with respect to the x-axis, we get an average value of zero when the average value above the baseline is averaged with the average value below the baseline. In these cases, we might use the average value of one half-cycle, or one alternation, of the sine wave. Either a positive or negative alternation can be used (see *Figure 2*).

**Figure 2.** The average value of one half-cycle is equal to the peak value multiplied by 0.636.

**Figure 2.**The average value of one half-cycle is equal to the peak value multiplied by 0.636.

The average voltage of a sine wave is the average of all instantaneous values for one alternation. That is, if the values of the sine function were added for each degree from 0° to 180°, with the sum divided by 180, an approximation of the average value would be calculated. The value of average voltage is calculated to be equal to the instantaneous peak value multiplied by a factor of 0.636. The average voltage can be calculated by applying the following formula:

V_{AVG} = V_{P} × 0.636

Where

V_{AVG} = average value of one half-cycle (in V)

V_{P}= peak value (in V)

0.636 = conversion factor from V_{P }to V_{AVG}

**Example:** What is the average value of one half-cycle of a sine wave if the peak voltage value is equal to 170 V?

V_{AVG} = V_{P} × 0.636

V_{AVG }= 170 × 0.636 = 108.12 V

If the half-cycle average value is known, the peak value can be calculated by applying the following formula:

V_{P}=V_{AVG}/0.636

Where

V_{AVG} = half-cycle average voltage value (in V)

V_{P}= peak value (in V)

0.636 = conversion factor from V_{P }to V_{AVG}

**Example:** What is the peak voltage value of an AC voltage if the half-cycle average value is equal to 108.12 V?

V_{P}=V_{AVG}/0.636

V_{P}=V_{AVG}/0.636= 108.12/0.636 =170V

## Root Mean Square (RMS) Value

The RMS value of an AC current or voltage gives us the equivalent DC current or voltage value that produces the same heating effect as the original alternating waveform when applied to a purely resistive load. For example, an AC current with an RMS value of 1 A causes the same heating effect as 1 A of DC. When AC flows through a resistance, the power dissipated is constantly changing with the AC sine wave. The power dissipated at any instant can be calculated by applying the following formula:

P_{D} = I^{2} × R

Where

P_{D} = power dissipated (in W)

I = circuit current (in A)

R = circuit resistance (in Ω)

**Example:** What is the power dissipated in a 4 A circuit with a resistance of 10 Ω?

P_{D} = I^{2} × R

P_{D} = I^{2} × R = 4^{2} × 10 = 160 Watts

The power dissipated is always a positive value, even when the current value is negative. The value of I^{2} is always positive because a negative value multiplied by a negative value yields a positive value.

When calculating the rms value, the peak value of the current or voltage sine wave is divided by the square root of 2. A current sine wave can be calculated by applying the following formula:

\[{{I}_{RMS}}=\frac{{{I}_{P}}}{\sqrt{2}}\]

where

I_{RMS} = current RMS value (in A)

I_{P} = current peak value (in A)

\(\sqrt{2}\)= 1.414

Since the reciprocal value of 1.414 is 0.707, the RMS value of a current sine wave can also be calculated by applying the following formula:

I_{RMS} = 0.707 × I_{P}

When calculating voltage sine waves, V_{RMS} is substituted for I_{RMS}.

**Example:** What is the RMS voltage if peak voltage is equal to 230 V?

V_{RMS} = 0.707 × V_{P}

V_{RMS} = 0.707 × 230 = 162.6 V

**Figure 3.** The effective value of voltage or current is equal to the peak value multiplied by 0.707.

**Figure 3.**The effective value of voltage or current is equal to the peak value multiplied by 0.707.

**Note: **

*When an AC voltage or current rating is given without any qualifiers such as peak or average, it is implied that the voltage or current rating is the RMS value. For example, an appliance is rated at 120VAC/ 10A/ 60Hz. The RMS values are 120VAC and 10A.*

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