The idea of *effective value* arises from the need to measure the effectiveness of a voltage or current source in delivering power to a resistive load. We will learn the term of rms voltage and current here.

**Contents**show

The

effective valueof a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.

*Make sure to read what is ac circuit first.*

There are several types of power in ac circuit:

- Maximum average power transfer
- Voltage and current RMS
- Power factor and apparent power
- Power triangle and power complex
- Power ac conservation

## How to Calculate RMS Voltage and Current

In Figure.(1), the circuit in (a) is ac while the circuit in (b) is dc. Our objective is to find *I _{eff}* that will transfer the same power to resistor R as the sinusoid

*i*.

Figure 1. Finding the effective current: (a) ac circuit, (b) dc circuit |

The average power absorbed by the resistor in the ac circuit is

(1) |

while the power absorbed by the resistor in the dc circuit is

(2) |

Equating the expressions in Equations.(1) and (2) and solving for *I _{eff}*, we obtain

(3) |

The effective value of the voltage is found in the same way as current; that is

(4) |

This indicates that the effective value is the (square) *root* of the *mean* (or average) of the *square* of the periodic signal.

Thus, the effective value is often known as the *root-mean-square* value, or *rms* value for short; and we write

(5) |

For any periodic function *x(t)* in general, the rms value is given by

(6) |

The

effective valueof a periodic signal is its root mean square (rms) value.

Equation.(6) states that to find the rms value of *x(t)*, we first find its *square x ^{2}* and then find the mean of that, or

(7) |

and the square root (√) of that mean. The rms value of a constant is the constant itself.

For the sinusoid *i(t)* = *I _{m}* cos ωt, the effective or rms value is

(8) |

Similarly, for *v(t)* = *V _{m}* cos ωt,

(9) |

Keep in mind that Equations.(8) and (9) are only valid for sinusoidal signals.

Before moving on, remember all the equations we had in Instantaneous Power and Average Power Formula.

The average power can be written in terms of the rms values

(10) |

Similarly, the average power absorbed by a resistor R can be written as

(11) |

When a sinusoidal voltage or current is specified, it is often in terms of its maximum (or peak) value or its rms value, since its average value is zero.

The power industries specify phasor magnitudes in terms of their rms values rather than peak values.

For instance, the 110 V available at every household is the rms value of the voltage from the power company.

It is convenient in power analysis to express voltage and current in their rms values.

Also, analogue voltmeters and ammeters are designed to read directly the rms value of voltage and current, respectively.

Read also : power factor and apparent power

## Root Mean Square Value Examples

For better understanding let us review examples below :

**1. Determine rms value of the current waveform in Figure.(2). If the current is passed through a 2 Ω resistor, find the average power absorbed by the resistor.**

Figure 2 |

*Solution :*

The period of the waveform is *T* = 4. Over a period, we can write the current waveform as

The rms value is

The power absorbed by a 2 Ω resistor is

**2. The waveform is shown in Figure.(3) is a half-wave rectified sine wave. Find the rms value and the amount of average power dissipated in a 10 Ω resistor.**

Figure 3 |

*Solution :*

The period of the voltage waveform is T = 2π, and

The rms value is obtained as

But sin^{2}*t* = ½(1 – cos 2*t*). Hence,

The average power absorbed is