11.4 Effective or RMS Value

← Back to Fundamentals of Electric Circuits Overview

11.4 Effective or RMS Value

The idea of effective value arises from the need to measure the efectiveness of a voltage or current source in delivering power to a resistive load.

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

Practice Problem 11.6

For Example 11.6.

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

In Fig. 11.13, the circuit in (a) is ac while that of (b) is dc. Our objecti ve is to find Ieff that will transfer the same po wer to resistor R as the sinusoid i. The average power absorbed by the resistor in the ac circuit is

$$P = \frac{1}{T} \int_0^T i^2 R \, dt = \frac{R}{T} \int_0^T i^2 \, dt \tag{11.22}$$

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

$$P = I_{\text{eff}}^2 R \tag{11.23}$$

Equating the expressions in Eqs. (11.22) and (11.23) and solving for Ieff, we obtain

$$I_{\text{eff}} = \sqrt{\frac{1}{T} \int_0^T i^2 dt}$$

(11.24)

The effective value of the v oltage is found in the same w ay as current; that is,

$$V_{\text{eff}} = \sqrt{\frac{1}{T} \int_0^T v^2 dt}$$

(11.25)

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

$$I_{\rm eff} = I_{\rm rms}, \qquad V_{\rm eff} = V_{\rm rms} \tag{11.26}$$

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

$$X_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T x^2 dt}$$

(11.27)

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

Equation 11.27 states that to find the rms value of x(t), we first find its square x² and then find the mean of that, or

$$\frac{1}{T} \int_0^T x^2 dt$$

and then the square root ( √ ______ ) of that mean. The rms value of a constant is the constant itself. For the sinusoid i(t) = Im cos ωt, the effective or rms value is

$$I_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T I_m^2 \cos^2 \omega t \, dt}$$ $$= \sqrt{\frac{I_m^2}{T} \int 0 \, T \, \frac{1}{2} (1 + \cos 2\omega t) \, dt} = \frac{I_m}{\sqrt{2}} \tag{11.28}$$

Similarly, for v(t) = Vm cos ωt,

$$V_{\rm rms} = \frac{V_m}{\sqrt{2}}\tag{11.29}$$

Keep in mind that Eqs. (11.28) and (11.29) are only valid for sinusoidal signals.

The average power in Eq. (11.8) can be written in terms of the rms values.

$$P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) = \frac{V_m}{\sqrt{2}} \frac{I_m}{\sqrt{2}} \cos(\theta_v - \theta_i)$$

= $V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i)$ (11.30)

Similarly, the average power absorbed by a resistor R in Eq. (11.11) can be written as

$$P = I_{\rm rms}^2 R = \frac{V_{\rm rms}^2}{R}$$

(11.31)

When a sinusoidal voltage or current is specified, it is often in terms of its maximum (or peak) v alue or its rms v alue, since its a verage value is zero. The power industries specify phasor magnitudes in terms of their rms values rather than peak v alues. 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, analog v oltmeters and ammeters are designed to read directly the rms value of voltage and current, respectively.

Determine the rms v alue of the current w aveform in Fig. 11.14. If the Example 11.7 current is passed through a 2- Ω resistor, find the average power absorbed by the resistor.

Solution:

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

$$i(t) = \begin{cases} 5t, & 0 < t < 2 \\ -10, & 2 < t < 4 \end{cases}$$

The rms value is

$$I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T i^2 dt} = \sqrt{\frac{1}{4} \left[ \int_0^2 (5t)^2 dt + \int_2^4 (-10)^2 dt \right]}$$ $$= \sqrt{\frac{1}{4} \left[ 25 \frac{t^3}{3} \right]_0^2 + 100t \Big|_2^4} = \sqrt{\frac{1}{4} \left( \frac{200}{3} + 200 \right)} = 8.165 \text{ A}$$

The power absorbed by a 2-Ω resistor is

$$P = I_{\text{rms}}^2 R = (8.165)^2 (2) = 133.3 \text{ W}$$

Find the rms value of the current waveform of Fig. 11.15. If the current flows through a 9-Ω resistor, calculate the average power absorbed by the resistor.

Answer: 9.238 A, 768 W.

Figure 11.14 For Example 11.7.

For Example 11.8.

Example 11.8 The waveform shown in Fig. 11.16 is a half-w ave rectified sine wave. Find the rms value and the amount of average power dissipated in a 10-Ω resistor.

Solution:

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

$$v(t) = \begin{cases} 10 \sin t, & 0 < t < \pi \\ 0, & \pi < t < 2\pi \end{cases}$$

The rms value is obtained as

$$V_{\text{rms}}^2 = \frac{1}{T} \int_0^T v^2(t) \, dt = \frac{1}{2\pi} \left[ \int_0^{\pi} (10 \sin t)^2 \, dt + \int_{\pi}^{2\pi} 0^2 \, dt \right]$$

But sin² t = __1 ² (1 − cos 2t). Hence,

$$V_{\text{rms}}^2 = \frac{1}{2\pi} \int_0^{\pi} \frac{100}{2} (1 - \cos 2t) dt = \frac{50}{2\pi} \left( t - \frac{\sin 2t}{2} \right) \Big|_0^{\pi}$$ $$= \frac{50}{2\pi} \left( \pi - \frac{1}{2} \frac{\sin 2\pi - 0}{2} \right) = 25, \qquad V_{\text{rms}} = 5 \text{ V}$$

The average power absorbed is

$$\sqrt{\pi} \alpha \int_0^{\pi} \alpha \sqrt{1 - \frac{1}{2} \sin 2\pi} dt$$ $$P = \frac{V_{\text{rms}}^2}{R} = \frac{5^2}{10} = 2.5 \text{ W}$$

Practice Problem 11.8

Find the rms value of the full-wave rectified sine wave in Fig. 11.17. Calculate the average power dissipated in a 6-Ω resistor.

Answer: 70.71 V, 833.3 W.