Mean Value of AC Signal- Formula, Practice Problems, FAQs

You must have come across the term mean value in mathematics. Let’s say there are 20 students in the class and you are asked to calculate the mean height of the students in the class. We know the mathematical formula to calculate the mean value. So that value gives us the general idea about the height of students in that class. Since the value of AC signal varies with respect to the time. So we use the mean value to represent the signal.

Table of Contents

Mean value
Mean value of sinusoidal function in complete cycle
Mean value of sinusoidal function in half cycle
Practice Problems
FAQs

Mean value

In an AC circuit the values of current and voltage are a function of time. So that means they keep changing with respect to time. If someone asks the value of a signal he/she needs to specify the particular time. So the one parameter that is used to define the AC signal is Mean Value or Average Value.

Average value of AC voltage is defined based on the charge transfer. It is the equivalent AC voltage at which the charge transfer in the AC circuit is equal to charge transfer in the DC circuit.

By using the concept of integration from mathematics, we can calculate the mean value of current which is the function of time as given by the formula. Below is the graphical representation of alternating current i=f(t) as the function of time from time instant t1 to t2.

So the average value of current over a certain interval of time is given by,

$i_{a v g} = \frac{\int_{t_{1}}^{t_{2}} i d t}{\int_{t_{1}}^{t_{2}} d t} = \frac{\int_{t_{1}}^{t_{2}} f (t) d t}{\int_{t_{1}}^{t_{2}} d t} = \frac{1}{Δt} \int_{t_{1}}^{t_{2}} i d t$

Between any two arbitrary interval x1 and x2, the average value of any function of 𝑥 is defined as:

$< f (x) > = \frac{\int_{x_{1}}^{x_{2}} f (x) d x}{\int_{x_{1}}^{x_{2}} d x} = \frac{A r e a u n d e r f (x) c u r v e f r o m x_{1} t o x_{2}}{L e n g t h o f i n t e r v a l f r o m x_{1} t o x_{2}}$

Mean value of sinusoidal function in complete cycle

$i_{a v g} = \frac{1}{Δt} \int_{t_{1}}^{t_{2}} i d t = \frac{1}{Δt} \int_{0}^{T} i_{o} s i n (ω t) d t$

$i_{a v g} = \frac{- 1}{ω T} i_{o} [c o s (ω t)] 0 T$

$i_{a v g} = \frac{- i_{o}}{ω T} [c o s (ω T) - c o s (0)] ⬚ ⬚$

We know, $T = \frac{2 Π}{ω} \Rightarrow ω T = 2 Π$

Therefore, $i_{a v g} = \frac{- i_{o}}{ω T} [c o s 2 Π - c o s 0] ⬚ ⬚$

$i_{a v g} = \frac{- i_{o}}{ω T} [1 - 1] ⬚ ⬚$

$i_{a v g} = 0$ ……for one period

So the average value of sinusoidal function is zero over one complete cycle. Even by observing we can conclude that the mean value of a perfectly sinusoidal signal is zero in the complete cycle, because the amount signal is positive is the same as negative so net will definitely be zero. So we calculate the average value for half a cycle.

Mean value of sinusoidal function in half cycle

$i_{a v g} = \frac{1}{Δt} \int_{t_{1}}^{t_{2}} i d t = \frac{1}{Δt} \int_{0}^{\frac{T}{2}} i_{o} s i n (ω t) d t$

$i_{a v g} = \frac{- 1}{ω T} i_{o} [c o s (ω t)] 0 T 2$

$i_{a v g} = \frac{- i_{o}}{ω T} [c o s (ω T 2) - c o s (0)] ⬚ ⬚$

We know, $T = \frac{2 Π}{ω} \Rightarrow \frac{ω T}{2} = Π$

Therefore, $i_{a v g} = \frac{- i_{o}}{ω T} [c o s Π - c o s 0] ⬚ ⬚$

$i_{a v g} = \frac{- i_{o}}{ω T} [- 1 - 1] ⬚ ⬚$

$i_{a v g} = \frac{2}{Π} i_{o}$ ……for one period

Practice Problems

Q. Find the average value of the current, the graph is shown below with respect to time.

A. According to the formula,

$i_{a v g} = \frac{1}{Δt} \int_{t_{1}}^{t_{2}} i d t$

$i_{a v g} = \frac{1}{T} \int_{0}^{T} i (t) d t$

We can write it as ,

$i_{a v g} = \frac{1}{T i m e P e r i o d} \times A r e a o f g r a p h$

In given graph time period is 2 s.

$A r e a = \frac{1}{2} \times b a s e \times h e i g h t$

$A r e a = \frac{1}{2} \times 2 \times 10$

Area=10

$i_{a v g} = \frac{1}{2} \times 10$

$i_{a v g} = 5 A$

Q. Find the average value of the current, the graph is shown below with respect to time.

A. According to the formula,

$i_{a v g} = \frac{1}{Δt} \int_{t_{1}}^{t_{2}} i d t$

$i_{a v g} = \frac{1}{T} \int_{0}^{T} i (t) d t$

We can write it as ,

$i_{a v g} = \frac{1}{T i m e P e r i o d} \times A r e a o f g r a p h$

Area of the rectangular part is 10T

Area of the triangular part is $\frac{1}{2} \times T \times 10 = 5 T$

So total area is 15T

$i_{a v g} = \frac{1}{T} \times 15 T$

$i_{a v g} = 15 A$

Q. The electric current in a circuit is given by i(t)=io(tT) for some time. Calculate the mean value or average value of current for the periods t=0 to t=T.

A. $i_{a v g} = \frac{1}{Δt} \int_{t_{1}}^{t_{2}} i d t$

$i_{a v g} = \frac{1}{T} \int_{0}^{T} i (t) d t$

$i_{a v g} = \frac{1}{T} \int_{0}^{T} i_{o} (\frac{t}{T}) d t$

$i_{a v g} = \frac{i_{o}}{T^{2}} \int_{0}^{T} t d t$

$i_{a v g} = \frac{i_{o}}{T^{2}} [t 22] 0 T$

$i_{a v g} = \frac{i_{o}}{T^{2}} \frac{T^{2}}{2}$

$i_{a v g} = \frac{i_{o}}{2}$

Q. Find the average value of the square waveform shown below till t=2T.

A. According to the formula,

$v_{a v g} = \frac{1}{Δt} \int_{t_{1}}^{t_{2}} v d t$

$v_{a v g} = \frac{1}{T} \int_{0}^{T} v (t) d t$

We can write it as ,

$v_{a v g} = \frac{1}{T i m e P e r i o d} \times A r e a o f g r a p h$

Area of the graph in one time period,

$A r e a = V_{o} \frac{T}{2} - V_{o} \frac{T}{2}$

Area=0

So $v_{a v g} = 0$

FAQs

Q. What is the significance of average value?
A. Average value of AC voltage is defined based on the charge transfer. It is the equivalent AC voltage at which the charge transfer in the AC circuit is equal to charge transfer in the DC circuit.

Q. What is the significance of RMS value?
A. RMS value is defined based on the heating effect of the waveform. It is the equivalent value of AC voltage at which heat dissipation in an AC circuit is equal to heat dissipation in a DC circuit.

Q. Can the form factor be less than unity?
A. No. As $F o r m F a c t o r = \frac{R M S V a l u e}{A v e r a g e V a l u e}$ . RMS value is always greater than or equal to one. So the form factor is always greater than or equal to one. Form factor is unity for the square waveform.

Q. Why is the average value of sinusoidal signal calculated in half cycle?
A. The average value of a whole sinusoidal waveform over one complete cycle is zero as the two halves cancel each other out, so the average value is taken over half a cycle.