Problem 51-1    Source:  Problem 29 - page 402 -   BOYLESTAD, Robert - Book: Introdução à Análise de Circuitos - 10ª edição - Ed. Pearson Education do Brasil - 2004.

Calculate the phase difference between the waveforms of each pair below:

a)  v = 4 sin (ωt + 50°)   and   i = 6 sin (ωt + 40°)

b)  v = 25 sin (ωt - 80°)   and   i = 5 x 10^-3 sin (ωt - 10°)

c)  v = 0.2 sin (ωt - 60°)   and   i = 0.1 sin (ωt + 20°)

d)  v = 200 sin (ωt - 210°)   and   i = 25 sin (ωt - 60°)

Solution of the Problem 51-1   

Item a   

Note that both v and i are written as a sine function. Therefore, the difference between the angles of the functions is the response. Soon:

Then:

Therefore, we conclude that the voltage is advanced by 10° in relation to the current.

Observation -    This conclusion is due to the definition adopted for the angle theta. Choosing the definition θ = θ_i - θ_v then θ = -10° and the conclusion would be that the current would be delayed by 10° in relation to the voltage. That is, the result is the same.

Item b   

In the same way as in the previous item we have:

Therefore, it is concluded that the voltage is delayed (due to the negative signal) of 70° in relation to the current.

Item c   

From the definition:

Thus, it is concluded that the voltage is delayed (due to the negative signal) of 80° in relation to the current.

Item d   

In the same way as in the previous item we have:

Therefore, it is concluded that the voltage is delayed (due to the negative signal) of 150 ° in relation to the current.

Notice that in this exercise there was no concern with the range of functions, since they are completely independent variables.

You can see the results graphically (not in scale) in the four figures below.

Phasors Graphics