# Thread: Current in circuit equation (trigonometry!)

1. ## Current in circuit equation (trigonometry!)

Hey this is my first post... been really stuck on this particular question from a paper:

In a particular circuit the current, I amperes, is given by:

I=4sinθ - 3cosθ θ>0

where θ is an angle related to the voltage.

Given that I=R x sin(θ-a), where R>0 and 0<=a<360degrees,

a) find the value of R, and the value of a to 1 decimal place.
b) Hence solve the equation 4sinθ - 3cosθ = 3 to find the values of θ between 0 and 360degrees.
c) Write down the greatest value for I.
d) Find the value of θ between 0 and 360degrees at which the greatest value of I occurs.

I think if someone could show me the steps to answering part a) I could be able to do the rest myself.
Thanks

Hey this is my first post... been really stuck on this particular question from a paper:

In a particular circuit the current, I amperes, is given by:

I=4sinθ - 3cosθ θ>0

where θ is an angle related to the voltage.

Given that I=R x sin(θ-a), where R>0 and 0<=a<360degrees,

a) find the value of R, and the value of a to 1 decimal place.
b) Hence solve the equation 4sinθ - 3cosθ = 3 to find the values of θ between 0 and 360degrees.
c) Write down the greatest value for I.
d) Find the value of θ between 0 and 360degrees at which the greatest value of I occurs.

I think if someone could show me the steps to answering part a) I could be able to do the rest myself.
Thanks
You have:

I=4 sin(θ) - 3 cos(θ) ..(1)

and

I=R sin(θ-a) ..(2)

where R>0 and θ in [0, 360).

Applying the trig identity sin(θ-a) = sin(θ) cos(a) - cos(θ) sin(a) and putting it into (2)
we have:

I = R sin(θ) cos(a) - R cos(θ) sin(a) ...(3)

Equating the coefficients of sin(θ) and cos(θ) in (1) and (3) we have:

R cos(a) = 4 ..(4)

and

R sin(a) = 3 ..(5)

So squaring and adding (4) and (5):

R^2 cos^2(a) + R^2 sin^2(a) = 4^2 + 3^2 = 25.

So as cos^2(a) + sin^2(a) = 1, R^2 = 25^2, and so R=5.

From (4) and (5) we deduce that cos(a)>0 and sin(a)>0, so a is in the first quadrant,
and dividing (5) by (4) that:

tan(a) = 3/4,

so a = arctan(3/4) ~= 36.87 degrees

RonL

3. cheers! didnt see the similarities between the two equations after using the addition rule before.