Question

If current i1 = 3a sin t and i2 = 4a cos wt then i3 is Advertisement

Answer 1

Answer:5A sin( wt+53°)

Explanation:I3 = I1 + I2

I3= 3A sin wt + 4A cos wt

nA = √(3)^2+(4)^2

= 5A

I3 = 5A (3/5 sin wt + 4/5 cos wt)

= 5A ( cos 53° sin wt + sin 53° cos wt)

We know that , sin (A+B ) = sin A cos B + cos A sin B

Therefore, I3 = 5A sin ( wt + 53°).

Answer 2

Answer:

Required value is

$5a \sin(wt + {53}^{o} )$

Explanation:

We know,

$I₃ = I₁ + I₂....................(i)$

Given,

I₁ = 3a sin t and I₂= 4a cos wt

From equation (i),

$I₃ = 3asin(wt) + 4a \cos(wt) \\ = 5( \frac{3a}{5} \sin(wt) + \frac{4a}{5} \cos(wt) )$

$= 5a( \cos( {53}^{o} ) \sin(wt) + \sin( {53}^{o} ) cos(wt) )$

$= 5a \sin(wt + {53}^{o} )$