Question
If $${i_1} = 3\sin \omega t$$ and $${i_2} = 4\cos \omega t,$$ then $${i_3}$$ is
If $${i_1} = 3\sin \omega t$$ and $${i_2} = 4\cos \omega t,$$ then $${i_3}$$ is
A.
$$5\sin \left( {\omega t + {{53}^ \circ }} \right)$$
B.
$$5\sin \left( {\omega t + {{37}^ \circ }} \right)$$
C.
$$5\sin \left( {\omega t + {{45}^ \circ }} \right)$$
D.
$$5\cos \left( {\omega t + {{53}^ \circ }} \right)$$
Answer :
$$5\sin \left( {\omega t + {{53}^ \circ }} \right)$$
Solution :
$$\eqalign{ & {i_3} = {i_1} + {i_2} = 3\sin \omega t + 4\cos \omega t \cr & = 5\sin \left( {\omega t + \phi } \right), \cr & {\text{where}}\,\tan \phi = \frac{3}{4}\,\,or\,\,\phi = {53^ \circ }. \cr} $$
$$\eqalign{ & {i_3} = {i_1} + {i_2} = 3\sin \omega t + 4\cos \omega t \cr & = 5\sin \left( {\omega t + \phi } \right), \cr & {\text{where}}\,\tan \phi = \frac{3}{4}\,\,or\,\,\phi = {53^ \circ }. \cr} $$
