Physics

Problem

Two currents enter a node:

I_1(t)=4\sin(\omega t)\,\text{A}

and

I_2(t)=2\,\text{A}

. One current leaves the node:

I_3(t)=3\cos(\omega t)\,\text{A}

. Which expression gives the current

I_4(t)

in the remaining branch that also leaves the node? a)

I_4(t)=4\sin(\omega t)+2+3\cos(\omega t)\,\text{A}

I_4(t)=4\sin(\omega t)+2-3\cos(\omega t)\,\text{A}

I_4(t)=-4\sin(\omega t)-2+3\cos(\omega t)\,\text{A}

I_4(t)=4\sin(\omega t)-2-3\cos(\omega t)\,\text{A}