Two currents enter a node: and

PHYSICS · Single correct · AP · Physics 2: Algebra-based · Electric Circuits · Electric Circuits

Problem

Two currents enter a node: I1(t)=4sin⁡(ωt) AI_1(t)=4\sin(\omega t)\,\text{A} and I2(t)=2 AI_2(t)=2\,\text{A}. One current leaves the node: I3(t)=3cos⁡(ωt) AI_3(t)=3\cos(\omega t)\,\text{A}. Which expression gives the current I4(t)I_4(t) in the remaining branch that also leaves the node? (a) I4(t)=4sin⁡(ωt)+2+3cos⁡(ωt) AI_4(t)=4\sin(\omega t)+2+3\cos(\omega t)\,\text{A} (b) I4(t)=4sin⁡(ωt)+2−3cos⁡(ωt) AI_4(t)=4\sin(\omega t)+2-3\cos(\omega t)\,\text{A} (c) I4(t)=−4sin⁡(ωt)−2+3cos⁡(ωt) AI_4(t)=-4\sin(\omega t)-2+3\cos(\omega t)\,\text{A} (d) I4(t)=4sin⁡(ωt)−2−3cos⁡(ωt) AI_4(t)=4\sin(\omega t)-2-3\cos(\omega t)\,\text{A}

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