IB Mathematics Applications & Interpretation Question Type 8: Applying Complex Numbers in the Context of Sinusoidal Functions Exercises

Three balanced three-phase voltages are given by the equations:

$$V_a=100\cos (60t),V_b=100\cos (60t-120^{\circ }),V_c=100\cos (60t+120^{\circ })$$

Show that their sum is zero by phasor addition.

Express $V(t)$ in the complex exponential form $V(t)=\text{Re}\{V{e}^{i40t}\}$ and state the complex number $V$ in the form $r{e}^{i\theta }$.

Sum three sinusoidal voltages of the same frequency: $$V_1=10\cos (100t),V_2=10\cos (100t+120^{\circ }),V_3=10\cos (100t+240^{\circ })$$ where $V_1,V_2,V_3$ are in volts and $t$ is in seconds.

Express the total voltage as $V(t)=A\cos (100t+B)$.

Calculate the sum $$V(t)=10\cos (40t+350^{\circ })+10\cos (40t-10^{\circ })$$ and express it in the form $A\cos (40t+B)$.

Two voltages are given by $$V_1=20\cos (40t+170^{\circ }),V_2=20\cos (40t-170^{\circ }).$$ Find the resultant voltage $V(t)=V_1+V_2$ in the form $A\cos (40t+B)$.

Derive an expression for the sum $$V(t)=V_m\cos (\omega t+\alpha )+V_m\cos (\omega t-\alpha )$$ in the form $$V(t)=A\cos (\omega t+\varphi )$$ and find $A$ in terms of $V_m$ and $\alpha$.

Given two sinusoidal voltages $$V_1(t)=10\cos (40t),V_2(t)=20\cos (40t+30^{\circ }),$$ find the resulting voltage in the form $$V(t)=A\cos (40t+B).$$

A sinusoid has phasor representation $V=12+5i$. Convert this to the time-domain form $$v(t)=A\cos (40t+\varphi )$$ giving the values of $A$ and $\varphi$, where $\varphi$ is in degrees.

Given $$V_1(t)=15\cos (50t-20^{\circ })\text{and}{V}_2(t)=25\cos (50t+40^{\circ }),$$ determine the amplitude $A$ and phase $B$ of $$V(t)=V_1(t)+V_2(t)=A\cos (50t+B).$$

Express the sinusoidal voltage $V(t)=10\cos (40t)$ as a complex phasor $V=A\angle \varphi$.

Given phasors $V_1=5\angle 30^{\circ }$, $V_2=12\angle (-45^{\circ })$, and $V_3=8\angle 60^{\circ }$ at $\omega =40$rad/s, find the time-domain sum $$v(t)=\Re \{(V_1+V_2+V_3)e^{j40t}\}$$ in the form $A\cos (40t+B)$.

Express the difference $$V(t)=30\cos (40t+45^{\circ })-50\cos (40t-15^{\circ })$$ in the form $A\cos (40t+B)$, where $A>0$ and $-180^{\circ }<B\le 180^{\circ }$.