Chapter 4 - AC Network Analysis - Part 1 Circuits - Homework Problems - Page 167: 4.3

a) $V=31.25\cos(25t)\ V$ b) $V=-125\cos(50t+\pi/2)=125\cos(50t-\pi/2)\ V$ c) $V=625\cos(50t+5\pi/6)\ V$ d) $V=-50\cos(50t+5\pi/6)=50\cos(50t-\pi/6)\ V$

given an inductor $L=250\ mH = 250\times 10^{-3}\ H$, we are required to find the voltages. a) for current $i_c(t)=5\sin(25t)\ A$ the frequency is $\omega = 25\ Hz$ the impedance of the inductor $Z=j\omega L=j\times 25\times 250\times 10^{-3}=j6.25\ \Omega $ the voltage is $V=I\times Z=(5\sin(25t))\times (j6.25)$ convert $sin$ to $cos$ $V=(5\cos(\pi/2-25t))\times (j6.25)$ using identity $cos(x)=cos(-x)$ $V=(5\cos(25t-\pi/2))\times (j6.25)$ convert to polar form $V=(5\angle- 90^{\circ})\times (6.25\angle90^{\circ})$ (note: remember that $ j = 1 \angle90^{\circ}$) $V=5\times 6.25 \ \angle(90-90)^{\circ}$ $V=31.25\ \angle0^{\circ}\ V$ convert back to Cartesian form and answer is $$V=31.25\cos(25t)\ V$$ b) for current $i_c(t)=-10\cos(50t)\ A$ the frequency is $\omega = 50 \ Hz$ the impedance of the inductor $Z=j\omega L=j\times 50\times 250\times 10^{-3}=j12.5\ \Omega $ the voltage is $$V=I\times Z=(-10\cos(50t))\times (j12.5)=-125\cos(50t+\pi/2)\ V$$ (note: $ j $ shifts a cosine by $90^{\circ}$) c) for current $i_c(t)=25\cos(100t+\pi/3)\ A$ the frequency is $\omega = 100 \ Hz$ the impedance of the inductor $Z=j\omega L=j\times 100\times 250\times 10^{-3}=j25\ \Omega $ the voltage is $V=I\times Z=(25\cos(100t+\pi/3))\times (j25)$ $$=625\cos(50t+5\pi/6)\ V$$ d) for current $i_c(t)=20\sin(10t-\pi/12)\ A$ the frequency is $\omega = 10 \ Hz$ the impedance of the inductor $Z=j\omega L=j\times 10\times 250\times 10^{-3}=j2.5\ \Omega $ the voltage is $V=I\times Z=(20\cos(10t-\pi/12)))\times (j2.5)$ $=-50j\sin(50t+5\pi/6)\ V$ convert to $cos$ $=-50j\cos(\pi/2-(50t+5\pi/6))$ $=-50j\cos(-\pi/3-50t)$ use identity $cos(x)=cos(-x)$ $V=-50j\cos(50t+pi/3)$ $=-50\cos(50t+pi/3+pi/2)$ (note: $j$ shifts $cos$ by $+90^{\circ}$) $$V=-50\cos(50t+5\pi/6)\ V$$ (note: even a minus sign shits a cosine by -180 so you can write the above answer as $50\cos(50t+5\pi/3-\pi)=50\cos(50t-\pi/6)\ V$ )