10 ms

a. $\left(20\text{V}\right)\text{sin}200\pi t\text{,}\left(0.20\text{A}\right)\text{sin}200\pi t$; b. $\left(20\text{V}\right)\text{sin}200\pi t\text{,}\left(0.13\text{A}\right)\text{sin}\left(200\pi t+\pi \text{/}2\right)$; c. $\left(20\text{V}\right)\text{sin}200\pi t\text{,}\left(2.1\text{A}\right)\text{sin}\left(200\pi t-\pi \text{/}2\right)$

$\begin{array}{c}{v}_R=\left(V_{0}R\text{/}Z\right)\text{sin}\left(\omega t-\varphi \right);v_C=\left(V_0{X}_C\text{/}Z\right)\text{sin}\left(\omega t-\varphi +\pi \text{/}2\right)=\text{−}\left(V_0{X}_C\text{/}Z\right)\text{cos}\left(\omega t-\varphi \right);\hfill \\ v_L=\left(V_0{X}_L\text{/}Z\right)\text{sin}\left(\omega t-\varphi –\pi \text{/}2\right)=\left(V_0{X}_L\text{/}Z\right)\text{cos}\left(\omega t-\varphi \right)\hfill \end{array}$

$v(t)=(10.0\text{V})\text{sin}90\pi t$

2.00 V; 10.01 V; 8.01 V

a. 160 Hz; b. $40\text{Ω}$; c. $\left(0.25\text{A}\right)\text{sin}{10}^{3}t$; d. 0.023 rad

a. halved; b. halved; c. same

$v(t)=(0.14\text{V})\text{sin}(4.0\times {10}^{2}t)$

a. 12:1; b. 0.042 A; c. $2.6\times {10}^3\text{Ω}$

Angular frequency is $2\pi$ times frequency.

yes for both

The instantaneous power is the power at a given instant. The average power is the power averaged over a cycle or number of cycles.

The instantaneous power can be negative, but the power output can’t be negative.

There is less thermal loss if the transmission lines operate at low currents and high voltages.

The adapter has a step-down transformer to have a lower voltage and possibly higher current at which the device can operate.

so each loop can experience the same changing magnetic flux

a. $530\text{Ω}$; b. $53\text{Ω}$; c. $5.3\text{Ω}$

a. $1.9\text{Ω}$; b. $19\text{Ω}$; c. $190\text{Ω}$

360 Hz

$i\left(t\right)=\left(3.2\text{A}\right)\text{sin}\left(120\pi t\right)$

a. $38\text{Ω}$; b. $i\left(t\right)=\left(4.24\text{A}\right)\text{sin}\left(120\pi t-\pi \text{/}2\right)$

a. $770\text{Ω}$; b. 0.16 A; c. $I=\left(0.16\text{A}\right)\text{cos}\left(120\pi t–0.33\pi \right)$; d. $v_R=62\text{cos}\left(120\pi t–0.33\pi \right)$; $v_C=103\text{cos}\left(120\pi t-0.83\pi \right)$

a. $690\text{Ω}$; b. 0.15 A; c. $I=\left(0.15\text{A}\right)\text{sin}\left(1000\pi t-0.753\right)$; d. $1100\text{Ω}$, 0.092 A, $I=\left(0.092\text{A}\right)\text{sin}\left(1000\pi t+1.09\right)$

a. $5.7\text{Ω}$; b. $29\text{°}$; c. $I\text{}=\left(30.\text{A}\right)\text{cos}\left(120\pi t+0.51\right)$

a. 0.89 A; b. 5.6A; c. 1.4 A

a. 5.3 W; b. 2.1 W

a. inductor; b. $X_L=52\text{Ω}$

$1.3\times {10}^{\mathrm{-6}}\text{F}$

a. 820 Hz; b. 7.8

a. 50 Hz; b. 50 W; c. 6.32; d. 50 rad/s

The reactance of the capacitor is larger than the reactance of the inductor because the current leads the voltage. The power usage is 30 W.

a. 45:1; b. 0.68 A, 0.015 A; c. $160\text{Ω}$

a. 41 turns; b. 40.9 mA

a. $i\left(t\right)=\left(1.26\text{A}\right)\text{sin}\left(200\pi t+\pi \text{/}2\right)$; b. $i\left(t\right)=\left(7.96\text{A}\right)\text{sin}\left(200\pi t-\pi \text{/}2\right)$; c. $i\left(t\right)=\left(2\text{A}\right)\text{sin}\left(200\pi t\right)$

a. $2.5\times {10}^3\text{Ω},3.6\times {10}^{\mathrm{-3}}\text{A}$; b. $7.5\text{Ω},1.2\text{A}$

a. 19 A; b. inductor leads by $90\text{°}$

$11.7\text{Ω}$

14 W

a. $5.9\times {10}^4\text{W}$; b. $1.64\times {10}^{11}\text{W}$

a. 335 MV; b. the result is way too high, well beyond the breakdown voltage of air over reasonable distances; c. the input voltage is too high

a. $20\text{Ω}$; b. 0.5 A; c. $5.4\text{°}$, lagging;
d. $\begin{array}{c}{V}_R=\left(9.96\text{V}\right)\text{cos}\left(250\pi t+5.4\text{°}\right)\text{,}{V}_C=\left(12.7\text{V}\right)\text{cos}\left(250\pi t+5.4\text{°}-90\text{°}\right)\text{,}\hfill \\ V_L=\left(11.8\text{V}\right)\text{cos}\left(250\pi t+5.4\text{°}+90\text{°}\right)\text{,}{V}_{\text{source}}=\left(10.0\text{}\text{V}\right)\text{cos}\left(250\pi t\right);\hfill \end{array}$ e. 0.995; f. 6.25 J

a. $0.75\text{Ω}$; b. $7.5\text{Ω}$; c. $0.75\text{Ω}$; d. $7.5\text{Ω}$; e. $1.3\text{Ω}$; f. $0.13\text{Ω}$

The units as written for inductive reactance Equation 15.8 are $\frac{\text{rad}}{\text{s}}\text{H}$. Radians can be ignored in unit analysis. The Henry can be defined as $\text{H}=\frac{\text{V}·\text{s}}{\text{A}}=\text{Ω}·\text{s}$. Combining these together results in a unit of $\text{Ω}$ for reactance.

a. 156 V; b. 42 V; c. 154 V

a. $\frac{v_{\text{out}}}{v_{\text{in}}}=\frac{1}{\sqrt{1+1\text{/}{\omega }^2{R}^2{C}^2}}$ and $\frac{v_{\text{out}}}{v_{\text{in}}}=\frac{\omega L}{\sqrt{R^2+{\omega }^2{L}^2}}$; b. $v_{\text{out}}\approx v_{\text{in}}$ and $v_{\text{out}}\approx 0$