For a pure resistor, $I_{\text{rms}}=V_{\text{rms}}/R=220/100=2.2\,\text{A}$. The average power over a full cycle is $P=V_{\text{rms}}I_{\text{rms}}=220(2.2)=484\,\text{W}$, equivalently $V^2/R$.
(a) $2.2\,\text{A}$. (b) $484\,\text{W}$.
For sinusoidal ac, $V_{\text{rms}}=V_0/\sqrt2$ and $I_0=\sqrt2 I_{\text{rms}}$. Thus $V_{\text{rms}}=300/\sqrt2=212\,\text{V}$ and $I_0=\sqrt2(10)=14.1\,\text{A}$.
(a) $212\,\text{V}$. (b) $14.1\,\text{A}$.
The inductive reactance is $X_L=\omega L=2\pi fL=2\pi(50)(44\times10^{-3})=13.8\,\Omega$. Hence $I_{\text{rms}}=V_{\text{rms}}/X_L=220/13.8=15.9\,\text{A}$.
$15.9\,\text{A}$.
The capacitive reactance is $X_C=1/(\omega C)=1/[2\pi(60)(60\times10^{-6})]=44.2\,\Omega$. Therefore $I_{\text{rms}}=V_{\text{rms}}/X_C=110/44.2=2.49\,\text{A}$.
$2.5\,\text{A}$.
An ideal inductor and an ideal capacitor do not dissipate energy over a full ac cycle. In each half-cycle, energy is alternately stored in and returned from the magnetic field of the inductor or electric field of the capacitor. Therefore the average power absorbed over a complete cycle is zero.
Zero in both circuits.
The angular frequency of free LC oscillations is $\omega_0=1/\sqrt{LC}$. With $L=27\times10^{-3}\,\text{H}$ and $C=30\times10^{-6}\,\text{F}$, $\omega_0=1/\sqrt{(27\times10^{-3})(30\times10^{-6})}=1.11\times10^3\,\text{rad s}^{-1}$.
$1.1\times10^3\,\text{rad s}^{-1}$.
At resonance, $X_L=X_C$, so the impedance is just $R$. The rms current is $I=V/R=200/20=10\,\text{A}$. Average power is $P=I^2R=10^2(20)=2.0\times10^3\,\text{W}$.
$2.0\times10^3\,\text{W}$.
At resonance, $\omega_0=1/\sqrt{LC}=1/\sqrt{5.0(80\times10^{-6})}=50\,\text{rad s}^{-1}$, so $f_0=\omega_0/(2\pi)=7.96\,\text{Hz}$. At resonance, $Z=R=40\,\Omega$. With $V_{\text{rms}}=230\,\text{V}$, $I_{\text{rms}}=230/40=5.75\,\text{A}$ and the current amplitude is $I_0=\sqrt2 I_{\text{rms}}=8.13\,\text{A}$. The reactances are $X_L=\omega_0L=250\,\Omega$ and $X_C=1/(\omega_0C)=250\,\Omega$. Hence $V_R=I_{\text{rms}}R=230\,\text{V}$, $V_L=I_{\text{rms}}X_L=1437.5\,\text{V}$ and $V_C=1437.5\,\text{V}$. Since $V_L$ leads the current by $90^\circ$ and $V_C$ lags by $90^\circ$, their phasor sum is zero.
(a) $7.96\,\text{Hz}$. (b) $Z=40\,\Omega$ and current amplitude $8.13\,\text{A}$. (c) $V_R=230\,\text{V}$, $V_L=1.44\times10^3\,\text{V}$, $V_C=1.44\times10^3\,\text{V}$; the LC drops cancel because they are equal and opposite in phase.