Chapter 7: Alternating Current (EMI and AC)

(a) $I_{\text{rms}}=\frac{V_{\text{rms}}}{R}=\frac{220}{100}=2.2$ A.

(b) For a pure resistor the current and voltage are in phase ($\cos \varphi =1$):

$P=V_{\text{rms}}{I}_{\text{rms}}=220\times 2.2=484$ W.

(a) 2.2 A. (b) 484 W.

(a) $V_{\text{rms}}=\frac{V_0}{\sqrt{2}}=\frac{300}{1.414}\approx 212$ V.

(b) $I_0=\sqrt{2}{I}_{\text{rms}}=1.414\times 10\approx 14.1$ A.

(a) About 212 V. (b) About 14.1 A.

$$X_L=2\pi \nu L=2\times 3.14\times 50\times 44\times 10^{-3}\approx 13.8\Omega$$

$$I_{\text{rms}}=\frac{V_{\text{rms}}}{X_L}=\frac{220}{13.8}\approx 15.9\text{A}$$

About 15.9 A.

$$X_C=\frac{1}{2\pi \nu C}=\frac{1}{2\times 3.14\times 60\times 60\times 10^{-6}}\approx 44.2\Omega$$

$$I_{\text{rms}}=\frac{V_{\text{rms}}}{X_C}=\frac{110}{44.2}\approx 2.49\text{A}$$

About 2.49 A.

Zero in both. In a pure inductor the current lags the voltage by 90°; in a pure capacitor it leads by 90°. So $\cos \varphi =\cos 90^{\circ }=0$ and

$$P=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi =0$$

In one quarter-cycle the inductor (or capacitor) stores energy from the source, and in the next quarter-cycle it gives the same energy back.

Zero for both; the phase difference is 90°, so $\cos \varphi =0$ (energy stored is returned every cycle).

$${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{27\times 10^{-3}\times 30\times 10^{-6}}}=\frac{1}{\sqrt{8.1\times 10^{-7}}}=\frac{1}{9\times 10^{-4}}\approx 1.1\times 10^3\text{rad/s}$$

${\omega }_0\approx 1.1\times 10^3$ rad s⁻¹.

At resonance $X_L=X_C$, so $Z=R=20$ Ω and $\cos \varphi =1$.

$I_{\text{rms}}=\frac{200}{20}=10$ A, and $P=I_{\text{rms}}^{2}R=10^2\times 20=2000$ W.

2000 W (2 kW).

(a)

$${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5.0\times 80\times 10^{-6}}}=\frac{1}{0.02}=50\text{rad/s},{\nu }_0=\frac{{\omega }_0}{2\pi }\approx 7.96\text{Hz}$$

(b) At resonance $Z=R=40$ Ω.

$I_{\text{rms}}=\frac{230}{40}=5.75$ A; amplitude $I_0=\sqrt{2}\times 5.75\approx 8.13$ A.

(c) $X_L={\omega }_{0}L=50\times 5=250$ Ω and $X_C=\frac{1}{{\omega }_{0}C}=\frac{1}{50\times 80\times 10^{-6}}=250$ Ω.

• $V_R=I_{\text{rms}}R=5.75\times 40=230$ V
• $V_L=I_{\text{rms}}{X}_L=5.75\times 250\approx 1437.5$ V
• $V_C=I_{\text{rms}}{X}_C\approx 1437.5$ V

The voltages across L and C are equal in size but 180° out of phase, so across the LC combination $V_{LC}=V_L-V_C=0$. The whole 230 V appears across R.

(a) About 7.96 Hz (50 rad/s). (b) Z = 40 Ω, current amplitude about 8.13 A. (c) $V_R$ = 230 V, $V_L$ = $V_C$ ≈ 1437.5 V, and $V_{LC}$ = 0.