Alternating Current

Given:
- Resistance, $R = 100\,\Omega$
- rms voltage, $V = 220\,\text{V}$
- Frequency, $f = 50\,\text{Hz}$

(a) rms value of current:

Using Ohm's law for ac circuits:
$$I = \frac{V}{R} = \frac{220}{100} = 2.2\,\text{A}$$

(b) Net power consumed over a full cycle:

For a purely resistive circuit, power factor $\cos\phi = 1$.

$$P = VI\cos\phi = VI = 220 \times 2.2 = 484\,\text{W}$$

Alternatively, $P = I^2 R = (2.2)^2 \times 100 = 484\,\text{W}$

Answer: (a) $I = 2.2\,\text{A}$, (b) $P = 484\,\text{W}$

(a) rms voltage from peak voltage:

Given: Peak voltage, $v_m = 300\,\text{V}$

The relation between rms voltage and peak voltage is:
$$V = \frac{v_m}{\sqrt{2}} = \frac{300}{\sqrt{2}} = \frac{300}{1.414} \approx 212.1\,\text{V}$$

(b) Peak current from rms current:

Given: rms current, $I = 10\,\text{A}$

The relation between peak current and rms current is:
$$i_m = \sqrt{2}\, I = \sqrt{2} \times 10 = 14.14\,\text{A}$$

Answer: (a) $V \approx 212.1\,\text{V}$, (b) $i_m \approx 14.14\,\text{A}$

Given:
- Inductance, $L = 44\,\text{mH} = 44 \times 10^{-3}\,\text{H}$
- rms voltage, $V = 220\,\text{V}$
- Frequency, $f = 50\,\text{Hz}$

Step 1: Calculate inductive reactance $X_L$:
$$X_L = \omega L = 2\pi f L = 2\pi \times 50 \times 44 \times 10^{-3}$$
$$X_L = 2 \times 3.14 \times 50 \times 0.044 = 13.82\,\Omega$$

Step 2: Calculate rms current:
$$I = \frac{V}{X_L} = \frac{220}{13.82} \approx 15.92\,\text{A}$$

Answer: The rms value of current in the circuit is $I \approx 15.92\,\text{A}$.

Given:
- Capacitance, $C = 60\,\mu\text{F} = 60 \times 10^{-6}\,\text{F}$
- rms voltage, $V = 110\,\text{V}$
- Frequency, $f = 60\,\text{Hz}$

Step 1: Calculate capacitive reactance $X_C$:
$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 60 \times 60 \times 10^{-6}}$$
$$X_C = \frac{1}{2 \times 3.14 \times 60 \times 60 \times 10^{-6}} = \frac{1}{0.02262} \approx 44.2\,\Omega$$

Step 2: Calculate rms current:
$$I = \frac{V}{X_C} = \frac{110}{44.2} \approx 2.49\,\text{A}$$

Answer: The rms value of current in the circuit is $I \approx 2.49\,\text{A}$.

For the inductor circuit (Exercise 7.3):

In a purely inductive circuit, the current lags the voltage by $\phi = \dfrac{\pi}{2}$.

$$P = VI\cos\phi = VI\cos\left(\frac{\pi}{2}\right) = VI \times 0 = 0\,\text{W}$$

For the capacitor circuit (Exercise 7.4):

In a purely capacitive circuit, the current leads the voltage by $\phi = \dfrac{\pi}{2}$.

$$P = VI\cos\phi = VI\cos\left(\frac{\pi}{2}\right) = VI \times 0 = 0\,\text{W}$$

Explanation:

In a pure inductor or pure capacitor, the phase difference between voltage and current is $90°$. The power factor $\cos\phi = \cos 90° = 0$. During one half of the cycle, energy is stored in the magnetic field (inductor) or electric field (capacitor), and during the other half, the same energy is returned to the source. Hence, the net power absorbed over a complete cycle is zero in both cases. The current flowing is called wattless current.

Answer: Net power absorbed = 0 W in both circuits.

Given:
- Capacitance, $C = 30\,\mu\text{F} = 30 \times 10^{-6}\,\text{F}$
- Inductance, $L = 27\,\text{mH} = 27 \times 10^{-3}\,\text{H}$

Formula for angular frequency of free (LC) oscillations:
$$\omega_0 = \frac{1}{\sqrt{LC}}$$

Calculation:
$$\omega_0 = \frac{1}{\sqrt{27 \times 10^{-3} \times 30 \times 10^{-6}}}$$
$$= \frac{1}{\sqrt{810 \times 10^{-9}}}$$
$$= \frac{1}{\sqrt{8.1 \times 10^{-7}}}$$
$$= \frac{1}{9 \times 10^{-4} \times \sqrt{1}}$$

Let us compute: $\sqrt{8.1 \times 10^{-7}} = \sqrt{8.1} \times 10^{-3.5} = 2.846 \times 10^{-3.5}$

More carefully:
$$LC = 27 \times 10^{-3} \times 30 \times 10^{-6} = 810 \times 10^{-9} = 8.1 \times 10^{-7}$$
$$\sqrt{LC} = \sqrt{8.1 \times 10^{-7}} = 9 \times 10^{-4} \times \sqrt{1} \approx 9 \times 10^{-4}\,\text{(since }\sqrt{8.1}\approx 2.846\text{)}$$

Actually: $\sqrt{8.1 \times 10^{-7}} = 2.846 \times 10^{-3.5}$. Let me redo:
$$\sqrt{8.1 \times 10^{-7}} = \sqrt{8.1} \times 10^{-3.5} = 2.846 \times 3.162 \times 10^{-4} = 8.997 \times 10^{-4} \approx 9 \times 10^{-4}$$

$$\omega_0 = \frac{1}{9 \times 10^{-4}} \approx 1.11 \times 10^{3}\,\text{rad/s}$$

Answer: The angular frequency of free oscillations is $\omega_0 \approx 1.11 \times 10^3\,\text{rad/s}$.

Given:
- $R = 20\,\Omega$
- $L = 1.5\,\text{H}$
- $C = 35\,\mu\text{F} = 35 \times 10^{-6}\,\text{F}$
- rms voltage, $V = 200\,\text{V}$
- Condition: supply frequency = natural frequency (resonance)

At resonance:

At resonance, $X_L = X_C$, so they cancel each other. The impedance of the circuit is:
$$Z = R = 20\,\Omega$$

The power factor at resonance:
$$\cos\phi = 1$$

rms current at resonance:
$$I = \frac{V}{Z} = \frac{V}{R} = \frac{200}{20} = 10\,\text{A}$$

Average power transferred:
$$P = VI\cos\phi = 200 \times 10 \times 1 = 2000\,\text{W}$$

Alternatively: $P = I^2 R = (10)^2 \times 20 = 2000\,\text{W}$

Answer: The average power transferred to the circuit at resonance is $P = 2000\,\text{W} = 2\,\text{kW}$.

Given:
- $L = 5.0\,\text{H}$
- $C = 80\,\mu\text{F} = 80 \times 10^{-6}\,\text{F}$
- $R = 40\,\Omega$
- rms voltage, $V = 230\,\text{V}$

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(a) Resonant frequency:

At resonance: $\omega_0 = \dfrac{1}{\sqrt{LC}}$

$$\omega_0 = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{\sqrt{4 \times 10^{-4}}} = \frac{1}{2 \times 10^{-2}} = 50\,\text{rad/s}$$

$$\boxed{\omega_0 = 50\,\text{rad/s}}$$

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(b) Impedance and amplitude of current at resonance:

At resonance, $X_L = X_C$, so:
$$Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{R^2 + 0} = R = 40\,\Omega$$

rms current:
$$I = \frac{V}{Z} = \frac{230}{40} = 5.75\,\text{A}$$

Amplitude (peak) of current:
$$i_m = \sqrt{2}\,I = \sqrt{2} \times 5.75 = 8.13\,\text{A}$$

$$\boxed{Z = 40\,\Omega, \quad i_m \approx 8.13\,\text{A}}$$

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(c) rms potential drops across each element:

Across resistor R:
$$V_R = IR = 5.75 \times 40 = 230\,\text{V}$$

Inductive reactance at resonance:
$$X_L = \omega_0 L = 50 \times 5.0 = 250\,\Omega$$

Across inductor L:
$$V_L = IX_L = 5.75 \times 250 = 1437.5\,\text{V}$$

Capacitive reactance at resonance:
$$X_C = \frac{1}{\omega_0 C} = \frac{1}{50 \times 80 \times 10^{-6}} = \frac{1}{4 \times 10^{-3}} = 250\,\Omega$$

Across capacitor C:
$$V_C = IX_C = 5.75 \times 250 = 1437.5\,\text{V}$$

Potential drop across LC combination:

Since $V_L$ and $V_C$ are equal in magnitude but opposite in phase (they differ by $180°$):
$$V_{LC} = V_L - V_C = 1437.5 - 1437.5 = 0\,\text{V}$$

This confirms that at resonance ($X_L = X_C$), the net voltage drop across the LC combination is zero, and the entire source voltage appears across the resistor ($V_R = 230\,\text{V} = V$).

Answer:
- $V_R = 230\,\text{V}$
- $V_L = V_C \approx 1437.5\,\text{V}$
- $V_{LC} = 0\,\text{V}$ at resonance.