Alternating Current - Power in AC Circuits

A series LCR circuit with $R = 80 \, \Omega$, $X_L = 100 \, \Omega$, and $X_C = 40 \, \Omega$ is connected to a $200 \, V, 50 \, Hz$ AC supply. Calculate the power factor and the average power dissipated in the circuit.

1. First, calculate the impedance $Z$: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{80^2 + (100 - 40)^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = 100 \, \Omega$.

2. Calculate the Power Factor: $\cos \phi = \frac{R}{Z} = \frac{80}{100} = 0.8$.

3. Calculate $I_{rms}$: $I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{100} = 2 \, A$.

4. Calculate Average Power: $P_{avg} = V_{rms} I_{rms} \cos \phi = 200 \times 2 \times 0.8 = 320 \, W$.

The power factor is the ratio of resistance to impedance. The average power is the product of the effective voltage, effective current, and the power factor. Alternatively, $P_{avg} = I_{rms}^2 R = 2^2 \times 80 = 320 \, W$ gives the same result.

Show that the average power consumed by a pure inductor over one complete cycle of AC is zero.

For a pure inductor, the current $i$ lags the voltage $v$ by a phase angle of $\frac{\pi}{2}$. If $v = V_m \sin(\omega t)$, then $i = I_m \sin(\omega t - \frac{\pi}{2}) = -I_m \cos(\omega t)$. $P_{avg} = \frac{1}{T} \int_0^T v \cdot i \, dt = \frac{1}{T} \int_0^T (V_m \sin \omega t) (-I_m \cos \omega t) \, dt$ $P_{avg} = -\frac{V_m I_m}{2T} \int_0^T \sin(2\omega t) \, dt$. Since the integral of a sine function over a full period is zero, $P_{avg} = 0$.

Mathematically, the phase difference $\phi$ is $90^\circ$. Using the formula $P_{avg} = V_{rms} I_{rms} \cos 90^\circ$, since $\cos 90^\circ = 0$, the power dissipation is zero.

An AC voltage $v = 282 \sin(100\pi t)$ is applied to a resistor of $100 \, \Omega$. Calculate (i) the RMS voltage, (ii) the RMS current, and (iii) the average power dissipated over a complete cycle.

From the equation $v = V_0 \sin(\omega t)$, we have: $V_0 = 282 \, V$

(i) RMS voltage: $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{282}{1.414} \approx 200 \, V$

(ii) RMS current: $I_{rms} = \frac{V_{rms}}{R} = \frac{200}{100} = 2 \, A$

(iii) Average power dissipated: $P_{avg} = V_{rms} I_{rms} = 200 \times 2 = 400 \, W$ Alternatively: $P_{avg} = I_{rms}^2 R = 2^2 \times 100 = 4 \times 100 = 400 \, W$

In a purely resistive circuit, the phase difference $\phi$ is zero, making the power factor $\cos \phi = 1$. Thus, the average power is simply the product of the RMS voltage and RMS current.

A capacitor of capacitance $C$ and a resistor of $R = 30 \, \Omega$ are connected in series to an AC source of $200 \, V, 50 \, Hz$. If the power factor of the circuit is $0.6$, calculate the capacitance of the capacitor.

Given: $V_{rms} = 200 \, V$ $f = 50 \, Hz$ $R = 30 \, \Omega$ $\cos \phi = 0.6$

Step 1: Find Impedance $Z$: $\cos \phi = \frac{R}{Z} \implies 0.6 = \frac{30}{Z}$ $Z = \frac{30}{0.6} = 50 \, \Omega$

Step 2: Find Capacitive Reactance $X_C$: $Z = \sqrt{R^2 + X_C^2} \implies 50^2 = 30^2 + X_C^2$ $2500 = 900 + X_C^2$ $X_C^2 = 1600 \implies X_C = 40 \, \Omega$

Step 3: Find Capacitance $C$: $X_C = \frac{1}{2\pi f C}$ $40 = \frac{1}{2 \times 3.14 \times 50 \times C}$ $C = \frac{1}{40 \times 314} \approx 7.96 \times 10^{-5} \, F = 79.6 \, \mu F$

The power factor determines the ratio of resistance to the total impedance. By finding the impedance first, we can isolate the capacitive reactance and subsequently calculate the capacitance using the source frequency.