Question

(a) Explain the significance of the root-mean-square value of an alternating current or voltage waveform: Define the form factor of such a wave form. (6)

(b) Draw the circuit of Half-wave rectifier and its output waveform. A diode whose internal resistance is 20Ω is to supply power to 1000 Ω load from 110 V (RMS) source. Calculate

(i) peak load current,

(ii) DC load current,

(iii) AC load current. (10)

Answer 1

Part (a)

The root-mean-square (RMS) value of an alternating current (AC) or voltage waveform represents the equivalent DC value that would produce the same heating effect in a resistive load. In simpler terms, it's the effective value of the varying AC signal. For a sinusoidal waveform, the RMS value is 0.707 times the maximum (peak) value (Irms = 0.707 * Imax or Irms = Imax / √2). Ammeters and voltmeters typically measure the RMS value of current and voltage, respectively. Unless otherwise stated, values of AC current and voltage are assumed to be RMS values in electrical engineering.

The form factor of an AC waveform is the ratio of its RMS value to its average value. For a perfect sine wave, the form factor is approximately 1.11 (RMS value/Average Value = 1.11). This factor indicates how closely a waveform resembles a pure sine wave; a form factor closer to 1.11 suggests a waveform that is more sinusoidal.

Part (b)

Given:

Supply voltage, $$V_{rms} = 110\ V$$

Diode internal resistance, $$R_D = 20\ \Omega$$

Load resistance, $$R_L = 1000\ \Omega$$

The total resistance in the conducting circuit is:

$$R_T = R_D + R_L$$

$$R_T = 20 + 1000 = 1020\ \Omega$$

(i) Peak Load Current

The peak value of the supply voltage is:

$$V_{peak} = \sqrt{2}\,V_{rms}$$

$$V_{peak} = 1.414 \times 110 = 155.56\ V$$

Therefore, the peak load current is:

$$I_{peak} = \frac{V_{peak}}{R_T}$$

$$I_{peak} = \frac{155.56}{1020} = 0.1525\ A$$

Peak load current:

$$I_{peak}\approx0.153\ A=153\ mA$$

(ii) DC Load Current

For a half-wave rectifier, the average or DC value of current is:

$$I_{DC} = \frac{I_{peak}}{\pi}$$

$$I_{DC} = \frac{0.1525}{3.142} = 0.0485\ A$$

DC load current:

$$I_{DC}\approx0.0485\ A=48.5\ mA$$

(iii) AC Load Current

For a half-wave rectified current, the RMS load current is:

$$I_{RMS} = \frac{I_{peak}}{2}$$

$$I_{RMS} = \frac{0.1525}{2} = 0.07625\ A$$

The AC component of the load current is:

$$I_{AC} = \sqrt{I_{RMS}^{2}-I_{DC}^{2}}$$

$$I_{AC} = \sqrt{(0.07625)^2-(0.0485)^2}$$

$$I_{AC} \approx 0.0588\ A$$

✓ Verified Model Answer (Text Solution)