Q6 (16 Marks) Electrical Circuits & Calculations 🔥 Repeated 7x in exams
MET • Written Exam

(a) Explain what is meant by the terms wave form, frequency and average value. (6)

(b) A moving coil ammeter, a thermal ammeter and a rectifier are connected in series with a resistor across a 110 V sinusoidal a.c. supply. The circuit has a resistance of 50 Ω to current in one direction and, due to the rectifier, an infinite resistance to current in the reverse direction. Calculate: (10)

(i) the readings on the ammeters.

(ii) the form and peak factors of the current wave.

Appeared In: Mar 2026Dec 2024Aug 2024Jun 2024Oct 2019Jul 2019Apr 2019

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
Part (a)

Explain the terms waveform, frequency and average value

1. Waveform

A waveform is the shape or pattern obtained when an alternating voltage or current is plotted against time.

For a sinusoidal alternating current, the waveform is a sine wave, in which the magnitude and direction of current vary continuously with time.

2. Frequency

Frequency is the number of complete cycles of an alternating quantity occurring in one second.

The unit of frequency is hertz (Hz).

$$f=\frac{1}{T}$$

where:

  • f = frequency in Hz
  • T = time period of one complete cycle in seconds

3. Average Value

The average value of an alternating quantity is the arithmetic mean of its instantaneous values over a specified period.

For a symmetrical sinusoidal AC waveform, the average value over a complete cycle is zero, because the positive and negative half-cycles cancel each other.

For a rectified waveform, the average value is obtained by considering the rectified current over the complete cycle.

Part (b)

Ammeter Readings, Form Factor and Peak Factor

Given:

  • AC supply voltage = 110 V RMS
  • Resistance = 50 Ω
  • Resistance to current in the reverse direction = infinite
  • Therefore, current flows through the circuit in only one direction.

Hence, the current is a half-wave rectified sine wave.

Step 1: Calculate the Peak Voltage

The given 110 V is the RMS value of the sinusoidal AC supply.

For a sinusoidal waveform:

$$V_m=\sqrt{2}\times V_{rms}$$

Therefore:

$$V_m=\sqrt{2}\times110$$

$$V_m=155.56\ V$$

Step 2: Calculate the Peak Current

Using Ohm's law:

$$I_m=\frac{V_m}{R}$$

$$I_m=\frac{155.56}{50}$$

$$I_m=3.11\ A$$

Therefore:

$$\boxed{I_m=3.11\ A}$$

This current flows only during one half-cycle because the rectifier blocks current in the opposite direction.

The current waveform is therefore a half-wave rectified sine wave.

(i) Ammeter Readings

Moving Coil Ammeter

A moving coil ammeter responds to the average value of current.

For a half-wave rectified sine wave:

$$I_{avg}=\frac{I_m}{\pi}$$

$$Substituting\:I_{m}=3.11\ A$$

$$I_{avg}=\frac{3.11}{\pi}$$

$$I_{avg}=0.99\ A$$

Therefore, the moving coil ammeter reads:

$$\boxed{I_{MC}=0.99\ A}$$

Thermal Ammeter

A thermal ammeter operates on the heating effect of current and therefore indicates the RMS value of current.

For a half-wave rectified sine wave:

$$I_{rms}=\frac{I_m}{2}$$

Therefore:

$$I_{rms}=\frac{3.11}{2}$$

$$I_{rms}=1.555\ A$$

Hence, the thermal ammeter reads:

$$\boxed{I_{thermal}=1.56\ A}$$

(ii) Form Factor and Peak Factor

Form Factor

The form factor is defined as:

$$Form\ Factor=\frac{RMS\ value}{Average\ value}$$

For a half-wave rectified sine wave:

$$Form\ Factor=\frac{I_m/2}{I_m/\pi}$$

Therefore:

$$Form\ Factor=\frac{\pi}{2}$$

$$\boxed{Form\ Factor=1.57}$$

Peak Factor

The peak factor is defined as:

$$Peak\ Factor=\frac{Maximum\ value}{RMS\ value}$$

For the half-wave rectified sine wave:

$$Peak\ Factor=\frac{I_m}{I_m/2}$$

Therefore:

$$\boxed{Peak\ Factor=2.0}$$

Final Answers

  • Supply voltage: 110 V RMS
  • Peak voltage: 155.56 V
  • Peak current: 3.11 A
  • Moving coil ammeter reading: 0.99 A
  • Thermal ammeter reading: 1.56 A
  • Current waveform: Half-wave rectified sine wave
  • Form factor: 1.57
  • Peak factor: 2.0

Therefore:

$$\boxed{I_{MC}=0.99\ A}$$

$$\boxed{I_{thermal}=1.56\ A}$$

$$\boxed{Form\ Factor=1.57}$$

$$\boxed{Peak\ Factor=2.0}$$

Note: The values 1.1 A, 1.11 and 1.414 are not applicable to the stated half-wave rectified circuit. For the given circuit, the correct values are 1.56 A, 1.57 and 2.0, respectively.

← Back to MET Question Bank Upload Recent Question Paper →