By the end of this chapter, you will be able to:
These skills will help you work confidently and safely with electrical systems, making you a valuable professional in the electrical trade.
In telecommunication engineering, understanding electrical quantities is essential for designing, installing, and troubleshooting equipment such as transmitters, receivers, amplifiers, and network infrastructure. Electrical principles underpin the operation of devices ranging from fiber optic transceivers to wireless base stations. Kenyan professionals must be adept at calculating, measuring, and applying these quantities to maintain efficient and safe communication systems in diverse contexts such as mobile networks, radio stations, and internet service providers.
Electrical quantities are the measurable properties that describe the behavior of electrical circuits and components. The three fundamental quantities in telecommunication engineering are voltage (\(V\)), current (\(I\)), and resistance (\(R\)). These are governed by Ohm’s Law, which relates the three in a simple equation.
$$V = I \times R$$
Example 1:
A radio transmitter is powered by a 12 V supply and draws a current of 2 A. What is the resistance of the transmitter circuit?
Given: \(V = 12\) V, \(I = 2\) A$$R = \frac{V}{I}$$$$R = \frac{12}{2}$$$$R = 6$$ Answer: \(6\,\Omega\)
Example 2:
A fiber optic repeater requires a current of 0.5 A to operate and has a resistance of 24 \(\Omega\). What supply voltage is required?
Given: \(I = 0.5\) A, \(R = 24\) \(\Omega\)$$V = I \times R$$$$V = 0.5 \times 24$$$$V = 12$$ Answer: \(12\,\text{V}\)
Example 3:
A telecommunication technician needs to calculate the current flowing through a router with a resistance of 10 \(\Omega\) and a supply voltage of 5 V.
Given: \(V = 5\) V, \(R = 10\) \(\Omega\)$$I = \frac{V}{R}$$$$I = \frac{5}{10}$$$$I = 0.5$$ Answer: \(0.5\,\text{A}\)
Example 4:
A base station power amplifier has a resistance of 50 \(\Omega\) and is supplied with 48 V. What is the current?
Given: \(V = 48\) V, \(R = 50\) \(\Omega\)$$I = \frac{V}{R}$$$$I = \frac{48}{50}$$$$I = 0.96$$ Answer: \(0.96\,\text{A}\)
Example 5:
A signal booster consumes 3 A from a 24 V source. What is the circuit resistance?
Given: \(I = 3\) A, \(V = 24\) V$$R = \frac{V}{I}$$$$R = \frac{24}{3}$$$$R = 8$$ Answer: \(8\,\Omega\)
Electrical quantities are measured using the International System of Units (SI). Voltage is measured in volts (\(V\)), current in amperes (\(A\)), and resistance in ohms (\(\Omega\)). Power, another important quantity, is measured in watts (\(W\)), and is calculated as the product of voltage and current.
$$P = V \times I$$
Example 1:
A network switch operates at 230 V and draws 0.2 A. What is its power consumption?
Given: \(V = 230\) V, \(I = 0.2\) A$$P = V \times I$$$$P = 230 \times 0.2$$$$P = 46$$ Answer: \(46\,\text{W}\)
Example 2:
A Wi-Fi access point consumes 12 W and operates at 12 V. What is the current?
Given: \(P = 12\) W, \(V = 12\) V$$I = \frac{P}{V}$$$$I = \frac{12}{12}$$$$I = 1$$ Answer: \(1\,\text{A}\)
Example 3:
An outdoor antenna amplifier has a resistance of 20 \(\Omega\) and is supplied with 10 V. What is the power consumed?
Given: \(R = 20\) \(\Omega\), \(V = 10\) V
First, find current:$$I = \frac{V}{R}$$$$I = \frac{10}{20}$$$$I = 0.5$$
Now, find power:$$P = V \times I$$$$P = 10 \times 0.5$$$$P = 5$$
Answer: \(5\,\text{W}\)
Example 4:
A telecommunication relay uses 0.3 A at 24 V. What is its power rating?
Given: \(I = 0.3\) A, \(V = 24\) V$$P = V \times I$$$$P = 24 \times 0.3$$$$P = 7.2$$ Answer: \(7.2\,\text{W}\)
Example 5:
A modem is rated at 5 V and 2 A. What is its power consumption?
Given: \(V = 5\) V, \(I = 2\) A$$P = V \times I$$$$P = 5 \times 2$$$$P = 10$$ Answer: \(10\,\text{W}\)
Telecommunication engineers routinely calculate voltage drops, circuit resistance, and power requirements to ensure reliable operation of communication equipment. Accurate calculations help prevent overloads, signal loss, and equipment failure.
The voltage drop across a circuit is given by:
$$V_{drop} = I \times R$$
Example 1:
A 100 m cable supplying a CCTV camera carries 0.5 A and has a resistance of 2 \(\Omega\). What is the voltage drop?
Given: \(I = 0.5\) A, \(R = 2\) \(\Omega\)$$V_{drop} = I \times R$$$$V_{drop} = 0.5 \times 2$$$$V_{drop} = 1$$ Answer: \(1\,\text{V}\)
Example 2:
A repeater station is powered by a cable with total resistance 5 \(\Omega\) and draws 1.5 A. Find the voltage drop.
Given: \(R = 5\) \(\Omega\), \(I = 1.5\) A$$V_{drop} = I \times R$$$$V_{drop} = 1.5 \times 5$$$$V_{drop} = 7.5$$ Answer: \(7.5\,\text{V}\)
Example 3:
A radio mast light draws 0.2 A through a cable with 10 \(\Omega\) resistance. What is the voltage drop?
Given: \(I = 0.2\) A, \(R = 10\) \(\Omega\)$$V_{drop} = 0.2 \times 10$$$$V_{drop} = 2$$ Answer: \(2\,\text{V}\)
Example 4:
A router is powered via a cable with 3 \(\Omega\) resistance and draws 2 A. Find the voltage drop.
Given: \(R = 3\) \(\Omega\), \(I = 2\) A$$V_{drop} = I \times R$$$$V_{drop} = 2 \times 3$$$$V_{drop} = 6$$ Answer: \(6\,\text{V}\)
Example 5:
A backup battery system supplies 1 A through a cable with 0.5 \(\Omega\) resistance. What is the voltage drop?
Given: \(I = 1\) A, \(R = 0.5\) \(\Omega\)$$V_{drop} = I \times R$$$$V_{drop} = 1 \times 0.5$$$$V_{drop} = 0.5$$ Answer: \(0.5\,\text{V}\)
Power loss in a cable is given by:
$$P_{loss} = I^2 \times R$$
Example 6:
A signal cable carries 0.3 A with resistance 4 \(\Omega\). What is the power loss?
Given: \(I = 0.3\) A, \(R = 4\) \(\Omega\)$$P_{loss} = I^2 \times R$$$$P_{loss} = (0.3)^2 \times 4$$$$P_{loss} = 0.09 \times 4$$$$P_{loss} = 0.36$$ Answer: \(0.36\,\text{W}\)
Example 7:
A telecom tower cable carries 2 A and has 1 \(\Omega\) resistance. What is the power loss?
Given: \(I = 2\) A, \(R = 1\) \(\Omega\)$$P_{loss} = I^2 \times R$$$$P_{loss} = (2)^2 \times 1$$$$P_{loss} = 4 \times 1$$$$P_{loss} = 4$$ Answer: \(4\,\text{W}\)
Electrical quantities are measured using specialized instruments to ensure safe and reliable operation of telecommunication equipment. The most common tools are multimeters, clamp meters, and insulation testers.
| Name | Specification | Use |
|---|---|---|
| Multimeter | Measures V, I, R | Testing circuit voltage, current, resistance |
| Clamp Meter | Measures AC current up to 1000 A | Measuring current without breaking circuit |
| Insulation Tester | Measures insulation resistance (up to 1000 M\(\Omega\)) | Checking cable insulation integrity |
| Oscilloscope | Measures voltage waveform | Analyzing signal integrity |
| Power Meter | Measures power (W, kW) | Assessing equipment power consumption |
Multimeter:
Use to measure voltage, current, and resistance in equipment and cables. Always check battery status and calibrate regularly for accurate readings. Clean probes to ensure proper contact.
Clamp Meter:
Clip around the conductor without breaking the circuit to measure current. Store in a dry place and avoid dropping to maintain accuracy.
Insulation Tester:
Apply to cable ends to check insulation resistance. Keep away from moisture and store with protective caps.
Oscilloscope:
Connect probes to test points on circuit boards to visualize voltage waveforms. Maintain probe tips and avoid exposure to dust.
Power Meter:
Plug into socket or connect to circuit to measure power consumption. Periodically check calibration and avoid overloading.
Example 1:
A multimeter measures a voltage of 48 V across a telecom equipment terminal and a current of 0.5 A. What is the resistance?
Given: \(V = 48\) V, \(I = 0.5\) A$$R = \frac{V}{I}$$$$R = \frac{48}{0.5}$$$$R = 96$$ Answer: \(96\,\Omega\)
Example 2:
A clamp meter shows a current of 1.2 A in a base station power cable. The voltage measured is 220 V. What is the power consumption?
Given: \(I = 1.2\) A, \(V = 220\) V$$P = V \times I$$$$P = 220 \times 1.2$$$$P = 264$$ Answer: \(264\,\text{W}\)
Example 3:
An insulation tester reads 200 M\(\Omega\) on a fiber optic cable. If a voltage of 100 V is applied, what is the leakage current?
Given: \(R = 200 \times 10^6\) \(\Omega\), \(V = 100\) V$$I = \frac{V}{R}$$$$I = \frac{100}{200 \times 10^6}$$$$I = \frac{100}{200000000}$$$$I = 0.0000005$$ Answer: \(0.5\,\mu\text{A}\)
Example 4:
A power meter records 30 W consumption by a network router running at 15 V. What is the current?
Given: \(P = 30\) W, \(V = 15\) V$$I = \frac{P}{V}$$$$I = \frac{30}{15}$$$$I = 2$$ Answer: \(2\,\text{A}\)
Example 5:
An oscilloscope measures a peak-to-peak voltage of 10 V at a frequency of 1 MHz in a signal line. What is the RMS voltage?
Given: \(V_{pp} = 10\) V$$V_{RMS} = \frac{V_{pp}}{2 \sqrt{2}}$$$$V_{RMS} = \frac{10}{2 \times 1.414}$$$$V_{RMS} = \frac{10}{2.828}$$$$V_{RMS} = 3.54$$ Answer: \(3.54\,\text{V}\)
This chapter explores the fundamental electrical concepts essential for understanding and applying electrical quantities in engineering contexts. It begins with a review of core electrical quantities such as charge, current, voltage, power, and energy, along with their definitions and relationships. The importance of SI units for accurate measurement and calculation is emphasized, covering the standard units for each electrical quantity. Practical calculations are demonstrated to show how these quantities interact in real-world scenarios, and the use of measuring instruments like voltmeters, ammeters, ohmmeters, multimeters, power meters, and oscilloscopes is discussed. The chapter also examines battery parameters including voltage, capacity, and energy density, and describes the roles of the anode, cathode, and electrolyte in battery operation, as well as the processes of discharge and recharge. Various electrical devices and systems are evaluated for their advantages, disadvantages, and typical applications, providing a comprehensive overview of how electrical quantities are measured, calculated, and applied in technical practice.
1. Total charge transferred
Given:
\( I = 0.5 \) A
\( t = 10 \) minutes \( = 10 \times 60 = 600 \) s
Formula:$$ Q = I \times t $$
Substitution:$$ Q = 0.5 \times 600 $$
Calculation:$$ Q = 300 $$
Final Answer:
\( Q = 300 \) C (coulombs)
2. Electrical power delivered
Given:
\( V = 12 \) V
\( I = 2 \) A
Formula:$$ P = V \times I $$
Substitution:$$ P = 12 \times 2 $$
Calculation:$$ P = 24 $$
Final Answer:
\( P = 24 \) W (watts)
3. Total energy consumed
Given:
\( P = 120 \) W
\( t = 6 \) hours
Formula:$$ E = P \times t $$
Convert to kWh:$$ E = \frac{120 \times 6}{1000} $$
Calculation:$$ E = \frac{720}{1000} $$$$ E = 0.72 $$
Final Answer:
\( E = 0.72 \) kWh
4. Resistance of the device
Given:
\( V = 5 \) V
\( I = 0.5 \) A
Formula:$$ R = \frac{V}{I} $$
Substitution:$$ R = \frac{5}{0.5} $$
Calculation:$$ R = 10 $$
Final Answer:
\( R = 10 \) \(\Omega\) (ohms)
5. Energy stored in battery
Given:
Capacity \( = 4000 \) mAh \( = 4 \) Ah
\( V = 3.7 \) V
Formula:$$ E = V \times Q $$
\( Q \) is in coulombs:$$ Q = 4 \times 3600 = 14400 $$
Substitution:$$ E = 3.7 \times 14400 $$
Calculation:$$ E = 53280 $$
Final Answer:
\( E = 53,280 \) J (joules)
6. Power consumed by amplifier
Given:
\( I = 120 \) mA \( = 0.12 \) A
\( V = 24 \) V
Formula:$$ P = V \times I $$
Substitution:$$ P = 24 \times 0.12 $$
Calculation:$$ P = 2.88 $$
Final Answer:
\( P = 2.88 \) W (watts)
7. RMS voltage
Given:
Peak voltage \( V_{peak} = 20 \) V
Formula:$$ V_{rms} = \frac{V_{peak}}{\sqrt{2}} $$
Substitution:$$ V_{rms} = \frac{20}{\sqrt{2}} $$
Calculation:
\( \sqrt{2} \approx 1.414 \)
$$ V_{rms} = \frac{20}{1.414} $$$$ V_{rms} \approx 14.14 $$
Final Answer:
\( V_{rms} \approx 14.14 \) V
8. Energy from battery
Given:
Energy density \( = 150 \) Wh/kg
Mass \( = 2 \) kg
Total energy:$$ E = 150 \times 2 = 300 \text{ Wh} $$
Convert Wh to joules:
\( 1 \) Wh \( = 3600 \) J
$$ E = 300 \times 3600 $$$$ E = 1,080,000 $$
Final Answer:
\( E = 1,080,000 \) J (joules)
9. Energy delivered by battery
Given:
\( I = 3 \) A
\( t = 4 \) hours
\( V = 48 \) V
Formula:$$ E = V \times I \times t $$
Substitution:$$ E = 48 \times 3 \times 4 $$$$ E = 48 \times 12 $$$$ E = 576 $$
Convert to kWh:$$ E_{kWh} = \frac{576}{1000} = 0.576 $$
Final Answer:
\( E = 0.576 \) kWh
10. Total energy storage capacity
Given:
Each battery: \( 200 \) Ah, \( 12 \) V
Number of batteries: \( 10 \)
Total charge:$$ Q_{total} = 200 \times 10 = 2000 \text{ Ah} $$
Convert to coulombs:$$ Q_{total} = 2000 \times 3600 = 7,200,000 \text{ C} $$
Total energy:$$ E = V \times Q $$
Since all batteries are \( 12 \) V:$$ E_{total} = 12 \times 7,200,000 $$$$ E_{total} = 86,400,000 \text{ J} $$
Convert to MJ:$$ E_{MJ} = \frac{86,400,000}{1,000,000} = 86.4 $$
Final Answer:
\( E = 86.4 \) MJ (megajoules)
A telecommunication engineer at Safaricom is tasked with installing a backup battery system for a mobile base station in Kisumu. The battery is rated at 48 V and 100 Ah. Calculate the total energy (in joules) that the battery can deliver when fully charged. (4 marks)
An optical fiber transmission system requires a current of 0.5 A to operate. If the supply voltage is 12 V, determine the power consumed by the system. (4 marks)
A technician uses a multimeter to measure the voltage across a resistor and finds it to be 9 V. If the current through the resistor is 0.3 A, calculate the resistance. (4 marks)
The charge stored in a capacitor used in a telecom repeater circuit is 0.002 C. If the voltage across the capacitor is 5 V, what is the capacitance of the capacitor? (4 marks)
A telecommunication tower uses a battery with an energy density of 150 Wh/kg and a capacity of 200 Ah at 24 V. What is the minimum mass of the battery required to provide this capacity? (4 marks)
A DC power supply in a server room at a university provides 24 V. If the equipment draws 2 A for 5 hours, calculate the total energy consumed in kilowatt-hours (kWh). (4 marks)
A backup battery has an anode made of lithium and a cathode made of cobalt oxide. Describe the role of the electrolyte during discharge and recharge cycles. (4 marks)
At a county government data center, a technician measures the current in a cable using an ammeter and finds it to be 1.5 A. If the cable voltage is 230 V, calculate the electrical power being transmitted. (4 marks)
A telecommunication relay station uses a nickel-cadmium battery with a capacity of 50 Ah and a voltage of 12 V. Calculate the total charge stored in the battery in coulombs. (4 marks)
Compare the advantages and disadvantages of using lithium-ion, nickel-cadmium, and lead-acid batteries in telecommunication applications. (4 marks)
Given: \( V = 48\, \text{V} \), \( Q = 100\, \text{Ah} \)
Answer: \( 17,280,000\, \text{J} \)
Power: \( P = V \times I \)
Answer: \( 6\, \text{W} \)
Resistance: \( R = \frac{V}{I} \)
Answer: \( 30\, \Omega \)
Capacitance: \( C = \frac{Q}{V} \)
Answer: \( 400\, \mu\text{F} \)
Capacity: \( 200\, \text{Ah} \times 24\, \text{V} = 4,800\, \text{Wh} \)
Answer: \( 32\, \text{kg} \)
Energy: \( E = V \times I \times t \)
Answer: \( 0.24\, \text{kWh} \)
The electrolyte facilitates the movement of ions between the anode and cathode. During discharge, lithium ions move from anode to cathode through the electrolyte, generating current. During recharge, ions move from cathode to anode, restoring battery capacity.
Power: \( P = V \times I \)
Answer: \( 345\, \text{W} \)
Charge: \( Q = \text{Capacity} \times 3600 \)
Answer: \( 180,000\, \text{C} \)
Lithium-ion: High energy density, lightweight, long cycle life; expensive, sensitive to overcharge.
Question 11 (Compulsory - 20 marks)
A telecommunication engineer at Telkom Kenya is designing a power backup system for a remote cellular site in Narok County. The site is expected to require 600 W of power for 18 hours during power outages.
a) Calculate the minimum battery capacity (in Ah) required if the system uses a 48 V battery. Show all steps. (10 marks)
b) If the battery selected has an energy density of 120 Wh/kg, determine the minimum mass of battery required. (10 marks)
Question 12 (20 marks)
A technician at a hotel installs a new radio transmitter. The transmitter operates at 24 V and draws a current of 2.5 A.
a) Calculate the total charge delivered to the transmitter in 4 hours.
b) Determine the total energy consumed in joules.
c) If the transmitter is powered by a battery rated at 24 V, 10 Ah, how long can the transmitter run continuously before the battery is fully discharged?
Question 13 (20 marks)
At a SACCO office, a security system uses a lead-acid battery. The battery has a capacity of 80 Ah and a voltage of 12 V.
a) Calculate the total energy stored in the battery in kWh.
b) If the system draws a current of 4 A, how long (in hours) will the battery last?
c) Discuss two advantages and two disadvantages of using lead-acid batteries in this application.
Question 14 (20 marks)
A university campus uses a solar-powered repeater station. The station requires 300 W, and the solar panel delivers 15 V at 20 A.
a) Calculate the current required for the repeater station.
b) Determine if the solar panel can supply sufficient current.
c) If the system uses a lithium-ion battery rated at 15 V, 40 Ah, calculate the total energy stored in Wh.
d) Compare the benefits of lithium-ion batteries versus nickel-cadmium batteries for this application.
Question 11
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Question 12
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Question 13
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Question 14
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