1.1.3inputcurrent
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A transformer uses electromagnetic induction to transfer electrical energy from one circuit to another. It has a primary coil with a supply voltage V1 that induces a voltage V'1 in the secondary coil. The secondary coil then outputs a voltage V'2 to power the load circuit, with currents Ie, Io, and Im flowing in the primary coil, secondary coil, and load respectively.
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The document calculates the electrical load of communal services in an apartment building. It shows that 24 lighting points will draw 240 watts and 6 10A sockets will draw up to 12A, for a total demand of 18A per phase. The total demand current per phase for communal services is 18A.
4.1.6 Example Domestic 2
The document calculates the electrical load and demand for 6 living units per phase. It lists the types of loads in each unit, the quantity and allowance per unit, and uses this to calculate the total demand current for phases L1, L2 and L3, which is 154.4 amps for each phase. The key loads included are lighting, power outlets, cooking ranges, air conditioners and hot water systems.
4.1.5 Example Domestic 1
This document discusses the number of living units per phase of a project. It calculates that for 18 total living units divided into 3 phases, there would be 6 living units per phase.
4.1.3 Example Solution Three
The document calculates the electrical load and demand current for 11 circuits in a home. It groups the loads into categories like lighting, outlets, appliances, and assigns each a description, allowance, and demand calculation. The total demand current per phase is summarized at the bottom, with the highest draw of 46.5 amps on phase L3.
4.1.2 Example Three
The document outlines the functions and ratings of 12 electrical circuits. Circuit 1 provides power for 13 indoor lighting points. Circuit 10 powers an air conditioner rated at 23.6 amps per phase and can draw power from circuits L1, L2, and L3. Circuits 11a and 11b each power a controlled load hot water unit rated at 3.6 kW.
4.1.1 Example Single
This document calculates the total demand current for an electrical installation consisting of:
- 21 lighting points and 12 double sockets, contributing 5A and 15A respectively
- 15 single sockets contributing 10A
- A 6kW oven contributing 0.5A
- A 2.4kW water heater contributing 0.33A
The total demand current calculated is 45.83A.
5.4.4 LR Circuits
The document discusses the time constant and final current value for an RL circuit. It states that:
1) The time constant for the circuit is 0.17 seconds based on the given inductance and resistance values.
2) It will take approximately 0.85 seconds (5 time constants) for the current to reach its final value.
3) Using Ohm's law, the approximate final current after 0.85 seconds is 2 amps.
14.5.4 Example 1 Part 1
This document calculates the apparent power, power factor, and phase angle for a circuit. It finds that the apparent power is 2.308 kVA by multiplying the current of 9.615 by the voltage of 240. This apparent power is larger than the actual power of 1.5 kW, indicating a poor power factor of 0.65 or a 49.46 degree phase angle between the current and voltage.
14.1.2 TRA Power
Reactive power (Q) and true power (P) combine to form apparent power (S). Apparent power is the combination of true power, which is the usable energy in a circuit, and reactive power, which is stored energy that results from the combination of voltage and current out of phase.
14.5.8 Example 1 Part 5
This document calculates the true power, apparent power, and total current for a circuit. It determines that the true power is 1.5 kW, the apparent power is 1.5009 kVA, and the power factor is 0.99994. It then calculates that with an apparent power of 1.5009 kVA at 240 Volts, the total current would be 6.25 amps.
14.5.7 Example 1 Part 4
This document calculates the capacitive reactance (Xc), current (Ic), and reactive power (Qc) of a capacitor with a capacitance of 80 microfarads operating at 60 Hz with a voltage of 240 V. It then calculates the difference between the inductive and capacitive reactive power as 17 VAR.
14.5.6 Example 1 Part 3
This document calculates the capacitive reactance and capacitance value needed for a reactive power of 1.754 kVAR at 240V. It shows that the capacitive reactance is 32.845 ohms and the required capacitance is 80.761 microfarads.
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- 2. I e I o I m
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