1) Answer the following questions. a) As an engineer, you have been given a task to determine whether the load is resistive (R), inductive (L), capacitive (C), resistive-inductive (RL), or resistive-capacitive (RC) based on the current measured at the source as depicted in Table 1. The voltage source is 24020° V. Table 1 No. Source Current Load Type i 52 – 90° ii 5435° ii 520° iv 52 – 45° 5290° b) Power factor correction is a method used to increase the power factor (p.f) of the inductive load (RL) to an allowable value typically above than 0.85 as prescribed by TNB. This can be done by installing a capacitor in parallel to the inductive load. Consequently, load current can be reduced therefore, reducing the losses. Sketch the equivalent circuit diagram of the inductive load with improved p.f and phasor diagram of the load current, hence show how this method can reduce the load current.
1) Answer the following questions. a) As an engineer, you have been given a task to determine whether the load is resistive (R), inductive (L), capacitive (C), resistive-inductive (RL), or resistive-capacitive (RC) based on the current measured at the source as depicted in Table 1. The voltage source is 24020° V. Table 1 No. Source Current Load Type i 52 – 90° ii 5435° ii 520° iv 52 – 45° 5290° b) Power factor correction is a method used to increase the power factor (p.f) of the inductive load (RL) to an allowable value typically above than 0.85 as prescribed by TNB. This can be done by installing a capacitor in parallel to the inductive load. Consequently, load current can be reduced therefore, reducing the losses. Sketch the equivalent circuit diagram of the inductive load with improved p.f and phasor diagram of the load current, hence show how this method can reduce the load current.
Chapter23: Special-purpose Outlets—electric Heating, Air Conditioning
Section23.1: Electric Heat
Problem 17R: Electric heating cable embedded in plaster, or sandwiched between layers of dry wall, creates heat....
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