A half-wave rectifier has a source of 120 V rms at 60 Hz and an R-L load with R=10 2 and L= 10 mH. Determine (a) an expression for the load current, (b) the average current, (c) the power absorbed by the resistor (d) the power factor. Hint: Beta=3.5 Rad

A half-wave rectifier has a source of 120 V rms at 60 Hz and an R-L load with R=10 2 and L= 10 mH. Determine (a) an expression for the load current, (b) the average current, (c) the power absorbed by the resistor (d) the power factor. Hint: Beta=3.5 Rad

EBK ELECTRICAL WIRING RESIDENTIAL

19th Edition

ISBN:9781337516549

Author:Simmons

Publisher:Simmons

Chapter19: Special-purpose Outlets-water Pump, Water Heater

Section19.2: Water Heater Branch Circuit

Problem 8R: For residential water heaters, the Consumer Product Safety Commission suggests a maximum temperature...

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A thermistor is a special type of resistor whose resistance is more temperature-dependent than standard resistors. The resistance of the thermistor decreases with increasing temperature. This is the basic principle of thermistors. This allows the thermistor circuit to detect very small temperature changes that cannot be observed with the Resistance temperature detectors (RTDs) or the thermocouples. Since the resistance of a thermistor is temperature dependent, it can be connected to an electrical circuit to measure body temperature. This high sensitivity to temperature changes makes thermistors very useful for accurately controlling and calibrating temperature measurements. In general, resistance increases with increasing temperature in most metals, but thermistors react negatively.

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A half-wave rectifier has a source of 120 V rms at 60 Hz and an R-L load with R=10 2 and L= 10 mH.
Determine
(a) an expression for the load current,
(b) the average current,
(c) the power absorbed by the resistor
(d) the power factor. Hint: Beta=3.5 Rad

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Step 2: Deriving the expression of the load current.

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Step 3: Calculating the average current.

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Step 4: Calculating the power absorbed by the resistor.

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