|Oct28-07, 01:51 PM||#1|
Inductance & Resistance in AC
1. Inductor & Resistor connected in series
A circuit comprising of an Inductor & Resistor connected in series.
The values are:
R1 resistor Value 680Ω
Resistance of the Inductor's Windings: (R of L) 98.2Ω
Supply Voltage: 13.2V
Circuit Current: 8.1mA
Supply Freq: 50Hz
Voltage across the resistor R1 5.5V
Voltage across the inductor 10.4V
1. Use Ohm's Law to find the voltage dropped across the resistance of the inductor's windings (R of L)
2. Add this value to the voltage dropped across the resistor R1
3. Draw a scaled phasor diagram of the resistive voltage and the supply voltage, to find the voltage dropped across the pure inductance.
My solution so far:
1. = 8.1mA * 98.2Ω = 0.8V
Voltage dropped across R1 = 8.1mA * 680Ω = 5.5V
2. = 0.8V + 5.5v = 6.3V
3. This is where I get stuck
Can anyone Help!!!!
|Oct29-07, 05:49 AM||#2|
Draw a line AB of length 13.2 cm, which represents the supply voltage. From A draw an arc of length 5.5 cm, which represents voltage across R. From B draw an arc of length 10.4 cm, which represent the voltage across L.. Find the point of intersection C. Join AC. Drop a perpendicular from B to AC produced. The length of the perpendicular is the voltage across the pure inductance.
|Oct30-07, 04:25 PM||#3|
Thanks rl bahat.
That works fine... I couldn't see the wood for the trees!
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