Circuits - Scenario 2
This post presents a simple circuit that can be thought of as a basic model for wire insulation. When insulators weaken, high voltages can push stray currents out, hence the need for insulation resistance tests.
SITUATION
A 40-picoFarad capacitor is connected in series with 40-ohm resistor. If the supply is 220 Volts at 60 Hertz, what is the current flowing through the circuit?
ANALYSIS
1.) Capacitive reactance, vector:
X-> = 1 / (jwC) = -j / (2 pi f C)
2.) Impedance, vector:
Z-> = R-> + X->
Z-> = 40 + [ -j / (2 pi 60 * 40p) ]
Z-> = (40 - j 66.315M) ohms
Z-> = 66.315M ohms < (-89.999965)
3.) Current, vector:
i-> = V-> / Z->
i-> = 220V<0 / [ 66.315M ohms < (-89.999965) ]
i-> = 3.3175uA < 89.999965
CONCLUSION
The current is 3.3176 microAmperes, leading.
SITUATION
A 40-picoFarad capacitor is connected in series with 40-ohm resistor. If the supply is 220 Volts at 60 Hertz, what is the current flowing through the circuit?
ANALYSIS
1.) Capacitive reactance, vector:
X-> = 1 / (jwC) = -j / (2 pi f C)
2.) Impedance, vector:
Z-> = R-> + X->
Z-> = 40 + [ -j / (2 pi 60 * 40p) ]
Z-> = (40 - j 66.315M) ohms
Z-> = 66.315M ohms < (-89.999965)
3.) Current, vector:
i-> = V-> / Z->
i-> = 220V<0 / [ 66.315M ohms < (-89.999965) ]
i-> = 3.3175uA < 89.999965
CONCLUSION
The current is 3.3176 microAmperes, leading.
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