PART66 Licensing Info
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• Licence extension
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Last comments in Q&A
tsukiyo92 left comment on Faiz0108's question:
Then it should write 10 ohm IMPEDANCE instead of resistance ? its kind of misleading.
Then it should write 10 ohm IMPEDANCE instead of resistance ? its kind of misleading.
tsukiyo92 left comment on arifadli8's question:
This as simple to state, is called 'acceptor circuit', which have minimum impedance at resonant frequency.
Thus at non-resonant frequency, the current is lower than at resonant circuit.
This as simple to state, is called 'acceptor circuit', which have minimum impedance at resonant frequency.
Thus at non-resonant frequency, the current is lower than at resonant circuit.
tsukiyo92 left comment on arifadli8's question:
If given that is PI, then it is not correct answer.
High pass filter using capacitor series into circuit, and inductor shunt into earth/ground/neutral.
Capacitor allow higher frequency to pass on (to circuit), blocking lower frequency.
While inductor prevent higher frequency escape(into ground), allowing most of lower frequency return to ground. (DC also)
If given that is PI, then it is not correct answer.
High pass filter using capacitor series into circuit, and inductor shunt into earth/ground/neutral.
Capacitor allow higher frequency to pass on (to circuit), blocking lower frequency.
While inductor prevent higher frequency escape(into ground), allowing most of lower frequency return to ground. (DC also)
tsukiyo92 left comment on arifadli8's question:
I think it is not part of electrical fundamental, maybe somewhere part of electronic fundamental or instruments chapter
I think it is not part of electrical fundamental, maybe somewhere part of electronic fundamental or instruments chapter
tsukiyo92 left comment on arifadli8's question:
First, we need to understand the total current flowing over the armature is due to 'effective voltage' which is sum of applied voltage and back emf. (difference)
Given that we ave actual current of 10A flowing over armature of 0.1ohm :
V=IR
V= 10 * 0.1
V = 1V
With only 1V of effective voltage, that means back emf of 199V.
First, we need to understand the total current flowing over the armature is due to 'effective voltage' which is sum of applied voltage and back emf. (difference)
Given that we ave actual current of 10A flowing over armature of 0.1ohm :
V=IR
V= 10 * 0.1
V = 1V
With only 1V of effective voltage, that means back emf of 199V.
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