walter
Well-known member
Why is there inrush current in a capacitor?
I recently had a job interview where I was asked about an RC circuit. I thought I did a good job at explaining what happens when the switch is thrown. I stated that there is inrush current as the cap charges to almost the supply in five time constants, mostly in the first time constant. Then I was asked “Why is there inrush current in a capacitor?” I said because the capacitor is at zero volts and wants to be at supply level, but that was not the correct answer, and the message was that I need to be constantly learning and brushing up on what I have learned in the past. It was a busy week and I didn’t think much about it. A few days later I had a follow up interview where I was asked of this round of interviews which was the hardest question and if I could give a better answer now. I said “Why is there inrush current in a capacitor?”, and when prompted, I responded “Ohms law.” The rest of the interview seemed like a flaming death spiral. I have asked Six different people this question and received Four different responses. I found this video that shows the math around the RC circuit:
https://www.youtube.com/watch?v=OwbQLrNBDpY
At the end of the video, the equation is simplified to Ohms law, but that is not the Why, that is the math supporting what happens at t=0, the instant the switch is thrown. At the beginning and the end of this video, it is stated “at t=0 the uncharged capacitor behaves like a wire so I total = VB/R”. I suspect that there is inrush current in a partially charged capacitor as well as a completely discharged capacitor, so it seems like I am barking up the wrong tree. I have also read that a capacitor is open to DC (low frequency) and short to High frequency (A\C) and that a transition from 0v to DC supply is a high frequency event, turning a switch on. So the reason that there is inrush current is due to the supply voltage transitioning from Zero (or something) to max in a very short period of time. And when this short period of time is plugged into the equation: I=Io e^-t/RC it yields a high current. Is this correct?
I recently had a job interview where I was asked about an RC circuit. I thought I did a good job at explaining what happens when the switch is thrown. I stated that there is inrush current as the cap charges to almost the supply in five time constants, mostly in the first time constant. Then I was asked “Why is there inrush current in a capacitor?” I said because the capacitor is at zero volts and wants to be at supply level, but that was not the correct answer, and the message was that I need to be constantly learning and brushing up on what I have learned in the past. It was a busy week and I didn’t think much about it. A few days later I had a follow up interview where I was asked of this round of interviews which was the hardest question and if I could give a better answer now. I said “Why is there inrush current in a capacitor?”, and when prompted, I responded “Ohms law.” The rest of the interview seemed like a flaming death spiral. I have asked Six different people this question and received Four different responses. I found this video that shows the math around the RC circuit:
https://www.youtube.com/watch?v=OwbQLrNBDpY
At the end of the video, the equation is simplified to Ohms law, but that is not the Why, that is the math supporting what happens at t=0, the instant the switch is thrown. At the beginning and the end of this video, it is stated “at t=0 the uncharged capacitor behaves like a wire so I total = VB/R”. I suspect that there is inrush current in a partially charged capacitor as well as a completely discharged capacitor, so it seems like I am barking up the wrong tree. I have also read that a capacitor is open to DC (low frequency) and short to High frequency (A\C) and that a transition from 0v to DC supply is a high frequency event, turning a switch on. So the reason that there is inrush current is due to the supply voltage transitioning from Zero (or something) to max in a very short period of time. And when this short period of time is plugged into the equation: I=Io e^-t/RC it yields a high current. Is this correct?