Do your lights dim when your amp powers up?

Hi , I had similar problems in my 40 year old house. 2 years ago I added Solar and during the process I had a brand new panel installed. Prior to that I had an issue where my pool motor ( single speed at the time ) was not running smoothly. The issue was a WEAK breaker that wouldn’t hold the current. While replacing it , I noticed the bus bar was pitted and moved it to a new location, problem solved. I also TIGHTENED all the connections. Now with a new box and a variable speed motor no issues. I just had the kitchen remodeled a few months ago and we did all new kitchen wiring for all new appliances. We went to a large induction stove that requires a 50 amp circuit. We went 6 gauge and it’s rated for 60 amps but has a 50 amp breaker. When we kick it on high , turn on the vacuum or use the tea pot I get a very slight startup flicker in other lights. Like mentioned above if is just a fraction of a second don’t worry. As mentioned above i strongly urge you to check for loose screws/nuts at the panel and check for tired breakers as they are cheap to replace. Regards, Mike B. 

I would add that the experiment of using a long extension cord can cause voltage drop, which in turn may increase current draw. 

What @larryi  said...

Look out your window and see if the neighbors lights dim on your power up. If so that’s some real current draw. 😀

I would add that the experiment of using a long extension cord can cause voltage drop, which in turn may increase current draw. 

Hey @whart - I'm afraid your understanding is a little inconsistent.  You are correct that voltage drop would occur due to the extension cord, but this would cause a decrease in current.  Let me work through the math with you. 

Let's assume a 300W at idle amplifier.  At 120VAC this is the equivalent of a 48 Ohm resistor which is drawing 2.5 Amps. 

Let's check the basic math: 

Watts = A x V , or Watts = AxAxR

so: 

300W = 2.5A x 120V

We also know that V / I = R

120 / 2.5 = 48 Ohms. 

So, summarizing: 

Wall voltage: 120V

Amp power : 300

Amp current: 2.5

Equivalent R : 48 Ohms. 

Putting this in terms of current: 

2.5A = 120V / 48 Ohms

So, everything is consistent no matter how you look at it.  What happens now when we add the resistance of an extension cord?  We should avoid using cheap lamp cord extensions except for lamps, but let's discuss a really long or really bad one which has a whopping 2 Ohms of resistance.  Honestly that's a lot even for a super cheap extension, but it makes our math easier. :) 

So now the total series resistance = 2 Ohms + 48 Ohms = 50 Ohms. 

As we showed above, A = V / Ohms, so 

120 / 50 = 2.4 Amps.  It's dropped, not increased! 

Now, in terms of "voltage drop" what we mean is that literally a voltage drops across a resistance due to current.  So, taking our really bad lamp cord: 

2.4A x 2 Ohms = 4.8V of drop! 

And here's the real problem of "voltage drop."  Instead of your amplifier getting 120V it sees: 

120 - 4.8 = 115.2 V ! 

It's the same if we calculate it using A and Ohms:

2.4 A x 48 Ohms (for the amp) = 115.2V. 

So, you were a little mistaken.  You were right about the voltage dropping but so does current.  What is true though is that on startup the amp won't finish charging for a longer period of time due to the reduced current flow. 

There is an exception!! If you used a voltage regulator, current would rise when voltages dropped, but what's happening then is the voltage regulator is a variable impedance, it's no longer a fixed R value like we modeled here.