# Current behavior after closing a switch

#### ats314

Joined May 19, 2014
22
Hi all,

Here is a pic of the problem I am working on.
http://imgur.com/sn04vnG

I believe that both current I1 and I2 are going to be decreasing, because of the extra resistance added from the switch. I also believe it will be continuously decreasing in both cases, I'm just not 100% sure. Any help would be appreciated. Thank you.

#### WBahn

Joined Mar 31, 2012
29,501
What happens to the total resistance, in steady state (a long time after the switch is closed) seen by the battery compared to what it was in steady state just before the switch was closed?

Try putting some numbers to it. Assume that, in steady state after the switch was closed, that the 100 Ω resistor has 1 A flowing through it. What are the other currents and what is the battery voltage? Now, with that battery voltage, what are the steady state currents before the switch is closed?

#### ats314

Joined May 19, 2014
22
What happens to the total resistance, in steady state (a long time after the switch is closed) seen by the battery compared to what it was in steady state just before the switch was closed?

Try putting some numbers to it. Assume that, in steady state after the switch was closed, that the 100 Ω resistor has 1 A flowing through it. What are the other currents and what is the battery voltage? Now, with that battery voltage, what are the steady state currents before the switch is closed?
OK, so I did the math and when the switch i closed, I2 is definitely decreasing. I'm a little confused on I1, as the values according to the the way the arrow is drawn are negative, and it increases in negative value when the switch is closed. So, I'm not sure if this would be considered increasing current, if you just ignore the way the arrow is drawn, or decreasing current since it gets more negative.

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#### WBahn

Joined Mar 31, 2012
29,501
OK, so I did the math and when the switch i closed, I1 is definitely decreasing. I'm a little confused on I2, as the values according to the the way the arrow is drawn are negative, and it increases in negative value when the switch is closed. So, I'm not sure if this would be considered increasing current, if you just ignore the way the arrow is drawn, or decreasing current since it gets more negative.
Huh?

If I1 is positive, how can I2 be negative?

#### ats314

Joined May 19, 2014
22

#### MikeML

Joined Oct 2, 2009
5,444
Here is a pic of the problem I am working on.
http://imgur.com/
I hate looking at imgur and all of the advertising it loads. Why don't you just post it here?

#### WBahn

Joined Mar 31, 2012
29,501
http://imgur.com/a/ASWtZ

These are the simulated values I get. I'm translating this as I1 increasing and I2 decreasing. Am I wrong?
I just tried to look at your results and imgur is down.

Please post images as part of your post so that they will be archived on the AAC server and will not be dependent on some third-party outfit that may or may not even exist ten years from now.

Without seeing your results, the question I have is why you are interpreting a negative current in a component in a simulation as indicating the current in the same direction as an arrow that is drawn on a schematic? How does the simulation know what direction you drew the arrow?

#### JoeJester

Joined Apr 26, 2005
4,390
Here are his pictures:

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#### WBahn

Joined Mar 31, 2012
29,501
@ats314: Consider your second simulation. You know that the devices are all in series, which means that they all have the same current. Yet some of your devices have positive currents and some have negative currents. That should tell you that there is something subtle about simulation results that you need to understand.

Here's the secret key: All SPICE-based simulators (and most, if not all, other major ones) define current flowing into a pin as positive and they define the device current for a two-terminal device as being the pin 1 current.