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Q) We have to determine the I. Magnitude of electron charge is 1.6 x 10^-19C. I have fundamental confusion in this problem. Should we consider the negative charge of electron while doing the problem. I have solved the problem is two different ways, still unable to decide which is the right approach to solve the problem.

Current Flow

Method1: In this adding magnitude of electron charge. I = dq/dt = no of electrons x | charge on a single e |/ time ((125x10^9) x (1.6x10^-19C))/ (1x10^-9s) = 20A Conventional current flow is opposite to the electron flow. Hence the current is out of the node. The picture is attached below.

enter image description here

Now if we apply the KCL, at the node. Current going out is positive. 20A-2A-1A+I = 0,

I = -17A

Method2: In this method adding electron sign & charge also. I = dq/dt = no of electrons x charge on single e/ time ; ((125x10^9) x (-1.6x10^-19C))/ (1x10^-9s) = -20A Conventional current flow is opposite to the electron flow. Hence the current is out of the node. The picture is attached below.

Fig. 3

Taking KCL at the node. We will get. Current going out of node is taken as positive. -20A-2A-1A+I = 0

I = 23A

Help me which is the right approach to solve the problem.

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2 Answers 2

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Neither is the correct approach to the problem, both appear to be confusing you.

I would leave all the arrows on the diagram as they are. It's your step of flipping the electron arrow 'because they carry -ve charge' that appears to me to be your main problem.

The electron current is a current of negative charge carriers, so the current is negative, in the direction of the arrow.

Now compute the outgoing I as the sum of all the incoming currents: -20 A + 1 A + 2 A = -17 A.

In method2, you've flipped the current arrow, which if you take it through consistently, handles the -ve electron charge. However, you don't seem to realise that this is what you've done, and so handle it again.

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You do need to take into account the fact that electrons are negatively charged. In your first attempt, you did that correctly, and in your second attempt, you did it incorrectly.

Suppose we have 125 billion electrons per nanosecond traveling to the right. The following are all accurate ways of describing this situation:

  • approximately 20 A of negative charge traveling to the right
  • approximately -20 A of positive charge traveling to the right
  • approximately -20 A of negative charge traveling to the left
  • approximately 20 A of positive charge traveling to the left

Notice that there are three things in this statement which can be "flipped." If you take a true statement and you flip exactly two of those three things, you still have a true statement. In your first attempt, you used the last statement that I listed here, so you got the right answer.

In your second attempt, however, you drew the diagram with

  • approximately -20 A of positive charge traveling to the left

You flipped three parts of the statement instead of two, so you ended up with a false statement, and the wrong answer.

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