Video Transcript
Which of the following represents
the intensity and the direction of the electric current 𝐼 in the figure shown
below? (A) Two amps, out of the node; (B)
two amps, towards the node; (C) six amps, out of the node; (D) six amps, towards the
node.
To find the intensity and direction
of the electric current 𝐼, we can use Kirchhoff’s first law. Recall that Kirchhoff’s first law
states that the sum of the currents into a junction, or node, in the circuit must be
the same as the sum of the currents out of the junction, or node.
We will begin by labelling the
circuit with the currents. We can see that the currents 𝐼 one
and 𝐼 three come into the node, and the currents 𝐼 two and 𝐼 four go out of the
node. We’ve also shown 𝐼 five going out
of the node. 𝐼 five represents the unknown
current, 𝐼 , that we’re trying to find, so we don’t actually know the direction of
this current. However, if we’ve labeled it
incorrectly, we’ll find out when we complete the maths later on. So, it doesn’t matter too much how
we draw this current for now.
Kirchhoff’s first law tells us that
the total current into the junction, 𝐼 one plus 𝐼 three, is equal to the total
current out of the junction, 𝐼 two plus 𝐼 four plus 𝐼 five. We are given the values 𝐼 one
equals six amps, 𝐼 two equals two amps, 𝐼 three equals five amps, 𝐼 four equals
three amps, and 𝐼 five equals 𝐼.
By substituting the values of these
currents into the equation we obtained from Kirchhoff’s first law, we find that six
amps plus five amps equals two amps plus three amps plus 𝐼. Simplifying this, we find that 11
amps equals five amps plus 𝐼. By subtracting five amps from both
sides of the equation, we find that the current 𝐼 equals 11 amps minus five amps
equals six amps.
We also note that the value of 𝐼
we have calculated is a positive number. Since we labeled the current 𝐼 as
going out of the node in the diagram, the positive value of 𝐼 means that we have
correctly labeled the direction of the current 𝐼 as out of the node. If it turned out that we’d labeled
this direction incorrectly and the current were actually into the node, then this
value would have come out as a negative number. So, we have found that the current
𝐼 has a magnitude of six amps, out of the node. This corresponds to option (C).
We will now go through why the
other answer options are incorrect. We note that Kirchhoff’s first law
is a consequence of charge conservation. A fundamental concept in physics is
that charge will always be conserved. This means that charge isn’t
created or destroyed. Since current is the rate of flow
of charge, the current flowing into a point must be the same as current flowing out
of that point.
If the intensity of the current 𝐼
was equal to two amps, then charge would not be conserved at the node. Therefore, options (A) and (B) are
incorrect. Similarly, if the current 𝐼 was
equal to six amps but directed towards the node, then charge would not be conserved,
so option (D) is incorrect.
And so, the correct answer to this
question is option (C). The intensity and the direction of
the electric current 𝐼 is six amps, out of the node.
