Lesson Explainer: Electric Current | Nagwa Lesson Explainer: Electric Current | Nagwa

Lesson Explainer: Electric Current Physics • Third Year of Secondary School

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In this explainer, we will learn how to use the formula 𝐼=𝑄𝑑 to calculate the current through a point in simple circuits given the charge moving past that point in a given time.

Electric current is the flow of electric charge. Recall that different parts of the atom have different values of electric charge, as seen below.

Together, the positively charged protons, shown in pink, and neutral neutrons, shown in green, make up the nucleus. The negatively charged electrons, shown in blue, are outside the nucleus.

In a wire, electrons move around, but protons and neutrons do not.

However, early scientists did not know this, nor did they know what an atom was. When they wrote about the flow of electric charge, they assumed that the flow was of positively charged particles. This convention is still used today as conventional current, even though we know that it is the electrons that flow.

Conventional current is the flow of charge assuming that the electric charge carriers are positive, meaning they move away from the positive terminal of a cell and toward the negative terminal. When the current in a circuit is indicated on a circuit diagram, we can assume that it is showing the direction of the conventional current, rather than the flow of electrons. This is shown in the diagram below.

Electric current is the rate of the total electric charge passing a point in coulombs per second and is defined using the following equation.

Definition: Electric Current

When measuring the amount of charge passing through a point, the current through that point can be expressed as 𝐼=𝑄𝑑, where 𝐼 is the current, 𝑄 is the total charge moving past a point, and 𝑑 is the time taken.

The SI unit for current is the ampere. One ampere is equal to one coulomb per second, 1=1/ACs.

Current is positive when its direction is the same as that of conventional current, the direction in which positive charges would flow. Current is negative when its direction is opposite to that of conventional current, the direction in which the electrons are flowing.

Let’s look at an example.

Example 1: Finding the Current through a Point in a Circuit

The diagram shows a circuit consisting of a battery and a light-emitting diode (LED). Over a period of 25 seconds, a charge of 50 coulombs passes through point P in the circuit. What is the current in the circuit during this period?

Answer

In this circuit, conventional current has a counterclockwise direction and passes through the LED on its way to point P. The electrons flowing through it are not used up; they just move.

Measuring the current just before the LED would give the same value as at point P and, indeed, at any point in the circuit. To find the current in the circuit during this period then, we can measure the current at point P.

Since we are given the time taken to pass point P, 𝑑=25seconds, and the total charge of the electrons, 𝑄=50coulombs, we can solve for the current. Looking at the definition of electric current, 𝐼=𝑄𝑑.

We just need to put the values for 𝑄 and 𝑑 in, 50 C and 25 s, respectively, as follows: 𝐼=(50)(25).Cs

The units of coulombs divided by seconds (C/s) are equal to the SI units of current, amperes, with the symbol A, so (50)(25)=2.CsA

Therefore, the current in the circuit during this period is 2 amperes.

Sometimes the current is known in a circuit but the electric charge flowing past a specific point is not. We can isolate this charge past a point, 𝑄, in the equation in order to find it by itself.

Let’s start with the defining equation for current: 𝐼=𝑄𝑑.

We want just the electric charge, 𝑄. To isolate this variable, we can multiply both sides by 𝑑: 𝐼×𝑑=𝑄𝑑×𝑑.

Multiplying through cancels 𝑑 on the right-hand side, leaving behind 𝐼𝑑=𝑄.

Therefore, the total charge passing a point 𝑄 is equal to the product of the current through that point, 𝐼, and the time taken to pass, 𝑑.

Let’s look at some examples using this form of the equation.

Example 2: Finding the Charge Flowing through a Point in a Circuit

The diagram shows a circuit consisting of a battery and a resistor. The current through the circuit is 2.0 A. Over a period of 45 seconds, how much charge flows past point P in the circuit?

Answer

The current at point P is the same as the current at any point right before the negative terminal of the battery.

We are given current and a period of time. The amount of charge that flows past point P can be found by using the equation 𝐼𝑑=𝑄.

The value of current is 2 A, and the value of time is 45 s. Hence, (2)(45)=𝑄.As

The units of amperes are equivalent to C/s. From this, we see that multiplying these two terms cancels the units of time: AsCssCΓ—=(/)Γ—=.

Multiplying through, thus, gives an answer in coulombs: (2/)(45)=90.CssC

Therefore, over a period of 45 seconds, there are 90 coulombs of charge that move past point P in this circuit.

Example 3: Finding the Charge Flowing through a Point in a Circuit

The diagram shows a circuit consisting of a battery and a resistor. The current through the circuit is 50 mA. Over a period of 1.5 hours, how much charge flows past point P in the circuit?

Answer

We can use the equation 𝐼𝑑=𝑄 to find the charge. However, current has a unit prefix, and the time period is given in hours. Let’s try to get the answer in terms of coulombs.

For every 1 ampere, there are 1β€Žβ€‰β€Ž000 milliamperes: 11000.AmA

This means that 50 mA is 11000Γ—50=0.05.AmAmAA

For the value of time that we have been given, we will need to convert it from hours to seconds. There are 60 minutes in 1 hour: 601.minh

So 1.5 hours gives us 601Γ—1.5=90.minhhmin

Similarly, there are 60 seconds in 1 minute. So 90 minutes gives us 601Γ—90=5400.sminmins

Now that we have the values for 𝐼 and 𝑑, 0.05 amperes and 5β€Žβ€‰β€Ž400 seconds, we can put them into the equation to obtain the total charge flowing past point P: (0.05)(5400)=𝑄.As

The units of amperes are equivalent to C/s. This means that multiplying these two terms together cancels the units of time: AsCssCΓ—=(/)Γ—=.

The units left behind after multiplication are coulombs. The answer is thus (0.05/)(5400)=270.CssC

So, over that period of time, 270 coulombs of charge have passed by point P.

We can also rearrange the equation for electric current to make time the subject. This will allow us to discern the amount of time it takes for a certain amount of charge to pass by a point. Using the previously modified equation, we have 𝐼𝑑=𝑄.

We can then divide both sides by 𝐼 to isolate 𝑑: 𝐼𝑑𝐼=𝑄𝐼.

The 𝐼’s on the left-hand side cancel, giving us 𝑑=𝑄𝐼.

The time taken for charge passing a certain point 𝑑 can be found by dividing the total charge, 𝑄, by the current, 𝐼.

Let’s look at some examples using this form of the equation.

Example 4: Finding the Time in Which a Charge Passes through a Point in a Circuit

A laptop charger passes a current of 5.0 A through a laptop battery. Over a period of time, 45β€Žβ€‰β€Ž000 C of charge is transferred to the battery. For how many hours was the laptop left to charge?

Answer

The amount of time it takes for this laptop to charge can be found by using the equation 𝑑=𝑄𝐼.

The charge, 𝑄, is 45β€Žβ€‰β€Ž000 C and the current, 𝐼, is 5 A. Putting these values into the equation gives us 𝑑=(45000)(5).CA

The units of amperes are equivalent to C/s: ACs=/.

This means that the terms in the equation can be written as (45000)(5/).CCs

Dividing a number by a fraction is the same as multiplying this number by the reciprocal of that fraction. We can use this relation to see how the units in the fraction cancel: CCsCsCs(/)=Γ—(/)=.

The units left behind are just seconds, so the answer will be in seconds: (45000)(5/)=9000.CCss

We still have to convert these seconds into hours. There are 60 seconds in 1 minute, so 160Γ—9000=150.minsmin

60 minutes in 1 hour means that 150 minutes is 160Γ—150=2.5.hminminh

So, it takes this battery 2.5 hours to gain 45β€Žβ€‰β€Ž000 C of charge.

Example 5: Finding the Time in Which a Charge Passes through a Point in a Circuit

A rechargeable battery is left to charge for a period of time. It is charged with a current of 10 mA. After it has finished, the battery has gained 180 C of charge. For how many hours was the battery left to charge?

Answer

To find the amount of time needed to charge the battery, we just need to divide the total charge gained by this battery by the charging current: 𝑑=𝑄𝐼.

For every 1 ampere, there are 1β€Žβ€‰β€Ž000 milliamperes: 11000,11000Γ—10=0.01.AmAAmAmAA

Now, we can use the equation by putting in the charge of 180 C and current of 0.01 A: 𝑑=(180)(0.01).CA

The units of amperes are equivalent to C/s: ACs=/.

This makes the terms look like 𝑑=(180)(0.01/).CCs

Dividing a number by a fraction is the same as multiplying this number by the reciprocal of that fraction. We can use this relation to see how the units in the fraction cancel: CCsCsCs(/)=Γ—(/)=.

The units of coulombs cancel, leaving behind units of seconds: (180)(0.01)=18000.ss

Now, we convert the seconds to hours, starting with converting to minutes first. There are 60 seconds in 1 minute, so 160Γ—18000=300.minssmin

There are 60 minutes in 1 hour, meaning that 300 minutes is thus 160Γ—300=5.hminminh

So, it takes this battery 5 hours to gain 180 C of charge.

Let’s summarize what we have learned in this explainer.

Key Points

  • Electric current is the flow of electric charge and is measured in amperes: 𝐼=𝑄𝑑, where 𝐼 is the current, 𝑄 is the total electric charge past a point, and 𝑑 is the time.
  • Conventional current assumes that the charge carriers are positive, and it flows away from positive terminals and toward negative terminals.
  • Electrons move along the wire of a circuit, but the nuclei of the atoms in the wire do not.
  • To find the total charge past a point, the electric current equation can be rearranged to be 𝑄=𝐼𝑑.
  • To find the time it takes for some amount of charge to move past a point, the electric current equation can be rearranged to 𝑑=𝑄𝐼.

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