Question Video: Finding the Current through an Electrical Component | Nagwa Question Video: Finding the Current through an Electrical Component | Nagwa

Question Video: Finding the Current through an Electrical Component Physics • Third Year of Secondary School

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A piece of connecting wire in a circuit is 20 cm long. It has a resistance of 0.02 Ω and dissipates energy to its environment as heat at a rate of 2 W. How much is the current passing through the wire?

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Video Transcript

A piece of connecting wire in a circuit is 20 centimeters long. It has a resistance of 0.02 ohms and dissipates energy to its environment as heat at a rate of two watts. How much is the current passing through the wire?

Okay, so in this question, we’ve got a 20-centimeter-long piece of wire with a resistance of 0.02 ohms. Let’s label this resistance as 𝑅. As an aside, we can notice that the value of this resistance is pretty small. Usually when we see resistance values, it’s the resistance of a component in a circuit. But this time, we’re talking about a connecting wire. So, that’s one of the wires that connects together the various components in a circuit. These wires are deliberately designed to make it easy for the charges to flow along them. In other words, they’re deliberately made to have a small value of resistance.

Getting back to the question, we’re also told that this wire dissipates energy to its environment as heat, and that it does this at a rate of two watts. We can recall that power is defined as the rate of energy transfer over time. So, by telling us that the wire dissipates energy at a rate of two watts, the question is telling us that the power dissipated by the wire is two watts. Let’s label this power as 𝑃. We’re being asked to find the value of the current through the wire. We’ll label this current as 𝐼. We can recall that as well as being defined as the rate of energy transfer over time, there’s also an equation which links power, current, and potential difference. Specifically, power 𝑃 is equal to current 𝐼 multiplied by potential difference 𝑉.

In the case of the 20-centimeter wire from this question, the quantity 𝐼 in this equation is the current through the wire and 𝑉 is the potential difference across the wire. That’s the potential difference that would be measured by connecting a voltmeter to both ends of the 20-centimeter length. In this equation, we want to find the value of the current 𝐼. We know the value of the power 𝑃, but we don’t know the potential difference 𝑉. However, we can recall that Ohm’s law tells us that potential difference 𝑉 is equal to current 𝐼 multiplied by resistance 𝑅. So, Ohm’s law tells us that the potential difference across this wire is equal to the current through the wire multiplied by the wire’s resistance.

We can use this Ohm’s law equation to substitute into this equation for the power. Specifically, we can use Ohm’s law to replace the 𝑉 by 𝐼 multiplied by 𝑅. If we take our equation 𝑃 is equal to 𝐼 multiplied by 𝑉 and replace the 𝑉 by 𝐼 multiplied by 𝑅, then we get an equation that says 𝑃 is equal to 𝐼 squared times 𝑅. So, the power 𝑃 that’s dissipated by the wire is equal to the square of the current 𝐼 through the wire multiplied by the wire’s resistance 𝑅. We know the value of 𝑃, and we know the value of 𝑅. We’re trying to find the value of the current 𝐼. So, if we take this equation and rearrange it to make 𝐼 the subject, then we can use our values for the power 𝑃 and the resistance 𝑅 in order to calculate the current through the wire.

To make 𝐼 the subject of this equation, we’ll begin by dividing both sides by the resistance 𝑅. On the right-hand side, the 𝑅 in the numerator cancels with the 𝑅 in the denominator, and this leaves us with an equation that says 𝑃 divided by 𝑅 is equal to 𝐼 squared. The next step is to take the square root of both sides of the equation. On the right-hand side, we have the square root of 𝐼 squared, which is simply 𝐼. And then writing this equation the other way around, we have that 𝐼 is equal to the square root of 𝑃 divided by 𝑅.

Okay, let’s clear ourselves some space so we can substitute our values for 𝑃 and 𝑅 into this equation. When we sub in the values, we find that 𝐼 is equal to the square root of two watts over 0.02 ohms. A power with units of watts and a resistance with units of ohms means that we’ll calculate a current 𝐼 with units of amperes. To work out the value of this expression, let’s first evaluate the fraction under the square root. We’ve got two divided by 0.02, and this works out as 100. So, we have that 𝐼 is equal to the square root of 100 with units of amperes. Finally, evaluating the square root of 100 gives a result of 10. And so, we have that the current 𝐼 is equal to 10 amperes.

Our answer to the question is that the current passing through the wire is 10 amperes. As one last remark, it’s worth noticing that we were actually given some information in the question that we didn’t need. Specifically, we were told that the length of this wire was 20 centimeters. But we found that we could calculate the current 𝐼 through the wire using just the power 𝑃 dissipated by it and the wire’s resistance 𝑅. This meant that we didn’t actually use the value of the wire’s length to calculate our answer of 10 amperes.

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