# Question Video: Combined Variation and Its Applications Mathematics

In an electric circuit, the current intensity 𝐼 going through a resistor varies directly with the voltage drop 𝑉 across the resistor and inversely with the resistance 𝑅. Given that the current intensity is 2.5 A with a voltage drop of 5 V across a resistor of resistance 2Ω, find the value of the current intensity when the voltage drop is 15 V and the resistance is 3Ω.

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### Video Transcript

In an electric circuit, the current intensity 𝐼 going through a resistor varies directly with the voltage drop 𝑉 across the resistor and inversely with the resistance 𝑅. Given that the current intensity is 2.5 amps with a voltage drop of five volts across a resistor of resistance two ohms, find the value of the current intensity when the voltage drop is 15 volts and the resistance is three ohms.

So to solve this problem, what we could use is the triangle here we’ve got, which is one of our formula triangles. And it’s 𝑉 at the top, 𝐼, and 𝑅 for voltage, current, and resistance. This is something that you come across in physics.

Well, actually, what I want to do first though, instead of straight away using the triangle, is to think about direct variation and inverse variation or direct proportion and inverse proportion. Well, if you take a look at what we’ve got, we’re told that the current 𝐼 varies directly with the voltage and inversely with the resistance. So we’ve got 𝐼. And then we’ve got our proportionality sign. Then we’ve got 𝑉 over 𝑅.

Well, what we can do is to start to form an equation. And we can do this using 𝐾, which is our proportionality or variation constant. Cause what we can say is that 𝐼 is equal to 𝐾 multiplied by 𝑉 over 𝑅. Well, the first thing we need to see is find 𝐾. And to do that, we can use the information we’ve been given.

So we’re told that the current of intensity 2.5 amps with a voltage drop five volts across the resistor of resistance two ohms is what we get. So therefore, we can plug these values in to find out what 𝐾 is. So when we do that, we get 2.5 is equal to 𝐾 multiplied by five over two. So therefore, if we multiply both sides of the equation by two, we get five is equal to 𝐾 multiplied by five. So therefore, we’re gonna get 𝐾 is equal to one. And we get this because if we divide both sides by five, we just get 𝐾 is equal to one. So what does this actually tell us?

Well, if we substitute back in our value of 𝐾 to our equation, what we’re gonna get is 𝐼 is equal to one multiplied by 𝑉 over 𝑅. That’s because our 𝐾 is equal to one. Well, this can be rewritten as 𝐼 is equal to 𝑉 over 𝑅. Well, great, cause if we check our triangle that we started with at the beginning, we’d see that this would also tell us that 𝐼 is equal to 𝑉 over 𝑅. So it does look like it’s correct. And we can now use this to find the value of the current intensity when the voltage drop is 15 volts and the resistance is three ohms.

So to do that, what we do is we substitute in our values. So we get 𝐼 is equal to 15 over three. So therefore, what we get is 𝐼 is equal to five amps. So we know that the current intensity when the voltage drop is 15 volts and the resistance is three ohms is gonna be five amps. And we did that by showing how we could use the proportionality or variation constant to form an equation between 𝐼, 𝑉, and 𝑅. Which we then checked with our triangle that we know from physics, which tells us that 𝐼 is in fact equal to 𝑉 over 𝑅.