A circuit you're building needs an ammeter that goes from 0 mA...

Question

A circuit you're building needs an ammeter that goes from 0 mA to a full-scale reading of 50 mA. Unfortunately, the only ammeter in the storeroom goes from 0 μA to a full-scale reading of only 500 μA. Fortunately, you've just finished a physics class, and you realize that you can make this ammeter work by putting a resistor in parallel with it, as shown in FIGURE P28.56. You've measured that the resistance of the ammeter is 50.0 Ω, not the 0 Ω of an ideal ammeter. What is the effective resistance of your ammeter?

Schematic showing a circuit with a 500 μA ammeter and a resistor labeled

Accepted Answer

Hello, fellow physicists today we solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. You are working on a circuit that requires an amateur with a full scale reading of 75.0 milliamps. However, the only ammeter available in the store room has a range from zero micro amps to a full scale reading of 300.0 micro amps. The resistance of the amateur is measured to be 60.0 ohms to achieve the desired 75.0 milliamps reading capacity. You consider adding a resistor in parallel to the amateur, calculate the effective resistance of the modified amateur, which is a combination of the amateur and the parallel resistor. Awesome. So that's our end goal is we're trying to figure out what the effective resistance of our new and a and improved modified amateur will be. So that's our end goal. So right below our problem, we have a diagram of the amateur that we designed and that's represented by this dotted box. It's in a little dotted little rectangular or square shaped box I should say. And it shows in a circle with the letter A is our 300 micro amp amateurs amateur. So, and then we also have our current eye flowing from the left going into our resistor in parallel with our amateur. Awesome. And then the resistor is represented by a black jagged line. Awesome. We're also given some multiple choice answers. They're all in the same units of ohms. So let's read them off to see what our final answer might be. So A is 0.240 B is 0.138 C is 0.236. And finally D is 0.158. Awesome. So moving right along, first off, let us note that the amateur provided. So our aim amateur that's provided to us in the prompt shows a full scale deflection with a current of 300 0.0 micro amps which is equal to 0.300 milliamps. Therefore, a maximum current of 0.300 milliamps can be allowed through the ammeter note that the problem is asking us to measure a maximum current of 75.0 milliamps. Therefore, we must split the current in a way where the maximum current of 0.300 milliamps flows to the amur and there remain 74.7 milliamps flows through the resistor or following the concept of parallel resistors. So where in the world did we get the 74.7 milliamps? Well, simply we have to take our 75.0 milli amps and we need to subtract 0.300 milliamps. And that's how we get our 74 0.7 milliamps. Awesome. So now that we're following the concept of parallel resistors, we need to recall. So in following that mindset, that logic, we need to recall and use the equation to determine the potential difference across the resistor. And the amateur is the same. So we're trying to find the resistance using the logic that the potential difference across the resistor and amateur is the same. So the potential difference is represented by V. So V subscript R. So the potential difference across the resistor is equal to the potential difference across the amateur, which I'm going to denote it as V subscript A for the amateur. So therefore, using this equation, we can find the resistance is equal to 0.300 but we need to convert milliamps to amps. So all we have to do is just multiply this by 10 to the power of negative three amps. And then we need to multiply it by 60.0 ohms. Awesome. And then we divide all of this by 74.7 which that is in milliamps, we need to convert it to amp. So just multiply it by 10 to the power of negative three amps. Awesome. So when we plug that into a calculator, we will get 0.2409 which rounds to three decimal places as 0.241 arms. So at this stage, we can now start to solve for the effective resistance. So let's recall that the equation for the effective resistance states that one divided by the effective resistance, which we're gonna do is R subscript eff is equal to one divided by zero point two four ohms plus one divided by 60.0 ohms. OK. OK. So that means that when you add one, divided by 0.24 so that's how we got for resistance value that we just solved whereby just shortened it down to two decimal places to keep things simple. And when we add that to 60 so one divided by 60.0 ohms. So that is gonna be equal to when we add those two per those two fractions together, we will get 4.18 33. But we need to isolate and solve for the effective resistance. So when we do that, we will get that, the effective resistance is equal to one divided by 4.18 33, which when you plug that into a calculator, you will get 0.2390 which then rounds 20.240 ohms, which is our final answer. Hooray, we did it. So that means the effective resistance has to be 0.240 ohms. So looking at our multiple choice answers, the correct answer has to be the letter A 0.240. Ohms. Thank you so much for watching. Hopefully that helped and I can't wait to see the next video. Bye.

Key Concepts

Ammeter Functionality

Parallel Resistance

Ohm's Law

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