Video: Using a Galvanometer as a Voltmeter

Analog meters use a galvanometer, which essentially consists of a coil of wire with a small resistance and a pointer with a scale attached. When there is current in the coil, the pointer turns; the amount the pointer turns is proportional to the amount of current in the coil. Galvanometers can be used to make a voltmeter if a resistor is placed in series with the galvanometer. Consider a galvanometer that has a resistance of 25.00 Ξ© and gives a full scale reading for a 50.00-πœ‡A current. The galvanometer is used to make a voltmeter that has a full scale reading of 10.00 V, as shown. Recall that a voltmeter is connected in parallel with the component of interest, so the meter must have a high resistance or it will change the magnitude of current in the component. What is the potential drop across the series resistor in the meter? What is the resistance of the parallel resistor?

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Video Transcript

Analog meters use a galvanometer, which essentially consists of a coil of wire with a small resistance and a pointer with a scale attached. When there is current in the coil, the pointer turns. The amount the pointer turns is proportional to the amount of current in the coil. Galvanometers can be used to make a voltmeter if a resistor is placed in series with the galvanometer. Consider a galvanometer that has a resistance of 25.00 ohms and gives a full scale reading for a 50.00 microamp current. The galvanometer is used to make a voltmeter that has a full scale reading of 10.00 volts, as shown. Recall that a voltmeter is connected in parallel with a component of interest. So the meter must have a high resistance or it will change the magnitude of current in the component. What is the potential drop across the series resistor in the meter? What is the resistance of the parallel resistor?

In part one, we want to solve for the potential drop across the series resistor. We’ll call this 𝑉 sub 𝑅. And in part two, we want to solve for the resistance value of the resistor in parallel. We’ll call this value 𝑅 sub 𝑃. To start on our solution, we can recall that the current in this circuit, which we can call 𝐼, is given as 50.00 microamps. And that the resistance of the galvanometer, which we can call 𝑅 sub 𝑀, is given as 25.00 ohms.

To solve first for our voltage drop over our resistor 𝑅 sub 𝑆, we can see from our diagram that a potential difference of 10.00 volts is applied across the circuit. We know that some of that voltage is lost over 𝑅 sub 𝑆 and some over 𝑅 sub 𝑀. When we recall Ohm’s law that the voltage dropped over a particular component is equal to the current through that component multiplied by its resistance, we see that 𝑉 sub 𝑅 is equal to Δ𝑉, the total potential difference applied across the circuit, minus the product of the current in the circuit times the resistance of the meter 𝑅 sub 𝑀. When we plug in for these three values, making sure to use units of amperes for our current 𝐼, and enter these values on our calculator; we find that 𝑉 sub 𝑅, to four significant figures, is 9.999 volts. So 𝑅 sub 𝑆, the resistor setup in series, uses up nearly all of this circuit’s potential difference.

Now we move on to solving for the value of that resistor over which 𝑉 sub 𝑅 volts drops. Again, working off of Ohm’s law, we can write that Δ𝑉 is equal to the current in the circuit, 𝐼, plus the sum of the two resistors, as shown in the diagram. When we rearrange this equation to solve for 𝑅 sub 𝑃, we find it’s equal to Δ𝑉, the potential difference across the whole circuit, divided by 𝐼 minus 𝑅 sub 𝑀. We’re now ready to plug in for these three values and solve for 𝑅 sub 𝑃.

When we do, again converting the units of our current to amps, and enter these values on our calculator, we find that 𝑅 sub 𝑃, the resistor placed in parallel with our meter, to four significant figures, has a resistance of 200.0 kiloohms. That’s the resistance of the resistor we add to make a voltmeter.

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