Question Video: Comparing the Current through a Galvanometer and a Shunt Resistor | Nagwa Question Video: Comparing the Current through a Galvanometer and a Shunt Resistor | Nagwa

Question Video: Comparing the Current through a Galvanometer and a Shunt Resistor Physics • Third Year of Secondary School

The current ๐ผ in the circuit shown is 2.5 mA, which is the greatest current that can be measured using the circuit as an ammeter. The resistance of the galvanometer is ten times the resistance of the shunt. What is the difference between ๐ผ_S and ๐ผ_G? Answer to the nearest ฮผA.

03:41

Video Transcript

The current ๐ผ in the circuit shown is 2.5 milliamps, which is the greatest current that can be measured using the circuit as an ammeter. The resistance of the galvanometer is 10 times the resistance of the shunt. What is the difference between ๐ผ S and ๐ผ G? Answer to the nearest microampere.

Letโ€™s start by writing down the information weโ€™ve been given in the question. First, we were told that the current ๐ผ equals 2.5 milliamps. We also know that the resistance of the galvanometer, which weโ€™ll call ๐‘… G, is 10 times the resistance of the shunt ๐‘… S. So in equation form, we can say that ๐‘… G equals 10๐‘… S, where ๐‘… G is the resistance of the galvanometer and ๐‘… S is the resistance of the shunt.

The question is asking us to find the difference between ๐ผ S and ๐ผ G. In other words, we want to find the current through the shunt minus the current through the galvanometer. To do this, we should also recall some more general equations that apply to this circuit. To start, letโ€™s think about the current at this node here. We can see that the current ๐ผ splits into the two other currents, ๐ผ G and ๐ผ S, as shown by the red labels on the diagram. Since charge is conserved, we know that total current ๐ผ must equal the sum of these two currents, so ๐ผ equals ๐ผ G plus ๐ผ S.

Next, because parallel branches in a circuit have the same potential difference across them, we can say that ๐‘‰ S, the potential difference across the shunt, is equal to ๐‘‰ G, the potential difference across the galvanometer. To expand upon this, letโ€™s apply Ohmโ€™s law, which states that potential difference equals current times resistance. This tells us that ๐‘‰ S equals ๐ผ S times ๐‘… S, and ๐‘‰ G equals ๐ผ G times ๐‘… G. Subbing these in, we have that ๐ผ S times ๐‘… S equals ๐ผ G times ๐‘… G.

Now, we know a useful relationship between these resistance values. So to minimize the number of different variables in this expression, letโ€™s substitute 10๐‘… S in for this ๐‘… G term. And since weโ€™re trying to find a relationship between these two current values, next we should make one of them the subject of this equation. We could choose either one. So for now, letโ€™s just choose ๐ผ S, the current through the shunt. Dividing both sides of the equation by ๐‘… S and then canceling the like terms out of the numerator and denominator of both sides, we have that ๐ผ S equals 10 times ๐ผ G.

Now that we have the current through the shunt expressed in terms of the current through the galvanometer, letโ€™s substitute this into our expression for the current into and out of the junction. So we have that the total current ๐ผ equals ๐ผ G plus 10๐ผ G or simply 11๐ผ G. And remember that we have an actual numeric value for the total current ๐ผ, which is 2.5 milliamps. So letโ€™s rearrange this to solve for ๐ผ G in terms of ๐ผ. Dividing both sides of the equation by 11, we know that the current through the galvanometer is one eleventh of the total current. Since we know that ๐ผ S equals 10๐ผ G, we can also say that ๐ผ S equals ten elevenths of the total current.

Finally, we have the current through both the shunt and galvanometer expressed in terms of a value that we know. So weโ€™re ready to find their difference, ๐ผ S minus ๐ผ G. Substituting in our expression for each term and simplifying, the difference can be written as simply nine elevenths of the total current ๐ผ. So, plugging in the value of the total current, 2.5 milliamps, and grabbing a calculator, we get a result of about 2.045 milliamps. All we have to do now is express this in microamperes, and weโ€™ve reached our final answer.

Thus, we found that the difference between the current through the shunt and the galvanometer equals 2045 microamperes.

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