# Video: Physics Past Exam • 2017/2018 • Pack 1 • Question 37

Physics Past Exam • 2017/2018 • Pack 1 • Question 37

02:20

### Video Transcript

A moving coil galvanometer has resistance 45 ohms. When the galvanometer is connected to a shunt resistor, the current passing through the galvanometer is one tenth of the total circuit current. Find the value of the shunt resistor, 𝑅 sub 𝑆.

A galvanometer is an electrical device that’s used for measuring currents. We can depict this galvanometer as positioned in series in an electrical circuit. We’re told that the galvanometer is put in parallel with a shunt resistor. And this means that the current flowing into this now parallel branch has an option to either go through the shunt resistor or continue on through the galvanometer. Which way the current flows when it reaches this branch in the circuit has to do with the relative values of the shunt resistance and the galvanometer resistance.

We can call the galvanometer’s resistance 𝑅 sub 𝐺, and we’re told that that’s equal to 45 ohms. Since the galvanometer is positioned in parallel with the shunt resistor, using our knowledge of circuits, we know that the same amount of potential difference drops over either of these two branches.

This means that, using Ohm’s law, which says that the potential difference across a circuit is equal to the current in the circuit multiplied by its resistance, we can say then that the current that goes through the shunt resistor times that resistance’s value is equal to the current that goes through the galvanometer times that resistance value.

If we label the total current that runs through this circuit capital 𝐼, the problem statement tells us that one tenth of this total value goes through the galvanometer; that’s 𝐼 sub 𝐺. Since the total current in this circuit only has two branches to choose from, if one tenth of it goes through the galvanometer, that means nine tenths goes through the shunt resistor. So 𝐼 sub 𝐺 is equal to 𝐼 over 10, and 𝐼 sub 𝑆 is equal to 𝐼 times nine tenths.

Plugging these representations for 𝐼 sub 𝐺 and 𝐼 sub 𝑆 into our equation for Ohm’s law, we can say that nine tenths 𝐼 times 𝑅 sub 𝑆 is equal to one tenth 𝐼 times 𝑅 sub 𝐺. We see that the factor of 𝐼 over 10 cancels out from both sides of this expression. So nine times 𝑅 sub 𝑆 is equal to 𝑅 sub 𝐺, or 𝑅 sub 𝑆 is equal to 45 ohms divided by nine. This simplifies to five ohms. That’s the resistance value of 𝑅 sub 𝑆, the shunt resistor.