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

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

02:38

### 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.

Let’s start off by sketching out the scenario. At the outset, we have our galvanometer which is a device that measures current in line with our circuit. We’re told that overall the galvanometer has a resistance of 45 ohms. We then connect our galvanometer to a shunt resistor we can call 𝑅 sub 𝑠 set up in parallel with it.

To begin solving for 𝑅 sub 𝑠, lets consider how current might act as it enters this part of the circuit. because the galvanometer and the shunt resistor are set up in parallel, that means the current has two different branches it can travel through. If we call the total current moving through the circuit capital 𝐼, then we’re told in the problem statement that one-tenth of that total passes through the galvanometer branch.

We can say then that 𝐼 divided by 10 follows this branch of the parallel circuit. And that must mean that nine-tenths multiplied by 𝐼, the total circuit current, moves through the shunt resistor branch.

Now that we know how current divides up over these two branches, let’s recall how the current in each branch relates to the resistance in the branch. This relationship is described by Ohm’s law, which tells us that the potential difference dropped across each of these two branches is equal to the current in each one multiplied by the resistance in each one.

In other words, the potential difference across the shunt resistor is equal to nine-tenths 𝐼 times the value of that resistor or the potential difference dropped across the galvanometer is equal to one-tenth 𝐼 times 𝑅 sub 𝐺.

Looking back at our sketch, we can recall one more helpful fact about the way that these parallel branches will behave. We know that the total voltage drop across each of these parallel branches must be the same. This tells us that 𝑉 sub 𝐺 is actually equal to 𝑉 sub 𝑠.

Knowing that fact about this circuit, we can set these two values equal to one another. We can write that nine-tenths 𝐼 times 𝑅 sub 𝑠 is equal to one-tenth 𝐼 times 𝑅 sub 𝐺. Looking at this equation, we see that the factor of 𝐼 divided by 10 is common to both sides and therefore cancels out. This leaves us with an expression nine times the value of the shunt resistance 𝑅 sub 𝑠 is equal to 𝑅 sub 𝐺.

But remember we know 𝑅 sub 𝐺; we’re given it in our problem statement. 𝑅 sub 𝑠 then, the value of the shunt resistor, is 45 ohms divided by nine or five ohms. Based on the circuits performance and the way it divides up current, this is the value of the shunt resistor.