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
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
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.