Find the resistance that must be placed in parallel with a 60-ohm galvanometer having a 1.00-milliampere sensitivity to allow it to be used as an ammeter with a 25-amp full-scale reading.
We can call the resistance of the galvanometer, 60 ohms, 𝑅 sub 𝑔. And we’ll call the sensitivity of the galvanometer, 1.00 milliamperes, 𝐼 sub 𝑠. We want to solve for a resistance value — we’ll call it 𝑅 — which when put in parallel with the galvanometer lets it read out a 25-ampere full-scale current reading, a scale that we’ll call 𝐼 sub 𝑓.
If we draw a sketch of this parallel branch, we see we have our galvanometer with its resistance of 60 ohms and the resistor 𝑅 we wanna put in a parallel branch. The point of the resistor in a parallel branch is to mediate and moderate the current that flows through either branch of this parallel circuit.
Ohm’s law tells us that 𝑣 is equal to 𝐼 times 𝑅 or 𝑅, resistance, equals 𝑣 divided by 𝐼. 𝑅, in our case the resistance of the other branch of the circuit, is equal to 𝑣, where 𝑣 is the potential drop across the galvanometer, divided by 𝐼, where 𝐼 is the amount of current that moves through the branch of the circuit containing 𝑅.
Using Ohm’s law to write out this 𝑣 in terms of 𝑅 sub 𝑔 and 𝐼 sub 𝑠 and recognizing that 𝐼 is equal to 𝐼 sub 𝑓 minus 𝐼 sub 𝑠, we can write that the resistance 𝑅 we want to solve for is equal to 𝐼 sub 𝑠 times 𝑅 sub 𝑔 divided by 𝐼 sub 𝑓 minus 𝐼 sub 𝑠.
Since we’re given all three of these values in the exercise statement, we can plug in and solve now for 𝑅. With these values entered in and making sure to write all of our currents in units of amperes, when we calculate this value for 𝑅, we find that, to two significant figures, it equals 2.4 times 10 to the negative third ohms. That’s the resistance needed in the branch parallel to the galvanometer.