# Video: Using a Galvanometer as an Ammeter

Find the resistance that must be placed in parallel with a 60 Ξ© galvanometer having a 1.00-mA sensitivity to allow it to be used as an ammeter with a 25-A full-scale reading.

02:21

### Video Transcript

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, π sub π. And weβll call the sensitivity of the galvanometer, 1.00 milliamperes, πΌ sub π . We want to solve for a resistance value β weβll call it π β which when put in parallel with the galvanometer lets it read out a 25-ampere full-scale current reading, a scale that weβll call πΌ sub π.

If we draw a sketch of this parallel branch, we see we have our galvanometer with its resistance of 60 ohms and the resistor π 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.

Ohmβs law tells us that π£ is equal to πΌ times π or π, resistance, equals π£ divided by πΌ. π, in our case the resistance of the other branch of the circuit, is equal to π£, where π£ is the potential drop across the galvanometer, divided by πΌ, where πΌ is the amount of current that moves through the branch of the circuit containing π.

Using Ohmβs law to write out this π£ in terms of π sub π and πΌ sub π  and recognizing that πΌ is equal to πΌ sub π minus πΌ sub π , we can write that the resistance π we want to solve for is equal to πΌ sub π  times π sub π divided by πΌ sub π minus πΌ sub π .

Since weβre given all three of these values in the exercise statement, we can plug in and solve now for π. With these values entered in and making sure to write all of our currents in units of amperes, when we calculate this value for π, we find that, to two significant figures, it equals 2.4 times 10 to the negative third ohms. Thatβs the resistance needed in the branch parallel to the galvanometer.