Question

The resistance of a voltmeter is $180\Omega$ . When there is $5V$ potential difference across its ends, it gives full-scale deflection. What resistance should be connected in series to obtain a full-scale deflection at $25V$ ?

Answer 1

A voltmeter device is used to measure the potential difference between two points. A galvanometer is a device that is used to measure a small amount of current. A galvanometer is converted into a voltmeter when we connect a large amount of resistance in a series connection with the galvanometer. It is this resistance that they have asked us to find. Let us see how we can find this resistance.

Complete Step By Step Answer:
We have seen that the galvanometer can be used as a voltmeter to find the potential difference across two points. But when we connect only the galvanometer to find the potential difference that will show an incorrect value. Because there will be a very small amount of resistance in the galvanometer. Since this resistance is smaller than the resistance that is present in the circuit a large amount of current will flow through the galvanometer that will result in an incorrect value of potential difference. Therefore we will connect another resistor with a large amount of resistance in series with a galvanometer that will block a large amount of current passing through it. Therefore a large amount of current will now pass through the circuit and the galvanometer shows the correct potential difference between the two points. The formula for finding the potential difference before the resistance is connected to the circuit we can use the ohm’s law that is given by,
$\Rightarrow V = IR$
Here, $V$ is the voltage across two points
$I$ is said to be the current passing in the circuit. In this case, for full-scale deflection to happen in the galvanometer I should be equal to ${I_g}$ . ${I_g}$ is the maximum current for full-scale deflection.
$R$ is the resistance.
Therefore, before the galvanometer is connected with a resistor in parallel the potential difference is given as $5V$ and the resistance is given as $180\Omega$ . Therefore substituting all this in the equation that is given above we get,
$\Rightarrow 5 = {I_g} \times 180$ ……. (1)
Now when the galvanometer is connected to the voltmeter then the formula will be changed as,
$\Rightarrow V = {I_g}(R + {R_v})$
Here, ${R_v}$ is the resistance that is connected in series with the galvanometer.
Now the potential difference is given as,
$\Rightarrow V = 25V$
Therefore we can substitute in the above equation,
$\Rightarrow 25 = {I_g} \times (180 + {R_v})$ …… (2)
Now we should the equation (1) by (2) we get,
$\Rightarrow \dfrac{1}{5} = \dfrac{{180}}{{(180 + {R_v})}}$
Simplify to get the answer,
$\Rightarrow 180 + {R_v} = 900$
$\Rightarrow {R_v} = 720\Omega$
Therefore, the value of the resistance that is connected in series with the galvanometer is given by $720\Omega$ .

Note:
We use electrical instruments to measure current, voltage, and resistance because our ohm’s law is not always convenient to measure these parameters. The electrical instruments that are used to measure the current, voltage, and resistance are an ammeter, voltmeter, and multimeter. Ammeter is used to measure only the current, voltmeter is used to measure only the voltage but the multimeter is used to measure multiple parameters like current, voltage, resistance, etc.