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Voltmeter is considered one of the most asked concept.
21 Questions around this concept.
A galvanometer of resistance $100 \Omega$ has 50 divisions on its scale and has sensitivity of $20 \mu A /$ division. It is to be converted to a voltmeter with three ranges, of $0-2 \mathrm{~V}, 0-10 \mathrm{~V}$ and $0-20 \mathrm{~V}$. The appropriate circuit to do so is :
A voltmeter having a resistance of $1800 \Omega$ is employed to measure the potential difference across $200 \Omega$ resistance which is connected to dc power supply of 50 V and internal resistance $20 \Omega$. What is the approximate percent change in the P.D across $200 \Omega$ resistance as a result of connecting the voltmeter across it?

A galvanometer has a resistance of $30 \Omega$, and a current of 2 mA is needed for a full scale deflection. The resistance (in $\Omega$ ) that should be used to convert it into a voltmeter in the 0.2 V range is
A 100V voltmeter has an internal resistance of 20k$\Omega$.. is connected in series with a large resistance R across a 110V line. If the voltmeter reads 5V then the magnitude of R (in $k \Omega$ ) is
A galvanometer , whose resistance is 50 ohm , has 25 divisions in it. When a current of 4 X 10-4 A passes through it , its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V , it should be connected to a resistance (in ohm) of :
The actual value of resistance $R$, shown in the figure is $30 \Omega$. This is measured in an experiment as shown using the standard formula $R=\frac{V}{I}$, Where } V and } I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is $5 \%$ less, then the internal resistance (in $\Omega$ ) of the voltmeter is: (Round the answer to the nearest integer)
An ideal voltmeter V is connected to a $20 \Omega$ resistor and a battery with emf 10V and internal resistance $0.5 \Omega$ as in the figure.

The ratio of current in $20 \Omega$ resistor with the voltage of the battery
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In the given circuit, the voltmeter and the electric cell are ideal. Find the reading of the voltmeter
A constant 60 V d.c. supply is connected across two resistors of resistance $400 \mathrm{k} \Omega$ and $200 \mathrm{k} \Omega$. What is the reading of the voltmeter, also of resistance $200 \mathrm{k} \Omega$, when connected across the second resistor as shown in Figure?
An ammeter A of finite resistance and a resistor R are joined in series to an ideal cell C. A Potentiometer P is joined in parallel to R. The ammeter reading is $\mathrm{I}_0$ and the potentiometer reading is $\mathrm{V}_0$. P is now replaced by a voltmeter of finite resistance. The ammeter reading now is I and the voltmeter reading is V. Then
Voltmeter
It is a device used to measure Potential difference and is always put in parallel with the circuit element whose voltage is to be measured

Conversion of galvanometer into voltmeter: by Connecting a large Resistance R in series.

- The equivalent resistance of the combination is $G+R$
- The required value of high resistance to be connected in series with the galvanometer is
$
R=\frac{V}{I_g}-G=\left[\frac{V}{V_g}-1\right] G_{. \mathrm{V} \text { is the total voltage applied across the circuit and }} V_g \text { is the total }
$
voltage drop across the galvanometer.
- if $n^{\text {th }}$ part of the applied voltage across galvanometer, that is $V_g=V / n$, then $R=(n-1) G$.
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