Voltmeter

Voltmeter

A voltmeter is a fundamental device used to measure the voltage, or electrical potential difference, between two points in an electrical circuit. Just as a thermometer measures temperature or a speedometer measures speed, a voltmeter provides a direct reading of the "pressure" that drives electric current through a circuit.

Convert the galvanometer into a voltmeter by connecting a large series of resistances.

The equivalent resistance of the combination is $G+R$

The required value of high resistance to be connected in series with the galvanometer is $R=\frac{V}{I_g}-G=\left[\frac{V}{V_g}-1\right] G$. V is the total voltage applied across the circuit and $V_g$ is the total voltage drop across the galvanometer.

if $n^{t h}$ part of the applied voltage across the galvanometer, that is $V_g=V / n$, then $R=(n-1) G$.

Recommended Topic Video

Solved Examples Based on Voltmeter

Example 1: A galvanometer has a resistance of $30 \Omega$ and a current of 2mA is needed for a full-scale deflection. The resistance $($ in $\Omega)$ that should be used to convert it into a voltmeter of 0.2V range is

1) 70

2) 60

3) 80

4) 90

Solution:

Conversion of galvanometer into voltameter

Connected a large Resistance R in a series

wherein

To convert galvanometer into voltmeter of 0.2V range

$
\begin{aligned}
& V=I_g \cdot(R+G) \\
& \Rightarrow 0.2=2 * 10^{-3} *(30+R)
\end{aligned}
$
or $R=70 \Omega$

Hence, the answer is the option (1).

Example 2: A moving coil galvanometer allows a full-scale current of $10^{-2}$ A. A series resistance of $200 \Omega$ is required to convert the above galvanometer into a voltmeter of range0-5 V . Therefore the value of shunt resistance $(\operatorname{in} \Omega)$ required to convert the above galvanometer into an ammeter of range0-10 mA is :

1) 300

2) 100

3) 200

4) 10

Solution:

Conversion of galvanometer into voltameter

Connected a large Resistance in a series

wherein

$\begin{aligned} & (\mathrm{R}+\mathrm{G}) \mathrm{I}=\mathrm{V} \\ & \left(2 \times 10^2+G\right) 10^{-2}=V=5 \\ & =2+10^{-2} G=5 \\ & \mathrm{G}=300 \Omega\end{aligned}$

Hence, the answer is the option (1).

Example 3: A moving coil galvanometer has resistance $50 \Omega$ and it indicates full deflection at $4 m A$ current. A voltmeter is made using this galvanometer and a $5 k \Omega$ resistance. The maximum voltage (in V), that can be measured using this voltmeter, will be close to :

1) 20.2

2) 15

3) 40

4) 10

Solution:

Conversion of galvanometer into voltameter

Connected a large Resistance R in a series

wherein

Given $G=50 \Omega$
$
i_g=4 m A
$

Series resistance $\quad S=5 k \Omega$
$
\begin{aligned}
V & =i_g(G+S) \\
& =4 \times 10^{-3}(5000+50) \\
& =20.2 \mathrm{~V}
\end{aligned}
$

Hence, the answer is the option (1).

Example 4: A 100V voltmeter having an internal resistance of $20 \mathrm{k} \Omega$.. 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

1) 420

2) 210

3) 110

4) 630

Solution:

Equivalent resistance

$\begin{aligned} & V_R=110-5=105 \mathrm{~V} \\ & \frac{V_R}{V_1}=\frac{R}{R_1} \text { i.e } \frac{105}{5}=\frac{R}{20 k \Omega} \\ & R=420 k \Omega\end{aligned}$

Hence, the answer is the option (1).

Example 5: Two resistors $400 \Omega$ and $800 \Omega$ are connected in series across a 6 V battery. The potential difference measured by a voltmeter of $10 k \Omega$ across $400 k \Omega$resistor is close to :

1) 2 V
2) 1.8 V
3) 2.05 V
4) 1.95 V

Solution:

From the question

The circuit diagram is drawn as shown below

$\begin{aligned} & i=\frac{6}{800+\left(\frac{40 \times 100000}{400+100000}\right)} \\ & i=\frac{6}{1184.61}=0.00506 \\ & V_{A B}=6-(800 \times 0.00506)=6-4.05=1.95\end{aligned}$

Hence, the answer is the option (4).

Summary

A voltmeter is a crucial device for measuring the voltage across components in an electrical circuit, providing insights into the electrical potential differences that drive current. It is often created by converting a galvanometer through the addition of a high resistance in series. This allows it to measure voltage accurately without significantly altering the circuit it is testing. Understanding how to convert a galvanometer into a voltmeter and solve related problems is essential for diagnosing and analyzing electrical circuits effectively.