A simple DC voltmeter can be constructed by placing a resistor (R_{S}), called a multiplier, in series with the ammeter meter movement, and marking the meter face to read voltage (as shown in Figure).

Voltmeters are connected in parallel with the load (R_{L}) being measured.

Figure : Simple DC Voltmeter

When constructing a voltmeter, the resistance of the multiplier must be determined to measure the desired voltage. The Equation is a mathematical representation of the voltmeter’s multiplier resistance.

##### V = I_{m}R_{s} + I_{m}R_{m}

##### I_{m}R_{s }= V – I_{m}R_{m}

where

V = voltage range desired

I_{m }= meter current

R_{m} = meter resistance

R_{s} = multiplier resistance or series resistance

**Example:**

**A 2 mA meter movement with internal resistance of 25 ohms is to be constructed as a voltmeter. What value must the series resistance be to measure full scale voltage of 100 volts?**

Solution :

Since R_{m} is negligibly low, then

R_{s} = V/I_{m}

R_{s} = 100 / (2×10^{-3})

R_{s} = 50kΩ

When a voltmeter is connected in a circuit, the voltmeter will draw current from that circuit. This current causes a voltage drop across the resistance of the meter, which is subtracted from the voltage being measured by the meter. This reduction in voltage is known as the loading effect and can have a serious effect on measurement accuracy, especially for low current circuits.

The accuracy of a voltmeter (K_{v} ) is defined as the ratio of measured voltage when the meter is in the circuit (V_{w}) to the voltage measured with the meter out of the circuit.

The below Equation is a mathematical representation of the accuracy of a voltmeter, or true voltage (V_{o}).

Meter accuracy can also be determined by comparing the relationship between the input and circuit resistances using Ohm’s Law as described below.

where

I_{m} = meter current

V_{o} = true voltage

R_{o} = circuit resistance

R_{in} = input resistance of the voltmeter

V_{w} = indicated voltage

K_{v} = meter accuracy

**Example:**

**A voltmeter in the 100 volt range with a sensitivity of 40 KΩ/V is to measure the voltage across terminals ab.**

Find :

- V
_{o}, true voltage - V
_{w}, indicated voltage - K
_{v}, meter accuracy

Solution :

**Find V _{o} , true voltage**

V_{o} = { 100 KΩ / (100 KΩ + 100 KΩ) } x 220 volts

V_{o} = 110 volts

**Find V _{w} , indicated voltage**

R_{o }= (100 x 100) / (100 + 100) = 50KΩ

R_{in }= S V = 40KΩ x 100 = 4.4 MΩ

V_{w }= R_{in }/ (R_{o + Rin )}

V_{w }= { 4.4MΩ / (50KΩ + 4.4MΩ) } 110 volts

V_{w }= 108.9 volts

**Find K _{v} , meter accuracy**

K_{v} = V_{w }/ V_{o}

K_{v} = 108.9/110

K_{v} = 0.99 or 99%

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