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Voltmeter Loading Effect
When a voltmeter is used to measure the voltage across a circuit component, the voltmeter is connected in parallel with the component
If the resistance of voltmeter is not high as compared to resistance of the circuit across which is connected, the measured voltage will be less than the actual
This called the loading effect because the voltmeter loads lower the effective resistance of the circuit and change the voltage reading
The voltmeter loading can be reduced by using a high sensitivity voltmeter
The parallel combination of two resistor will reduce the total equivalent resistance.
The voltage across the component is less whenever the voltmeter is connected.
The effect is called the loading effect of an instrument.
EXAMPLE 2.5.1
Two different voltmeters are used to measure the voltage across Rb in the circuit. The meters are as follows:
Meter 1: S =1 k/V, Rm = 0.2k, range 10V
Meter 2: S = 20 k/V, Rm = 1.5k , range 10 V
Calculate:
i. Voltage across Rb without any meter across it
ii. Voltage across Rb when meter 1 is used
iii. Voltage across Rb when meter 2 is used
iv. Error in the voltmeters
SOLUTION
I. Voltage across Rb without any meter across it
ππ π = 5(30)//5π+25π = 5π
II. Voltage across Rb when meter 1 is used
π ππ1 = ππ1ππππππ = 1π 10 = 10 πΞ©
π πππ1 = 10π // 5k = 3.333 KΟ
ππ π = 3.333(30)//3.333π+25π = 3.526V
III. Voltage across Rb when meter 2 is used
π ππ2 = ππ2ππππππ = 20π 10 = 200 πΞ©
π πππ1 = 200π // 5k = 4.878 kΞ©
ππ π = 4.878(30)//4.878π+25π = 4.879 V
IV. Error in the voltmeters
%ππ1 = 5 β 3.526 5 Γ 100% = 29.47%,
%ππ1 = 5 β 4.879 5 Γ 100% = 2.06%
Ammeter Insertion Effect
Inserting an ammeter in a circuit always increase the resistance of the circuit, therefore, always reduces the current in the circuit.
The error caused by ammeter in a circuit to obtain current reading called ammeter insertion effect.
The error caused by ammeter depends on relationship between value of resistance in the original circuit (Iy) and value of resistance in the ammeter (Ix)
Inserting Ammeter in a circuit always increases the resistance of the circuit and, thus always reduces the current in the circuit.
πΌπ¦ = π// π 1
Placing the meter in series with R1 causes the current to reduce to a value equal to:
πΌπ₯ = V/(π 1 + π π)
% error = |(actual-measurement)/actual| x100
EXAMPLE 2.5.2
Figure shows a simple series circuit of Ra and Rb connected to a 100Vdc source. If the current across Rb is measured by an ammeter having Rm = 50β¦, calculate the insertion error of the meter.
SOLUTION
πΌπ = 100 10π = 0.01 π΄
πΌπ = 100 10π + 50 = 9.95 ππ΄
%π = 0.01 β 9.95 π 0.01 Γ 100% = 0.5%
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