8/11/2019 EEE Lab - Worksheet 1

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Manuel Matthew Chanco V EEE1 lab WFUS2

Renz Marion Dela Cruz

Experiment 1

1.

The resistance of the ammeter may interfere with the reading of the current by giving a reading

lower than that of the actual. This is because aside from the resistor connected to the circuit, theresistance of the ammeter adds to the resistance of the resistor since the ammeter is

connected in series with the resistor. The result can also be seen in the following mathematical

relationship. Let Ii be the theoretical current and If be the actual current, and resistance in

reality is always positive.

2. To make an ammeter of bigger magnitude that the one in the measurements, one must insert a

resistor across the ammeter to form a parallel circuit. For example, an ammeter of current at

full scale (1 mA) and a resistance of 100 ohms must be adjusted to a current of 1 A. The voltage

is 0.1 V, so the overall resistance must be 0.1 ohms. To compute for the added resistor,

1/0.1 = 1/100 + 1/R

9.99 = 1/R

R=0.1 ohms

3.

The constructed 10V-voltmeter is supposed to be linear because the relationship between the

current and the voltage (Ohms Law) is linear.The constructed voltmeter shows that a 100%

deflection of the ammeter reading, which means that the current that passed through is equal to

1 mA, is equivalent to a voltage of 10V. A 20% deflection would mean that the voltage is equal to

20% of 10V, or 2V. By using Ohms Law and knowing that the resistance of the potentiometer is

equal to 10000 ohms and also the current to be equal to 0.2mA, it is verified that the voltage isreally equal to 2V.

Our constructed voltmeter shows an average error of 5.42%. Major sources of error would be

instrumental errors such as internal resistance of the ammeter, inaccurate display of voltage

source, and losses from wires and connections.

4.

The voltmeter has an internal resistance; therefore it has a current passing through. When it is

inserted between a resistor (if the resistor is part of a series of similar resistors), they form a

parallel circuit. This makes the resistor measurement seem lower than the theoretical value

because in a parallel circuit:

1/Rt = 1/R1 + 1/Rvoltmeter

In order to lessen the error, a voltmeter should have a big enough resistance to make the error

negligible.

5. Let Imbe the meter current or the maximum current that can be read within the gauge (which is

1ma in the setup), I be the circuit current (actual current in the circuit, R obe the resistance of

the ohmmeter, Ru be the unknown resistance, and D be the deflection

8/11/2019 EEE Lab - Worksheet 1

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( )

6.

The measurements in Procedure F are more precise than procedure E. In procedure F, the

measurements are done in a multimeter, and in procedure E, the measurements are done

through a makeshift ohmmeter, using a potentiometer. The errors accounted for are the margin

of measurements of the ammeter, the internal resistance of the makeshift ohmmeter and the

potentiometer calibration.

7.

Figure 5 shows that the voltmeter measures the actual voltage across the unknown resistance

but the ammeter actually measures the current across both the voltmeter and the unknown

resistance, while Figure 6 shows that the ammeter measures the actual current across the

unknown resistance but the voltmeter the voltage across both the ammeter and the unknown

resistance. If the resistance is simply computed as V/I, then Figure 6 will generate lesser error if

the unknown resistance is small while Figure 5 will generate lesser error if the unknown

resistance is large. This is because ammeters are generally designed with small internal

resistance; hence, at large unknown resistances, the voltmeter will practically measure the

voltage only across the resistance itself and the voltage across the ammeter can be neglected

(Ohms Law). This can be seen in the following relationship.

On the other hand, voltmeters are generally designed with large internal resistance; hence at

small unknown resistances, the ammeter will practically measure the current only through the

unknown resistance and the resistance across the voltmeter can be neglected.