HARDNESS, FRACTURE TOUGHNESS AND STRENGTH OF CERAMICS
Verification of ohm's law in parallel circuit
1. Introduction
Aim
|page 1 of 7
To verify the ohm’s law and hence determine the
unknown resistance of the given material of the
wire.
In beginning ,we were
introduced multimeter and
resistors. We connected
resistor in parallel with
ammeter and changed the
Voltage supplied and took
note of current in the
circuit.
3. Procedure
1-Disconnect the circuit from Ground
terminal.
2-using the voltmeter connected across the
power supply change the Voltage according
to the values in the table.
3-Connect ground to circuit.
4-Connect the ammeter in parallel with the
resistor and measure the current.
5-We Repeated these steps again until we
had the required values in the table.
|page 3 of 7
5. |page 7 of 7
Discussion
The purpose of experimentwas to verify ohm’s
law, which states that the potential difference
across a conductor and the current through it are
directly proportional. The constant R represents
the opposition to a flow of electrical charges in a
conductor. A conductor has a resistance of one
ohm if a potential difference of one volt across the
conductor results in a current of one ampere in the
conductor. Furthermore, cross section area makes
a difference. If something has a bigger cross
section area and equal length to something with a
smaller cross section area, then it is bound to have
less resistance than a rod of smaller area and
same length, because there is more area for the
flow of electrons and therefore less resistance in
the object ) it should be mentioned that this is only
true if temperature is kept constant, as was the
case in this lab ( To measure the average
resistance in the wire keeping length constant, we
used the formula R = V/I (taking the formula V = IR
and manipulating it algebraically(; the potential
difference across a conductor and the current sent
through it are directly proportional.
6. Calculation
|page 5 of 7
𝑉𝑇 =3.05 v When 𝑉𝑆=3
𝐼𝑇 = 18.50 mA
𝑅 = 𝑉/𝐼 𝑅 =
3.05𝑣
18.50𝑚𝐴
= 164.86Ω
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𝑉𝑇 =4.07 v When 𝑉𝑆=4
𝐼𝑇 = 24.50 mA
𝑅 = 𝑉/𝐼 𝑅 =
4.07𝑣
24.50𝑚𝐴
= 163.96Ω
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𝑉𝑇 = 5 v When 𝑉𝑆=5
𝐼𝑇 = 30.4 mA
𝑅 = 𝑉/𝐼 𝑅 =
5𝑣
30.4𝑚𝐴
= 164.47Ω
7. Calculation
|page 4 of 7
Theoretical resistance
𝑹𝑻 =𝑹𝟏+ 𝑹𝟐
When 𝑹𝟏=200Ω and 𝑹𝟐=1000Ω
So 𝑹𝑻 = 200Ω + 1000Ω =1200Ω
Practical resistance
𝑉𝑇 =1.49 v When 𝑉𝑆 =1
𝐼𝑇 =6.35 mA
𝑅 = 𝑉/𝐼 𝑅 =
1.49𝑣
6.35𝑚𝐴
= 164.88Ω
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𝑉𝑇 =2.03 v When 𝑉𝑆 =2
𝐼𝑇 =12.30 mA
𝑅 = 𝑉/𝐼 𝑅 =
2.03𝑣
12.30𝑚𝐴
= 165.04Ω
9. Name: Ahmed bahri omar
Experimental number ( 1 )
Date: 9 / 3 / 2021
Subject: Electricity and
electronics
Verificationof ohm's
law in parallel circuit
Supervisor: Ms issa Mahdi