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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.
Breadboard
DC power supply
Digital Multimeter
Resistors
Apparatus
1 2
3 4
|page 2 of 7
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
Calculation
|page 6 of 7
|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.
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
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
𝑉𝑇 =4.07 v When 𝑉𝑆=4
𝐼𝑇 = 24.50 mA
𝑅 = 𝑉/𝐼 𝑅 =
4.07𝑣
24.50𝑚𝐴
= 163.96Ω
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
𝑉𝑇 = 5 v When 𝑉𝑆=5
𝐼𝑇 = 30.4 mA
𝑅 = 𝑉/𝐼 𝑅 =
5𝑣
30.4𝑚𝐴
= 164.47Ω
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
‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
‫ـــــــــــــــــــــــــــــ‬
𝑉𝑇 =2.03 v When 𝑉𝑆 =2
𝐼𝑇 =12.30 mA
𝑅 = 𝑉/𝐼 𝑅 =
2.03𝑣
12.30𝑚𝐴
= 165.04Ω
Introduction
Aim
Apparatus
Procedure
Calculation
Discussion
1
2
3
4
5
6
Contents
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

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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.
  • 2. Breadboard DC power supply Digital Multimeter Resistors Apparatus 1 2 3 4 |page 2 of 7
  • 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 ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 𝑉𝑇 =4.07 v When 𝑉𝑆=4 𝐼𝑇 = 24.50 mA 𝑅 = 𝑉/𝐼 𝑅 = 4.07𝑣 24.50𝑚𝐴 = 163.96Ω ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ 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 ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬ ‫ـــــــــــــــــــــــــــــ‬ 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