PHY 103 - Group.3
Dear Students,
**please note that the answers is in the second page**
Please submit your homework under the following conditions:
(Make sure you will lose marks if you miss any step from the following)
1- You have only one try and make sure for all your calculations.
2- Don’t forget the units?
3- Solve the exercise by computer and use A4 paper.
4- Make your steps very clear and add suitable comments.
5- Submit your homework in one file as PDF (only).
6- Submit by blackboard and before deadline.
Best of Luck
Dr.Huda
-----------------------------------------------------------------------------------------------------------Ex.1 Find the total current in the circuit.1?
Circuit.1
Ex.2 A rod of metal has a resistance of (0.68 Ω), length of (200m) and cross section area of
(5 𝑥 10−6 𝑚2), find the resistivity and conductivity? (add comment)
Ex.3 An Aluminum wire has a current of (1.3A) and cross-section area of (2.54 𝑥 10−6 𝑚2 ),
what is the electric field?
Ex.4 Two materials with resistivity (2.44 𝑥 10−8 Ω. 𝑚) and (0.46 Ω. 𝑚), which material
suitable for carrying current? (Discuss you answer)
Ex.5 Two capacitors, the capacitance for each (𝐶1 = 200 𝑃𝐹) and (𝐶2 = 600 𝑃𝐹) connected
in parallel were charged to (120V).
(A) Find the charge on each capacitor?
(B) The equivalent capacitance?
Ex.1 Find the total current in the circuit.1?
** To find the total current in a circuit, you need to calculate the total resistance of
the circuit and then apply Ohm's Law (𝐼=𝑉/𝑅 ), where 𝑉 is the total voltage and 𝑅
is the total resistanceTo find the total current in the circuit is: to reduce the circuits in
simple circuits (make it smaller)
First the 7 ,1 and 10 ohms are in series
We know that THE RESISTENC IN series = 7+1+10 = 18 ohms
So, 6 and 18 are IN parallel CIRCUITS. 6*8/6+18 = 4.5 ohms so its less than the old
resistance
And THE REST IS IN SERIES TOO 2 +4.5+8 = 14.5 ohms
To find the current 𝑰=V/R
20/14.5 =
1.38 Amper
END OF QUISTION 1
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Ex.2 A rod of metal has a resistance of (0.68 Ω), length of (200m) and cross section area of
(5 𝑥 10−6 𝑚2 ), find the resistivity and conductivity? (add comment)
**To find the resistivity (𝜌) and conductivity (𝜎) of the metal rod, we can use the formula
relating resistance (R), resistivity (𝜌), length (𝐿), and cross-sectional area (𝐴):**
𝑳
𝑨
From the law we can know R=𝜹 𝑨 and 𝜹 = 𝑹 𝑳
find the resistivity 𝜹 =
𝟎.𝟔𝟖 ∗𝟓∗𝟏𝟎−𝟔
= 1.7*𝟏𝟎−𝟖 Ω.m
𝟐𝟎𝟎
*** this is the resistivity***
1
For the conductivity we know that conductivity is the inverse of resistivity 𝜎 = 𝛿 = 5.88 ∗ 10−9 ( Ω. m )
−1
End of quistion 2
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Ex.3 An Aluminum wire has a current of (1.3A) and cross-section area of (2.54 𝑥 10−6 𝑚2 ),
what is the electric field?
**To find the electric field generated by the current flowing through the aluminium
wire, you can use Ohm's Law, which states that the electric field (𝐸) is equal to the
current (𝐼) divided by the cross-sectional area (𝐴) of the wire multiplied by the
material's conductivity (𝜎)**
** We find the Aluminum resistivity: 𝜹 = 𝟐. 𝟖 ∗ 𝟏𝟎−𝟖 Ω 𝒎
𝑰
𝟏.𝟑 𝑨
From I and A we can know the current density J= 𝑨 = 𝟐.𝟓𝟒 𝒙 𝟏𝟎−𝟔 𝒎𝟐 ), = 5.12∗ 𝟏𝟎𝟓 A/𝒎𝟐
From this we can know the electric field from the law that says J=
𝐄
𝛅
E= 𝜹 ∗ 𝐽 = 𝟐. 𝟖 ∗ 𝟏𝟎−𝟖 𝒙 5.12∗ 𝟏𝟎𝟓
E= 0.0144 V/m
End of question 3
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Ex.4 Two materials with resistivity (2.44 𝑥 10−8 Ω. 𝑚) and (0.46 Ω. 𝑚), which material
suitable for carrying current? (Discuss you answer)
The resistivity of a material indicates how strongly it opposes the flow of
electric current. A lower resistivity means the material offers less
opposition to the flow of current, making it more conductive.
**To determine which material is suitable for carrying current, we need to consider
their resistivities. Resistivity is a measure of a material's ability to resist the flow of
electric current. A lower resistivity indicates that the material offers less resistance to
the flow of current, making it more conductive.
Comparing the resistivities provided
1. Material A: Resistivity = 2.44×10−8 Ω⋅
2. Material B: Resistivity = 0.46 Ω⋅m
Material B has a much lower resistivity compared to Material A. This means that Material B
offers less resistance to the flow of electric current compared to Material ATherefore
Material B is more suitable for carrying current because it will allow the current to flow more
easily with less resistance. It would be a better choice for applications where low resistance
and efficient current flow are important, such as in electrical wiring or electronic circuits.
End of question 4
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Ex.5 Two capacitors, the capacitance for each (𝐶1 = 200 𝑃𝐹) and (𝐶2 = 600 𝑃𝐹) connected
in parallel were charged to (120V).
(A) Find the charge on each capacitor?
(B) The equivalent capacitance?
(A)
**When capacitors are connected in parallel, the voltage across each capacitor is the
same. Therefore, both capacitors in this are charged to 120V
To find the charge on each capacitor, we'll use the formula 𝑄=𝐶×𝑉, where Q represents the
charge stored on the capacitor, C is the capacitance of the capacitor, and V is the voltage
across the capacitor.
(since they are connected in parallel)we'll calculate the charge on each capacitor
individually.
𝑪=
𝑸
𝑽
from this equation we can find the charge which is
Q=C * V
For Capacitor 1: 𝑄1=𝐶1×𝑉=200 pF×120 V= 24×10−9 C
For Capacitor 2: 𝑄2=𝐶2×𝑉=600 pF×120 V=72×10−9 C
----------------------------------------------------------------------------------------(B)
**When capacitors are connected in parallel, the total capacitance 𝐶eq of the combination is
the sum of the individual capacitances. **
Ceq=C1+C2
Ceq=200pF+600pF
𝐶eq=800 pF