
Section 1
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Section 2
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Section 3
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Section 4
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Section 5
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Section 6
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Section 7
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Section 8
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Section 9
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Lesson 1
Basically here we are starting with electrostatics.Which means we are looking at forces between two or more charges.
Just like a body have mass ‘m” , it also have a charge ‘Q”
Electron have negative charge and Proton have positive charge.
Charge comes as a result of transfer of electrons from one body to another.
Electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.Field helps in pumping energy..energy flow is actually causing a bulb to glow.
Question:1
Define the terms : Electric flux , Electric flux density and electric charge density?
Answer:
Electric flux is the rate of flow of the electric field through a given area.
Product of electric field intensity (E) and area of surface ( A)gives the value of electric flux (ϕ)
ϕ = E.A
Electric Flux Density is the amount of electric flux, the number of “lines,” passing through a given area.If flux is passing per unit surface of area A.then ϕ / A is called electric flux density
Charge density = Total charge / Volume
There are different dimensions to charge density.For example it can be linear,surface or volume
linear charge density = charge /length
surface charge density = charge /surface
volume charge density = charge /volume
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Now when the same electric charge flows through a conducting medium ,we say there is current
To have current,a potential difference is needed because charge flows from a higher potential to lower potential.
Charge / time = current
q/t = I
Current in one direction is DC and current in both direction is AC
But in a circuit when current is passed,we say it comes from a “source” to do work on a “load”.
Since it pass through a conductor in a circuit, it have reistance to the flow.The opposite to resistance is conductance.There maybe a switch in the circuit which opens and closes the circuit.The dependency of current,voltage,resitance are explained by ohms law which says
Voltage = current x resistance
V= I*R
Power = voltage x current
P= V * I
Refer the video for next question answer
Question :2
Find currents in all branches of the circuit in figure using Mesh current analysis
Answer:
Let us redraw the picture and make it more clear
So this is the mesh current analysis circuit
By convention, electron flow is from negative to positive and current flow from positive to negative.
A mesh current is the current passing through elements which are not shared by other loops.For example the current of R1 is I1 , current of R2 is I2.But you can see that there are elements shared by two meshes such as R3..Current of such elements is the algebraic sum of both meshes .So current of R3 is (I1+I2).You need to check the direction of current to finally find the algebraic sum.
Now, lets write the equation for mesh of I1,I2 and I3
A mesh equation is a KVL equation using mesh currents.
We start from a point and calculate the algebraic sum of voltage drops around the loop
Whereever I3 comes we use I1+I2
For resistors the voltage drop equals to the resistance multiplied b mesh currents
Let us apply KCL and KVL laws
Kirchhoffs Voltage Law or KVL, states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop”
Using Kirchhoffs Current Law, KCL the equations are given as:
At node A : I1 + I2 = I3
At node B : I3 = I1 + I2
Using Kirchhoffs Voltage Law, KVL the equations are given as
Loop 1 is given as : V1 = R1 I1 + R3 I3
12 = 2 I1 + 1 I3
Loop 2 is given as : V2 = R2 I2 + R3 I3
6 = 3 I2 + 1 I3
Loop 3 is given as : V3= V1V2
12 – 6 = 12I1 – 3 I2
As I3 is the sum of I1 + I2 we can rewrite the equations as;
Eq. No 1 : 12 = 2I1 + 1(I1 + I2) = 3I1 + I2
Eq. No 2 : 6 = 3I2 + (I1 + I2) = 4I1 + I2
We now have two “Simultaneous Equations” that can be reduced to give us the values of I1 and I2
3I1 +I2 =12
4I1 +I2 =6
Substitution of I1 in terms of I2 gives us the value of I1 as 6 Amps
Substitution of I2 in terms of I1 gives us the value of I2 as 18 Amps
As : I3 = I1 + I2 = 12 amp
The current flowing in resistor R3 is given as : 6 – 18 = 12 Amps
and the voltage across the resistor R3 is given as : I3R3 = 12 x 1 = 12 volts
Question :3
State and explain Coulumbs law of Electrostatics.
Answer:
There is always a force between two charges even when two charges are at rest.This force is the electrostatic force.
The electrostatic force between charges increases when the magnitude of the charges increases or the distance between the charges decreases.
Coulomb found that the magnitude of the electrostatic force between two pointlike charges is inversely proportional to the square of the distance between the charges. He also discovered that the magnitude of the force is proportional to the product of the charges.
That is:
F∝Q1*Q2/r^2
Coulomb’s law
Coulomb’s law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
F=k*Q1*Q2/r^2,
The proportionality constant k is called the electrostatic constant and has the value:
9,0×10^9 N⋅m^2⋅C^−2
in free space.