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SANJAY PANDEY
1. Electric current
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4. Electric Current in Conductors
"Electric current is defined as the amount of electric charge passing through a cross section of a conductor in unit time." In other words
Electric Current".
Mathematically
"The rate of flow of electric charge through a cross section of a conductor is called
ππ = βππ/βπ‘π‘
π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄ πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ ππππππππππππππ = ππππππππππππππππ ππβππππππππ / π‘π‘π‘π‘π‘π‘π‘π‘
Q is charge, t is time
ππβππ ππππππππππππππ ππππ π‘π‘π‘π‘π‘π‘π‘π‘ π‘π‘ = ππ = limΞπ‘π‘β0 ( βππ/βπ‘π‘) = ππππ/ππππ
Electric current is a scalar quantity. β’
β’
Unit: AMPERE. Ampere
1 ππππππππππππ = 1 ππππππππππππππ / 1 π π π π π π
In S.I system unit of electric current is ampere.
Ampere is defined as:
βCurrent through a conductor will be 1 ampere if one coulomb of electric charge passes through any cross section of conductor in 1 second.β β’
Types of current
1 ππππππππππππ = 1 ππππππππππππππ / 1 π π π π π π
There are two types of current. 1. 2.
Electronic Current
Electronic current flows from negative to positive terminal. Conventional Current
Direction of conventional current is taken from higher potential to the lower potential.
2.
Current density
Average current density ππ = Ξππ/Ξππ
ππ is current and A is area of the conductor
The current density at a point P is
2 ππ = limβπ‘π‘β0 (Ξππ/Ξππ) = ππππ/ππππ
If current i is uniformly distributed over an area S and is perpendicular to it, then ππ = ππ/ππ
For a finite area
ππ = β« ππβ. οΏ½οΏ½οΏ½οΏ½β ππππ
Where
ππβ = density of current (vector) οΏ½οΏ½οΏ½οΏ½β ππππ = area (vector) 3.
Drift speed
A conductor contains free electrons moving randomly in a lattice of positive ions. Electrons collide with positive ions and their direction changes randomly. In such a random movement, from any area equal numbers of electrons go in opposite directions and due to that no net charge moves and there is no current. But when there is an electric field inside the conductor a force acts on each electron in the direction opposite to the field. The electrons get biased in their random motion in favor of the force. As a result electrons drift slowly in the direction opposite to the field.
If ππ be the average time between successive collisions(, the distance drifted during this period is
The drift speed is
ππ =
1 1 ππ(ππ)2 = (ππππ/ππ)( ππ) Β² 2 2
π£π£ππ =
ππ 1 = (ππππ/ππ)ππ = ππππ ππ 2
ππ the average time between successive collisions, is constant for a given material at a given temperature. It is called the Relaxation time 4.
Relaxation Time
Suppose ππ π‘π‘β electron suffered its last collision π‘π‘ππ time ago, then relaxation time
3
5.
6.
βπ‘π‘ ππ
ππ =
ππ
Relation between current density and drift speed
ππ = ππ/π΄π΄ = ππππ π£π£ππ
Ohm's law
It states that the current density in a conductor is directly proportional to electric field across the conductor. ππβ β πΈπΈοΏ½β β ππβ = πππΈπΈοΏ½β
Proof:
πΈπΈοΏ½β is field and ππ is electrical conductivity of the material. We know that
And β΄
Or
β΄
β
or
ππ = ππππ οΏ½
ππ = ππ/π΄π΄ = ππππ π£π£ππ 1 2
π£π£ππ =
1 2
(ππππ/ππ)πποΏ½ =
(ππππ/ππ)ππ
ππππ 2 ππ 2ππ
πΈπΈ
ππ = ππππ (proved)
The resistivity of the material is defined as and
ππ =
ππ
π΄π΄
1
1 ππ
= πΈπΈ = . ππ
ππ
ππ
ππ =
1
ππ
ππ
π΄π΄
ππ
ππ = ππ ππ π΄π΄
ππ = π
π
π
π
(other form of Ohmβs law)
4 where
π
π
= ππ
ππ
π΄π΄
is called the resistance of the given conductor. The quantity 1/R is called the conductance.
Unit of resistivity Ο is ohm-meter (or β¦-m). The unit of conductivity Ο is (ππβππ β ππ)β1 written as ππβππ/ππ.
β’
Resistivity of a material ππ = 1/ππ Another form of Ohm's law
ππ = π£π£π£π£π£π£π£π£π£π£π£π£π£π£ ππππππππππππππππππππ ππππππππππππππ π‘π‘βππ ππππππππ ππππ ππ ππππππππππππππππππ = πΈπΈπΈπΈ (ππ = length of the conductor) ππ = π
π
π
π
π
π
= ππππππππππππππππππππ ππππ π‘π‘βππ ππππππππππππππππππ = ππ Γ
1/π
π
is called conductance 7.
ππ
π΄π΄
Temperature dependence of resistivity
As temperature of a resistor increases its resistance increases. The relation can be expressed as π
π
(ππ) = π
π
(ππ0 )[1 + πΌπΌ(ππ β ππ0 )]
πΌπΌ is called temperature coefficient of resistivity. 8. 9.
Thermistors: Measure small changes in temperatures Superconductors:
For these materials resistivity suddenly drops to zero below a certain temperature. For Mercury it is 4.2 K. For the super conducting material if an emf is applied the current will exist for long periods of time even for years without any further application of emf. Scientists have achieved superconductivity at 125 K so far. 10. Battery
5 Battery is a device which maintains a potential difference between its two terminals A and B.
In the battery some internal mechanism exerts forces on the charges and drives the positive charges of the battery towards one side (terminal A) and negative charges of towards another side (terminal B).
Let force on a positive charge ππ is πΉπΉβππ (a vector quantity). As positive charge accumulates on A and negative charge on B, a potential difference develops and grows between A and B. An electric field πΈπΈοΏ½β is
developed in the battery material from A to B and exerts a force πΉπΉβππ = πππΈπΈοΏ½β on a charge ππ. The direction of this force is opposite to πΉπΉβππ . In steady state, the charge accumulation on A and B is such that πΉπΉππ = πΉπΉππ . No further accumulation takes place. If a charge ππ is moved from one terminal (say B) to the other terminal say A, the work done by the battery force is ππ = πΉπΉππ ππ where ππ is distance between A and B. The work done by the battery force per unit charge is E = ππ/ππ = πΉπΉππ ππ /ππ
This E is called the emf of the battery. Please note that emf is not a force it is π€π€π€π€π€π€π€π€ ππππππππ/ππβππππππππ. If nothing is externally connected between A and B, then πΉπΉππ = πΉπΉππ = ππππ
Or
πΉπΉππ ππ = ππππππ = ππππ
As
ππ = ππ/ππ = πΉπΉππ ππ/ππ = ππππ/ππ = ππ
(because ππ = πΈπΈπΈπΈ)
ππ = potential difference between the terminals Therefore
ππ = ππ
Thus, the emf of a battery equals the potential difference between its terminals when the terminals are not connected externally. 11. Energy transfer in an electric circuit
When an electric charge ππ = ππππ goes through the circuit having resistance R the electric potential energy decreases by ππ = ππππ = (ππππ)(ππππ) = ππ 2 π
π
π
π
This loss in electric potential energy appears as increased thermal energy of the resistor. Thus a current ππ for a time π‘π‘ through a resistance π
π
increases the thermal energy by ππ 2 π
π
π
π
ππππππππππ ππππππππππππππππππ = ππ = ππ/π‘π‘ = ππΒ²π
π
= ππππ
12. Effect of internal resistance of a battery
Internal resistance of a battery is due mainly to the resistance of the electrolyte between electrodes. It is denoted by ππ and for ideal battery ππ = 0.
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Since the internal resistance of the battery is in series with the load the equivalent resistance of the circuit is π
π
ππππ = π
π
+ ππ. The current is thus reduced owing to the internal resistance, ππ = β / (ππ + π
π
), from what it would be in its absence. The potential difference across the load, equivalent to that across the battery, is less than the full emf of the battery because of the voltage drop across the internal resistance. β° = ππππ + ππππ ππππππππππ = ππππ = β° β ππππ
ππππππππππππππππππ ππππππππππππππππππππ ππππππππππππππ ππππππππππππ ππππ ππππππππππππππππ ππππππππππππππππ
= ππππππ ππππ π‘π‘βππ ππππππππππππππ β ππππ
(ππ = internal resistance)
The so-called terminal voltage of a battery is lower than the emf when it is discharging because of the voltage drop across the internal resistance. If, on the other hand, the battery is being charged by an external source such as a recharger, the current will be forced through the battery in the opposite direction; the terminal voltage will then be higher than the emf by the amount of the voltage difference across its internal resistance.
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β’
Polarity of a Resistor Assign a positive (+) sign for the voltage to the terminal of the element where the current enters and negative (β) sign to the terminal of the element where the current leaves it. 13. Kirchhoff's laws: Kirchhoffβs current law (KCL) or the junction law
The sum of all currents directed towards a point in a circuit is equal to the sum of all the currents directed away from the point.
Kirchhoffβs voltage law (KVL) or the loop law
The algebraic sum of all the potential differences along a closed loop in a circuit is zero.
β’
Strategy for multi-loop circuits:
1. 2. 3. 4. 5.
Sketch circuit. Replace resistor combinations with their equivalents. Label the positive direction of current in each branch of the circuit. Apply the junction rule. Apply the loop rule.
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β’ (a) (b) (c)
(e) (f) (g)
Writing The KVL Equations Pick a starting point on the loop you want to write KVL for. Imagine walking around the loop - clockwise or counterclockwise. When you enter an element there will be a voltage defined across that element. One end will be positive and the other negative. Pick the sign of the voltage definition on the end of the element that you enter. Conversely, you could choose the sign of the end you leave, except that you have to be consistent all the way around the loop. Write down the voltage across the element using the sign you got in the previous step. Keep doing that until you have gone completely around the loop returning to your starting point. Set your result equal to zero.
β’
Now, let's write KVL for each of the three loops in adjoining Fig.
β’
βπππ΅π΅ + ππ1 + ππ2 = 0
(d)
β’
For the first loop (Battery, Element 1, Element 2)
For the second loop (Element 2, Element 3, Element 4). Note, you have to be careful with this one because you might not expect the voltage across Element 3 to be defined the way it is. βππ2 β ππ3 + ππ4 = 0
For the third loop (Battery, Element 1, Element 3, Element 4) So, we get three equations - right?
βπππ΅π΅ + ππ1 β ππ3 + ππ4 = 0
Actually, that's not right, because we do not get three independent equations. There are only two independent equations we can write.
That's not immediately obvious, so write the three equations as shown below. We'll put a horizontal line between the first two and the third equation. βπππ΅π΅ + ππ1 + ππ2 = 0
βππ2 β ππ3 + ππ4 = 0 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. βπππ΅π΅ + ππ1 β ππ3 + ππ4 = 0 Can you see that you can add the first two equations to get the third? (Actually, there is a βππ2 and a +ππ2 , and those are the only things that cancel out when you add.) The third equation can be obtained from the first two equations, so it is not an independent equation. When you have the first two equations you can get the third from them!
9 What this means is that you have to be careful when you write KVL. You can write too many equations, and in being careful you might not write enough. Fortunately, if you look at a circuit you can almost always see how many independent loops there are by inspection. 14. Combination of resistors in series
For resistors in series, the current through each resistor is identical. πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ ππππππππππππππππππππ = π
π
β + π
π
β + π
π
β+. ..
15. Combination of resistors in parallel
For resistors in parallel, the voltage drop across each resistor is identical πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ ππππππππππππππππππππ = 1/π
π
β + 1/π
π
β + 1/π
π
β+. ..
16. Division of current in resistors joined in parallel ππβ/ππβ = π
π
β/π
π
β
ππβ = ππππβ/(π
π
β + π
π
β)
17. Batteries connected in series Where π
π
= external resistance
ππ = (ππβ + ππβ)/(π
π
+ ππβ)
ππβ = ππβ + ππβ
ππβ, ππβ are internal resistances of two batteries
18. Batteries connected in parallel πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ ππππππ = ππβ =
ππβππβ+ππ 2 ππ1 (ππβ+ππβ)
where ππβ, ππβ are emfs of of batteries , and ππβ, ππβ are internal resistances. So
ππππππππππππππππππππ ππππππππππππππππ ππππππππππππππππ = ππβ = ππβππβ/(ππβ + ππβ) ππ ππππ π‘π‘βππ ππππππππππππππππ = ππβ/(π
π
+ ππβ)
10 19. Wheatstone bridge
It is an arrangement of four resistances, and one of them can be measured if the other are known resistances. Relation Among Resistances In Balanced Condition
Rβ/Rβ = Rβ/Rβ
From figure:
Rβ & Rβ are connected in series.
Reason: (only one path for the flow of current)
Rβ & Rβ are connected in parallel.
Reason: (two paths for the flow of current)
Rβ & Rβ are connected in series.
Rβ & Rβ are connected in parallel.
If the there is no deflection in the galvanometer, then π
π
β/π
π
β = π
π
β/π
π
4
20. Ammeter
β π
π
4 = π
π
βπ
π
β/π
π
β
Used to measure current in a circuit. A small resistance is connected in parallel to the coil measuring current in an ammeter to reduce the overall resistance of ammeter. 21. Voltmeter
A resistor with a large resistance is connected in series with the coil.
When a volt meter is connected in parallel to the point between which the potential is to be measured, if a large resistance is connected, the equivalent resistance is less than the small resistance. 22.
Charging of the capacitor
It takes time to charge a capacitor and it takes time This time is dependent on the sizes of the capacitor and the resistor in the circuit. The case for charging a capacitor is described first, then discharging a capacitor.
to
discharge
one.
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Figure shows charging of a capacitor. Just before the switch is closed the charge on the capacitor is zero. When the switch is closed (at time t= 0), the charging starts. By KVL
πΈπΈ β πππ
π
β πππΆπΆ = 0 ππ
β β β β β β β β
πΈπΈ β ππππ β = 0 π
π
πΆπΆ
ππ
ππππ = πΈπΈ β ππππ ππππ
=
ππππ
πΈπΈπΈπΈβππ ππ
β«0
πΈπΈπΈπΈβππ
=
πΆπΆ
ππππ
πΆπΆπΆπΆ
ππππ
πΈπΈπΈπΈβππ
βππππ 1β
πΈπΈπΈπΈ
ππ
π‘π‘ ππππ
= β«0
πΈπΈπΈπΈβππ
πΈπΈπΈπΈ
ππ
=
πΆπΆπΆπΆ
π‘π‘
πΆπΆπΆπΆ π‘π‘
= ππ βπΆπΆπΆπΆ
ππ = πΈπΈ πΆπΆ(1 β ππ βπ‘π‘/πΆπΆπΆπΆ )
ππ is charge on the capacitor, π‘π‘ is time, πΈπΈ = emf of the battery, πΆπΆ = capacitance, π
π
is resistance of battery and connecting wires, πΆπΆπΆπΆ has units of time and is termed time constant.
In one time constant ππ (= πΆπΆπΆπΆ) 1
ππ = πΈπΈπΈπΈ οΏ½1 β οΏ½ = 0.63πΈπΈπΈπΈ ππ
12 Thus, 63% of the maximum charge is deposited in one time constant.
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23. Discharging
Just before the switch is closed (at t= 0) the charge on the capacitor is ππ0 , and the current is zero. At the time t after the switch is closed, the charge on the capacitor is q, the current is i.
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By KVL, Here
ππ
πΆπΆ
β΄
π
π
β
β«Q
β β β
β π
π
π
π
= 0
ππ = β
ππππ ππππ
ππππ ππ
=β
q dq 0 q
ln
ππππ ππππ
ππ
πΆπΆ
=β
1
πΆπΆπΆπΆ
π‘π‘
ππππ
= β«0 β
ππ
ππ0
=β
1
πΆπΆπΆπΆ
π‘π‘
πΆπΆπΆπΆ
ππππ
ππ = ππ0 ππ βπ‘π‘/πΆπΆπΆπΆ
Where, ππ is charge remaining on the capacitor and ππ0 is the initial charge. The constant πΆπΆπΆπΆ is the time 1
constant. At π‘π‘ = πΆπΆπΆπΆ, the remaining charge is ππ = ππ0 = 0.37ππ0 . Thus in one time constant 0.63% discharging is complete.
ππ
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24. Atmospheric electricity
At about 50 km above the earthβs surface, the air becomes highly conducting and thus there is perfectly conducting surface having potential of 400 kV with respect to earth and current (positive charge) comes down from this surface to earth. 25. Thunderstorms and lightning bring negative charge to earth.
Water vapour condenses to form small water droplets and tiny ice particles. A parcel of air (cloud) with these droplets and ice particles forms a thunderstorm. A matured thunderstorm is formed with its lower end at a height of 3-4 km above the earthβs surface and the upper end at about 6-7 km above the earthβs surface. Negative charge is at the lower end and positive charge is at the upper end of this thunderstorm. This negative charge creates a potential difference of 20 to 100 MV between these clouds and the earth. This cause dielectric breakdown of air and air becomes conducting.
There are number of thunderstorms every day throughout the earth. They charge the atmospheric battery by supplying negative charge to the earth and positive charge to the upper atmosphere.