4. Electric Current In Counductors By Sanjay Pandey

1

SANJAY PANDEY

1. Electric current

Mail add: [email protected] Contact No.: 09415416895, 09721573337, 09453763058

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

𝑖𝑖 = ∆𝑄𝑄/∆𝑡𝑡

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 / 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

Q is charge, t is time

𝑇𝑇ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑎𝑎𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡 = 𝑖𝑖 = limΔ𝑡𝑡→0( ∆𝑄𝑄/∆𝑡𝑡) = 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑

Electric current is a scalar quantity. •

•

Unit: AMPERE. Ampere

1 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 1 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 / 1 𝑠𝑠𝑠𝑠𝑠𝑠

In S.I system uni t of electric current is ampere.

Ampere is defined as:

“Current through a conductor will be 1 ampere if one coulomb of electric charge passes through any cross section of conductor in 1 second.” •

Types of current

1 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 1 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 / 1 𝑠𝑠𝑠𝑠𝑠𝑠

There are two types of current. 1.

Electronic Current

2.

Conventional Current

Electronic current flows from negative to positive terminal. Direction of conventional current is taken from higher potential to the lower potenti al.

2.

Current density

Average current density 𝑗𝑗 = Δ𝑖𝑖/Δ𝑆𝑆

𝑖𝑖 is current and A is area of the conductor

The current density at a point P is

2 𝑗𝑗 = lim∆𝑡𝑡→0 (Δ𝑖𝑖/Δ𝑆𝑆) = 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑

If current i is uniformly distributed over an area S and is perpendicular to i t, then 𝑗𝑗 = 𝑖𝑖/𝑆𝑆

For a fini te area

����⃗ 𝑖𝑖 = ∫ 𝑗𝑗⃗. 𝑑𝑑𝑑𝑑

Where

𝑗𝑗⃗ = density of current (vector) ����⃗ = area (v ector) 𝑑𝑑𝑑𝑑 3.

Drift speed

A conductor contains free electrons moving randomly in a lattice of positive ions. Electrons collide with positive ions and thei r 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 bias ed in their random motion in favor of the force. As a result electrons drift slowly in the direction opposite to the fi eld.

If 𝜏𝜏 be the average time between successive collisions, the distance drifted during this period is

The drift speed is

𝑙𝑙 =

1 2

𝑣𝑣𝑑𝑑 =

𝑎𝑎(𝜏𝜏)2 = 𝑙𝑙

𝜏𝜏

=

1 2

1 2

(𝑒𝑒𝑒𝑒/𝑚𝑚)( 𝜏𝜏) ²

(𝑒𝑒𝑒𝑒/𝑚𝑚)𝜏𝜏 = 𝑘𝑘𝑘𝑘

𝜏𝜏 the average time between successive collisions, is constant for a given material at a given temperature. It is called the Rel axation time 4.

Relaxation Time

Suppose 𝑖𝑖 𝑡𝑡ℎ electron suffered its last collision 𝑡𝑡𝑖𝑖 time ago, then rel axation time

3

5.

6.

∑𝑡𝑡 𝑖𝑖

𝜏𝜏 =

𝑛𝑛

Relation between current density and drift speed

𝑗𝑗 = 𝑖𝑖/𝐴𝐴 = 𝑛𝑛𝑛𝑛 𝑣𝑣𝑑𝑑

Ohm's law

It states that the current density in a conductor is directly proportional to electric field across the conductor. 𝑗𝑗⃗ ∝ 𝐸𝐸�⃗ ⇒ 𝑗𝑗⃗ = 𝜎𝜎𝐸𝐸�⃗

Proof:

𝐸𝐸�⃗ is field and 𝜎𝜎 is electrical conductivity of the material. We know that

And

𝑗𝑗 = 𝑛𝑛𝑛𝑛 �

∴

Or

∴

⇒

or

𝑗𝑗 = 𝑖𝑖/𝐴𝐴 = 𝑛𝑛𝑛𝑛 𝑣𝑣𝑑𝑑 1 2

𝑣𝑣𝑑𝑑 =

(𝑒𝑒𝑒𝑒/𝑚𝑚)𝜏𝜏� =

1 2

(𝑒𝑒𝑒𝑒 /𝑚𝑚)𝜏𝜏

𝑛𝑛𝑒𝑒 2 𝜏𝜏 2𝑚𝑚

𝐸𝐸

𝑗𝑗 = 𝜎𝜎𝜎𝜎 (prov ed)

The resistivity of the material is defi ned as and

𝜌𝜌 =

𝑖𝑖

𝐴𝐴

1

1 𝑉𝑉

= 𝐸𝐸 = . 𝜌𝜌

𝜌𝜌 𝑙𝑙

𝑗𝑗 =

1

𝜎𝜎

𝑖𝑖

𝐴𝐴

𝑙𝑙

𝑉𝑉 = 𝜌𝜌 𝑖𝑖 𝐴𝐴

𝑉𝑉 = 𝑅𝑅𝑅𝑅 (other form of Ohm’s law)

4 where

𝑅𝑅 = 𝜌𝜌

𝑙𝑙

𝐴𝐴

is called the resistance of the given conductor. The quantity 1/R is called the conductance.

Unit of resistivity ρ is ohm-meter (or Ω-m). The unit of conductivity σ is (𝑜𝑜ℎ𝑚𝑚 − 𝑚𝑚)−1 written as 𝑚𝑚ℎ𝑜𝑜/𝑚𝑚.

•

Resistivity of a material 𝜌𝜌 = 1/𝜎𝜎

Another form of Ohm's law

𝑉𝑉 = 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑡𝑡ℎ𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝐸𝐸𝐸𝐸 (𝑙𝑙 = length of the conductor) 𝑉𝑉 = 𝑅𝑅𝑅𝑅

𝑅𝑅 = 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝜌𝜌 ×

1/𝑅𝑅 is called conductance 7.

𝑙𝑙

𝐴𝐴

Temperature dependence of resistivity

As temperature of a resistor increases its resistance increases. The relation can be expressed as 𝑅𝑅(𝑇𝑇) = 𝑅𝑅(𝑇𝑇0 )[1 + 𝛼𝛼(𝑇𝑇 − 𝑇𝑇0 )]

𝛼𝛼 is called temperature coefficient of resistivity. 8. 9.

Thermistors: Measure small changes in temperatures Superconductors:

For these materi als resistivity suddenly drops to zero below a certain temperature. For Mercury it is 4.2 K. For the s uper 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 achiev ed superconductivity at 125 K so far. 10. Battery

5 Battery is a device which maintains a potential difference between i ts 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 (termi nal A) and negative charges of towards another side (terminal B).

Let force on a positive charge 𝑞𝑞 is 𝐹𝐹⃗𝑏𝑏 (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 𝐸𝐸�⃗ is

developed in the battery material from A to B and exerts a force 𝐹𝐹⃗𝑒𝑒 = 𝑞𝑞𝐸𝐸�⃗ on a charge 𝑞𝑞. The direction of this force is opposite to 𝐹𝐹⃗𝑏𝑏 . In steady state, the charge accumulation on A and B is such that 𝐹𝐹𝑏𝑏 = 𝐹𝐹𝑒𝑒 . No further accumulation takes place.

If a charge 𝑞𝑞 is moved from one terminal (say B) to the other terminal say A, the work done by the battery force is 𝑊𝑊 = 𝐹𝐹𝑏𝑏 𝑑𝑑 where 𝑑𝑑 is distance between A and B. The work done by the battery force per unit charge is E = 𝑊𝑊/𝑞𝑞 = 𝐹𝐹𝑏𝑏 𝑑𝑑 /𝑞𝑞

This E is called the emf of the battery. Pleas e note that emf is not a force it is 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑/𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 . If nothing is externally connected between A and B, then 𝐹𝐹𝑏𝑏 = 𝐹𝐹𝑒𝑒 = 𝑞𝑞𝑞𝑞

Or

𝐹𝐹𝑏𝑏 𝑑𝑑 = 𝑞𝑞𝑞𝑞𝑞𝑞 = 𝑞𝑞𝑞𝑞

(because 𝑉𝑉 = 𝐸𝐸𝐸𝐸 )

As

𝜀𝜀 = 𝑊𝑊/𝑞𝑞 = 𝐹𝐹𝑏𝑏 𝑑𝑑/𝑞𝑞 = 𝑞𝑞𝑞𝑞/𝑞𝑞 = 𝑉𝑉

𝑉𝑉 = potenti al difference between the terminals Therefore

𝜀𝜀 = 𝑉𝑉

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 𝑞𝑞 = 𝑖𝑖𝑡𝑡 goes through the circuit having resistance R the electric potential energy decreases by 𝑈𝑈 = 𝑞𝑞𝑉𝑉 = (𝑖𝑖𝑡𝑡 )(𝑖𝑖𝑅𝑅 ) = 𝑖𝑖 2 𝑅𝑅𝑡𝑡

This loss in electric potential energy appears as increased thermal energy of the resistor. Thus a current 𝑖𝑖 for a time 𝑡𝑡 through a resistance 𝑅𝑅 increases the thermal energy by 𝑖𝑖 2 𝑅𝑅𝑡𝑡 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑃𝑃 = 𝑈𝑈/𝑡𝑡 = 𝑖𝑖²𝑅𝑅 = 𝑉𝑉𝑉𝑉

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 𝑟𝑟 and for ideal battery 𝑟𝑟 = 0.

6

Since the internal resistance of the battery is in series with the load the equivalent resistance of the

circuit is 𝑅𝑅𝑒𝑒𝑞𝑞 = 𝑅𝑅 + 𝑟𝑟. The current is thus reduced owing to the internal resistance, 𝑖𝑖 = ℇ / (𝑟𝑟 + 𝑅𝑅), from

what it would be in i ts 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. ℰ = 𝑖𝑖𝑖𝑖 + 𝑖𝑖𝑖𝑖

𝑉𝑉𝑙𝑙𝑜𝑜 𝑎𝑎 𝑑𝑑 = 𝑖𝑖𝑖𝑖 = ℰ − 𝑖𝑖𝑖𝑖

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

= 𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑒𝑒 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 – 𝑖𝑖𝑖𝑖

(𝑟𝑟 = internal resistance)

The so-called terminal voltage of a battery is lower than the emf when i t 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.

7

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Polarity of a Resistor Assign a positive (+) sign for the voltage to the termi nal of the element where the current enters and negative (−) sign to the termi nal of the el ement where the current leaves it. 13. Kirchhoff's laws: Kirchhoff’s current law (KCL) or the junction law

The sum of all currents directed towards a point i n a circui t is equal to the sum of all the currents directed away from the point.

Kirchhoff’s voltage law (KVL) or the loop law

The algebraic sum of all the potenti al differences along a closed loop i n a circui t is zero.

•

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.

8

• (a) (b) (c)

(e) (f) (g)

Wri ting 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 defini tion on the end of the el ement that you enter. Convers ely, you could choose the sign of the end you leav e, 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 hav e gone completely around the loop returning to your s tarting poi nt. Set your result equal to zero.

•

Now, let's wri te KVL for each of the three loops in adjoining Fig.

•

−𝑉𝑉𝐵𝐵 + 𝑉𝑉1 + 𝑉𝑉2 = 0

(d)

•

For the firs t 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. −𝑉𝑉2 − 𝑉𝑉3 + 𝑉𝑉4 = 0

For the third loop (Battery, Element 1, Element 3, Element 4) So, we get three equations - right?

−𝑉𝑉𝐵𝐵 + 𝑉𝑉1 − 𝑉𝑉3 + 𝑉𝑉4 = 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 thi rd equation. −𝑉𝑉𝐵𝐵 + 𝑉𝑉1 + 𝑉𝑉2 = 0

−𝑉𝑉2 − 𝑉𝑉3 + 𝑉𝑉4 = 0 ………………………….…………………………………….. −𝑉𝑉𝐵𝐵 + 𝑉𝑉1 − 𝑉𝑉3 + 𝑉𝑉4 = 0 Can you see that you can add the first two equations to get the third? (Actually, there is a −𝑉𝑉2 and a +𝑉𝑉2 , 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 wri te 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. 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝑅𝑅₁ + 𝑅𝑅₂ + 𝑅𝑅₃+. ..

15. Combination of resistors in parallel

For resistors in parallel, the voltage drop across each resistor is identical 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 1/𝑅𝑅₁ + 1/𝑅𝑅₂ + 1/𝑅𝑅₃+. ..

16. Division of current in resistors joined in parallel 𝑖𝑖₁/𝑖𝑖₂ = 𝑅𝑅₂/𝑅𝑅₁

𝑖𝑖₁ = 𝑖𝑖𝑖𝑖₂/(𝑅𝑅₁ + 𝑅𝑅₂)

17. Batteries connected in series Where 𝑅𝑅 = external resistance

𝑖𝑖 = (𝜀𝜀₁ + 𝜀𝜀₂)/(𝑅𝑅 + 𝑟𝑟₀)

𝑟𝑟₀ = 𝑟𝑟₁ + 𝑟𝑟₂

𝑟𝑟₁, 𝑟𝑟₂ are i nternal resistances of two batteries

18. Batteries connected in parallel 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑒𝑒𝑒𝑒𝑒𝑒 = 𝜀𝜀₀ =

𝜀𝜀 ₁𝑟𝑟₂+𝜀𝜀2 𝑟𝑟1 (𝑟𝑟₁+𝑟𝑟₂)

where 𝜀𝜀₁, 𝜀𝜀₂ are emfs of of batteries , and 𝑟𝑟₁, 𝑟𝑟₂ are internal resistances. So

𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑐𝑐𝑐𝑐 = 𝑟𝑟₀ = 𝑟𝑟₁𝑟𝑟₂/(𝑟𝑟₁ + 𝑟𝑟₂) 𝑖𝑖 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝜀𝜀₀/(𝑅𝑅 + 𝑟𝑟₀)

10 19. Wheats tone 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

R₁/R₂ = R₃/R₄

From figure:

R₁ & R₂ are connected in seri es.

Reason: (only one path for the flow of current)

R₁ & R₃ are connected in parallel.

Reason: (two paths for the flow of current)

R₃ & R₄ are connected in seri es.

R₂ & R₄ are connected in parallel.

If the there is no deflection in the galvanometer, then 𝑅𝑅₁/𝑅𝑅₂ = 𝑅𝑅₃/𝑅𝑅4

20. Ammeter

⇒ 𝑅𝑅4 = 𝑅𝑅₃𝑅𝑅₂ /𝑅𝑅₁

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 seri es 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 circui t.

The case for charging a capacitor is described firs t, then discharging a capacitor.

to

discharge

one.

11

Figure shows charging of a capaci tor. Just before the switch is closed the charge on the capaci tor is zero. When the switch is closed (at time t= 0), the charging starts. By KVL

𝐸𝐸 − 𝑉𝑉𝑅𝑅 − 𝑉𝑉𝐶𝐶 = 0

⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒

𝑞𝑞

𝐸𝐸 − 𝑖𝑖𝑖𝑖 − 𝑅𝑅

𝑞𝑞

𝑖𝑖𝑖𝑖 = 𝐸𝐸 − 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑

=

𝐸𝐸𝐸𝐸 −𝑞𝑞 𝑞𝑞

∫0

𝐸𝐸𝐸𝐸 −𝑞𝑞

=

1−

𝑑𝑑𝑑𝑑

𝐶𝐶𝐶𝐶

𝑑𝑑𝑑𝑑

𝐸𝐸𝐸𝐸

𝑞𝑞

𝑡𝑡 𝑑𝑑𝑑𝑑

= ∫0

𝐸𝐸𝐸𝐸 −𝑞𝑞

𝐸𝐸𝐸𝐸

𝑐𝑐

𝐶𝐶

𝐸𝐸𝐸𝐸 −𝑞𝑞

−𝑙𝑙𝑙𝑙

=0

𝐶𝐶

=

= 𝑒𝑒

𝐶𝐶𝐶𝐶

𝑡𝑡

𝐶𝐶𝐶𝐶

−

𝑡𝑡 𝐶𝐶𝐶𝐶

𝑞𝑞 = 𝐸𝐸 𝐶𝐶(1 − 𝑒𝑒 −𝑡𝑡/𝐶𝐶𝐶𝐶 )

𝑞𝑞 is charge on the capacitor, 𝑡𝑡 is time, 𝐸𝐸 = emf of the battery, 𝐶𝐶 = capacitance, 𝑅𝑅 is resistance of battery and connecting wires, 𝐶𝐶𝐶𝐶 has uni ts of time and is termed time cons tant. In one time constant 𝜏𝜏 (= 𝐶𝐶𝐶𝐶) 1

𝑞𝑞 = 𝐸𝐸𝐸𝐸 �1 − � = 0.63𝐸𝐸𝐸𝐸 𝑒𝑒

12 Thus, 63% of the maximum charge is deposited in one time constant.

13

23. Discharging

Just before the switch is closed (at t= 0) the charge on the capacitor is 𝑄𝑄0 , 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.

14

By KVL, Here

𝑞𝑞

𝐶𝐶

∴

𝑅𝑅

⇒

∫Q

⇒ ⇒ ⇒

− 𝑅𝑅𝑅𝑅 = 0

𝑖𝑖 = −

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 𝑞𝑞

=−

q dq 0 q

ln

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑞𝑞

𝐶𝐶

=−

1

𝐶𝐶𝐶𝐶

𝑡𝑡

= ∫0 −

𝑞𝑞

𝑄𝑄0

=−

𝑑𝑑𝑑𝑑 1

𝐶𝐶𝐶𝐶

𝑡𝑡

𝑑𝑑𝑑𝑑

𝐶𝐶𝐶𝐶

𝑞𝑞 = 𝑄𝑄0 𝑒𝑒 −𝑡𝑡/𝐶𝐶𝐶𝐶

Where, 𝑞𝑞 is charge remai ning on the capacitor and 𝑄𝑄0 is the initi al charge. The constant 𝐶𝐶𝐶𝐶 is the time 1

constant. At 𝑡𝑡 = 𝐶𝐶𝐶𝐶 , the remaining charge is 𝑞𝑞 = 𝑄𝑄0 = 0.37𝑄𝑄0 . Thus in one time cons tant 0.63% discharging

is complete.

𝑒𝑒

15

24. Atmospheric electricity

At about 50 km above the earth’s surface, the ai r becomes highly conducting and thus there is perfectly conducti ng 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 v apour condenses to form small water droplets and tiny ice particles. A parcel of air (cloud) wi th 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 earth’s surface and the upper end at about 6-7 km above the earth’s 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 di electric breakdown of air and air becomes conducting.

There are number of thunders torms every day throughout the earth. They charge the atmospheric battery by supplying negative charge to the earth and positive charge to the upper atmosphere.