Electricity

In SI unit system, charge is measured in Coulomb(C)

Charge on electron = -1.6 × 10^-19C

As 1C = Electrons

μA = 10^-6A

mA = 10^-3A

As metals has most number of free electrons and from the choice above copper and iron are metals.

But when we know that more the current flowing in a conductor more are the free electrons. So, copper being a very good conductor has more number of free electrons.

The Ohm’s Law state that, the potential difference (voltage) across an ideal conductor is proportional to the current through it. The constant of proportionality is called the "resistance", R. Ohm's Law is given by: V = I R where V is the potential difference between two points which include a resistance R.

Electric current means rate of flow of an electric charge.

1 minute = 60 sec.

I = & I = 2A.

2 = . So, Q = 120C

Given I = 4.8 T = 1sec

Using I = we get Q = 4.8C

Since 1C = electrons

So, 4.8C = = 3 x 10¹⁹ electrons

Voltage is defined as Work done per unit charge.

V =

Charge = I x T

Voltage is defined as Work Done per unit charge.

V =

We know that potential difference b/w any two points is work done to carry charge from one point to another. So,

V = I x R

V = 1.5V & R = 10 Ω

Hence 1.5V = 10 Ω x I

I = 150 mA

Resistivity depends only on material of wire.

When resistance is connected in parallel we use the following formula

R(Ω) = ρ x

Since all the resistance are in series, so total resistance will be sum of all the resistance.

The two 4Ω resistance are in series & are in parallel to 8Ω resistance.

R1 = 8Ω & R2 = 8Ω

The two 5Ω resistors are in parallel which is results in 2.5Ω in series with the other two 5Ω resistors.

Parallel

It is called kilowatt hour and is a unit for energy and is especially used in electricity bills.

1 kilowatt hour

1 hour = 3600 seconds

1killo = 1000

1 kilowatt hour = 3600000 = A. 3.6 x 10⁶

Power = Voltage x Current; Power = 1.1kW = 1100W

Voltage = 220 V

Hence, Current

The solution that conducts electricity are called “electrolytes”. In electrolytes, the electric current flows due to both positive and negative ions.

In distilled water there is no free positive or negative ions hence it acts as an insulator.

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. Electric charges are of two types: Positive electric charge (proton) & negative electric charge (electron). Electric charge is an intrinsic (independent) property of electron and proton like mass.

The SI unit of electric charge is Coulomb.

In atoms, electrons move around the nucleus where protons remain bound. In some material like metals, under normal circumstances, the attractive force between the valence electrons (outermost ones) and nucleus is comparatively very small.

During the formation of metallic materials, these electrons get separated from their parent atom and move in a random manner. Such electrons are known as Free electrons.

These free electrons are responsible for their electricity conducting property.

The net quantity of an electric charge flowing through any cross-section of conductor is defined as electric current.

Thus,

The SI unit of current is Ampere (A).

The total resistance of the circuit increases by connecting the resistors in series hence, the current decreases. Thus, to control the current in the circuit, series connection of resistors is useful. Moreover, the fuse is always connected in series with the mains. Therefore, whenever there is short circuit in any electrical appliance, the fuse wire melts and stops electric current. As a result, damage to electrical appliance can be prevented.

Whereas, the advantage of parallel connection is that when we connect three bulbs in parallel all will get equal voltage and therefore will give equal amount of light irrespective of each other. Secondly, if any appliance is affected/fused, it will not affect the other appliance. When we need more current we use parallel connection as resistance gets decreased resulting in more amount of current.

On passing the electric current through electrolytic solution, negative charge moves towards positive terminal and positive ions towards negative terminal. This process of separating the ions is called electrolysis.

Faraday’s First Law – The mass of substance (metal) deposited at cathode on passing electric current through electrolytic solution is proportional to charge passing through it, M ∝ Q

Faraday’s Second Law – For a given amount of current passed, the masses of different elements deposited on cathode is proportional to their chemical equivalent (e).

On bringing some electric charge near any other charge, an attractive or repulsive force exerts on it. Thus, the work is to be done against this force keeping a charge in equilibrium and to move another near or far from it. This work is stored in the form of potential energy. This work done on the charge is called as an electric potential.

The Work required to bring the unit positive charge from infinity to any point against electric force is known as electrical potential at that point.

The SI unit is joule/coulomb or volt (V).

The resistors are connected across two points in the circuit in such a way that the current flowing through each resistor is the same and only one path is available for it to flow, then the resistors are said to be connected in series.

From the figure, if voltage drops across R₁, R₂ & R₃ are V₁, V₂ & V₃ respectively.

Moreover, same current I is flowing to all the three resistors.

Using Ohm’s Law

R₁ voltage drops across V₁ = IR₁

R₂ voltage drops across V₂ = IR₂

R₃ voltage drops across V₃ = IR₃

From above three equations

IR = IR₁ + IR₂ + IR₃

R = R₁ + R₂ + R₃

When more than one resistances are connected across two points in the circuit such that more than one paths are available for the current to flow and voltage drops across two ends of each resistor are same, then the resistors are said to be connected in parallel between these two points.

From the figure above, suppose the current flowing through resistors R₁, R₂ & R₃ are I_1, I₂ & I₃ respectively.

Hence, I = I₁ + I₂ + I₃

Using Ohm’s Law,

And

From the equations we get,

Electrical energy is the energy newly derived from electric potential energy or kinetic energy.

Suppose the electric current is flowing through some resistor (R). To flow this current continuously the battery has to provide energy to every electric charge. Now, the work required to keep the charge Q in motion by the battery of voltage of V is

W = VQ

From the definition of electric current,

Q = It

.: W = VIt

According to Ohm’s Law, V = IR

W = (IR)(I)(t)

The above equation is called Joule’s Law.

A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy.

As shown in figure, take solution of dilute sulphuric acid (H₂SO₄) in a beaker. Dip one copper plate and another zinc plate in the solution in such a way that they do not touch each other. These two plates get electrically charged due to the process between these two plates and the solution. The positive charge at copper plate and negative charge at zinc plate get deposited. Thus, electric potential difference is produced between two plates. Positive charge plate is called positive pole of battery or anode and the negatively charged plate is called negative pole of battery or cathode.

Such a simple battery was invented by Italian Scientist Alexandro Volte (1745-1827). Therefore, it is called Volta’s Cell.

The electrons flow from cathode towards anode.

An electrolyte is a substance that produces an electrically conducting solution when dissolved in a polar solvent, such as water. The dissolved electrolyte separates into cations and anions, which disperse uniformly through the solvent. Electrically, such a solution is neutral. If an electric potential is applied to such a solution, the cations of the solution are drawn to the electrode that has an abundance of electrons, while the anions are drawn to the electrode that has a deficit of electrons

Setting up the experiment as shown in the above figure. When we switch ON the circuit we observe that the bulb does not glow.

On adding some salt in the distilled water, we observe that the bulb has started glowing.

•In the definite physical situation, the electric current flowing through the conductor is directly proportional to the potential difference applied across it is known as Ohm’s Law.

•To verify Ohm’s Law experimentally we set up the following circuit

In the following circuit, we have taken 0.5 meter long nichrome wire, four to five battery of 1.5V each, Ammeter and key.

•Now, when the key is ON, the current will flow through the wire. Measure the current in ammeter and potential difference across the two ends using voltmeter and note In the following table

•After this, plot the V vs I on a graph paper, we will observe something like this

•We will observe the following points

1. The electric current in the conductor increases in same proportion with increase in voltage.

2. I-V is a straight line

3. The ratio of V and I remains constant every time.

Electroplating is a process that uses an electric current to reduce dissolved metal cations so that they form a thin coherent metal coating on an electrode.

To understand experimentally, let’s take a solution of copper sulphate in a beaker. In this solution, the iron spoon which is to be electroplated and a copper plate are taken as electrodes and are connected with battery and key as shown in the figure below.

On passing the current, the CuSO₄ which electrolyte in the above case is decomposed into Cu^{2 +} and SO₄^2-. As Cu^{2 +} is positively charged, so it moves towards negative terminal i.e. Metal spoon and deposits on iron spoon. Thus, on iron spoon the plating.

Given

Current (I) = 400mA = 0.4A

Time (t) = 1min = 60sec

Formula

I = current

Q = charge

T = time

Q = I x t

Q = 0.4A x 60 sec

Q = 24 C

Since 1 coulomb contains 0.625 x 10¹⁹ electrons

Hence 24 C contains 24 x 0.625 x 10¹⁹ electrons

= 15 x 10¹⁹ electrons

Given

Charge (Q) = 1800C

Time (t) = 1 hour = 3600 sec

Formula

I = current

Q = charge

T = time

Calculation

Current = 0.5A

Given

Battery = 12V

Resistance (parallel) = 5 Ω, 10 Ω and 30 Ω

Formula used

Parallel Resistance formula

Ohm’s Law

V = IR

V = voltage

I = current

R = resistance

Calculation

To solve part (a) we first have to solve part (b)

(b) Equivalent resistance

Hence, R = 3 Ω

(a) Total current

I = 4 A

1. Equivalent Resistance

Formula used

Parallel Resistance formula

Series resistance formula

R_a = 8Ω

R_b = 10Ω

Since R_a & R_b are in series, so total resistance is

R_Total = R_a + R_b

R_Total = 18Ω

2. Total current

Formula used

Ohm’s Law

V = voltage

I = current

R = resistance

I = 0.66 A

(a) First, we will find equivalent resistance.

Formula used

Parallel Resistance formula

Series resistance formula

The highlighted (yellow colour) resistance are in series with each other and is in parallel with the other one.

R_a = 30Ω + 30Ω

R_a = 60Ω

Total resistance is 20Ω

(b) Total current

Formula used

Ohm’s Law

V = voltage

I = current

R = resistance

2V = 20Ω x I

I = 0.1A

Formula used

Parallel Resistance formula

Series resistance formula

The two resistors highlighted with yellow are in series to each other & both of them are parallel to the 10 Ω resistor.

Equivalent resistance = 5Ω

These 3 resistances are in series.

Hence, total resistance is 15 Ω

Given

Lamp 1 = 100W

Lamp 2 = 60W

Voltage (V) = 220V

Formula Used

Ohm’s Law

V = voltage

I = Current

R = resistance

Power Formula

P = power

V = voltage

R = resistance

Parallel Resistance formula

Resistance of lamp 1

R = 484Ω

Resistance of lamp 2

R = 806.7Ω

Since, both the lamps are joined in parallel; Hence total resistance will be

R = 302.5Ω

Using Ohm’s Law

Total current = 0.73 A

i. Current passing

Formula used

Power formula è P = VI

P = power

V = Voltage

I = Current

I = 20 A

ii. Resistance of heater

Formula used

Power Formula

P = power

V = voltage

R = resistance

R = 11Ω

iii. Energy consumed in 2 hrs

The power is 4.4 kW; It means that 4400 joules is consumed in 1 sec

In 2 hrs è 3600 x 2 secs

The energy consumed will be

= 4400 x 3600 x 2

= 3.168 x 10⁷ joules