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Electrical Formulas deal with electricity, electronics, and electromagnetics. Voltage, current, power, and resistance are amongst the most widely used formulas for electrical physics. The fundamental components of electricity covered in the article are Ohm's Law, Volts, Watts and Ampere.
Fundamentals of Electricity
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Understanding how the various components of electricity might function together can undoubtedly benefit us in getting a comprehensive picture of electricity as a whole. The system can be understood by visualizing a system of water pipes where voltage represents the water pressure, the current indicates the rate of flow, and resistance represents the pipe size in this example.
- The voltage is equal to the current flowing in a circuit multiplied by the resistance of that circuit, according to Ohm's Law.
- The base unit of resistance in an electrical system is the ohm.
- One ohm is defined as an electrical Resistance between two points of a conductor that results in the creation of a current of one Ampere in the conductor when a constant potential difference of one volt is applied to the points.
- Similarly, the definition of one volt is given as the amount of potential required to send one ampere of current through one ohm of resistance.
- In physics, the product of one ampere and one volt is called power in watts.
- Therefore, one ampere of current flowing under the force of one volt produces one watt of energy.
Fundamentals of Electricity
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Common Electrical Formulas
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The most widely recognized formulas in electrical physics are:
Electrical Charge Formula
When subatomic particles are placed in an electromagnetic field, their electric charge leads them to experience a force. Coulomb is the SI unit of electric charge, and its symbol is Q.
Electrical Charge Formula
The formula for electric charge is as: \(Q=I\times T\)
Where,
- Q refers to electric charge,
- I refer to an electric current; and
- t is time
Electric Current Formula
In an electric circuit, an electric current is the steady flow of electrons. Electrons move when a potential difference is applied across a wire or terminal. Electric current is the rate at which an electric charge changes in a circuit. This current is equal to the voltage and resistance of the circuit. I is the symbol for electric current, and Amperes is the SI unit for it. The electric current is related to the electric charge and time.
Electric Current Formula
According to Ohm’s law, the general formula for electric current will be
I (current) =\(\frac{V(voltage)}{R(Resistance)}\)
Potential Difference Formula
The amount of work (W) done by an external agent in transporting a unit charge (Q) from one point to another is defined as the potential difference between two points (E) in an electrical circuit. The formula for calculating the potential difference is as follows: \(E= \frac{W}{Q}\)
Here,
- Potential difference is denoted as E,
- W is the work done in moving a charge from one point to another
- Q is the charge quantity in coulomb
Electric Power Formula
The rate at which work gets done can be defined as electric power. The watt, abbreviated as P, is the SI unit for power. The power formula links time, voltage, and charge together. The formula can be changed using Ohm's law. The electric power formula is as follows:
| \(P=V\times I\) Also, \(P=I^2\times R\) (According to Ohm’s Law) \(P=\frac{V^2}{R}\) |
Where power is denoted as P,
- I is the current,
- V is the voltage, and
- R is the resistance offered
Important Questions on Electric Power Formula |
Electric Field Formula
An electric field is a region generated by an electric charge around it, the influence of which may be recognized when another charge is placed into the field's territory.
The formula for the Electric Field is as follows:\(E=\frac{F}{q}\)
Where,
- Electric field is denoted as E,
- F is the force applied, and
- q is the charge
Electric Potential Formula
The two elements that give an object its electric potential energy are the charge it possesses and its relative position in relation to other electrically charged things. When an object moves in the presence of an electric field, it gains electric potential energy.
For each charge, the electric potential is computed by dividing the potential energy by the charge quantity. At each position around a point charge, the electric potential energy formula is given by: \(V= K\times\frac{q}{r}\)
As V is the electric potential and q is the point charge, while r is the distance from any place in the vicinity of the charge to the point charge and k is the coulomb constant where the value of k is 9 x 109 N.
Electric Flux Formula
The passage of the electric field through a particular region is measured by electric flux. The electric flux is determined by the number of electric field lines traveling through a generally perpendicular surface.
So, the formula for electric flux will be:
| \(\varphi\)p = EA |
The projected area is A cos θ when the same plane is slanted at an angle θ, and the total flux through the surface is:
| \(\varphi_p\)= EA cos\(\theta\) |
where E is the magnitude of the electric field, A is the area surface and \(\theta\) is the angle projected by the plane.
Table of Electrical Formulas
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Important Electrical Formulas are listed below:
| Formula | Description |
|---|---|
| Electric Charge | Formula: Q = I x t; Unit: Colulomb |
| Electric Current | Formula: I (current) = V/R; Unit: Ampere |
| Potential Difference | Formula: E = W/Q; Unit: Volts |
| Electric Power | Formula: P = VI; Unit: Watt |
| Electric Field | Formula: E = F/q; Unit: Vm-1 |
| Electric Potential | Formula: V = k x [q/r]; Unit: Volts |
| Electric Flux | Formula: \(\varphi_p\) = EA; Unit: Nm2/C |
| Ohm’s law | V ∝ I or V = RI |
| Specific resistance | \(\rho= \frac{RA}{l} ohm-m\) |
| Drift Velocity | \(V_d = \frac{eE r}{m}\) |
| Current Density (J) | \(\sigma = \frac{J}{E} = \frac{1}{\rho} mho/m\) |
| Mobility | \(\mu = \frac{V_d}{E} = \frac{er}{m} m^2 /V^{-s}\) |
| Kirchhoff’s Rules | First law (Current law or Junction law): At each node Σi = 0 (i.e. i1 + i2 + i3 + ……… + in = 0) Second law (Voltage law or loop law): ΣiR = Σ emf. |
| Wheatstone Bridge | (Ig = o) If P/Q>R/S, VD > VB ; If P/Q<R/S, VB > VD |
| Internal resistance (r) | \(r = \frac{E-V}{R}R\) |
Previous Year Questions
- An electron moves in a circle of radius 1.0cm with a constant speed of…
- An electric motor runs on DC source of emf 200V and draws...
- In the Wheatstone's network given…
- A current of 5A is passing through a metallic wire of cross-sectional area…
- A galvanometer of resistance 20Ω is to be converted into…
- A wire of length l and resistance R is stretched to get…
- In a good conductor of electricity, the type of bonding that exists is…
- Which of the following graph represents the variation of resistivity…
- A square loop ABCD, carrying a current I, is placed near…
- With increase in temperature the conductivity of…
- Wire bound resistors are made by…
- Which of the following is secondary cell…
- Which of the following is NOT the name of a secondary cell…
- Which of the following is correct for…
- Which of the following I-V graph represents…
- A 6V battery is connected to the terminals of a three metre long…
- Five equal resistances each of resistance…
- A galvanometer of 50Ω resistance has 25 divisions…
- A cell of constant emf first connected to a resistance…
Things to Remember
- A free-electron leaves a vacancy that can be filled by an electron from another atom that has been driven out of its orbit. An electron flow is created when free electrons migrate from one atom to another. The flow of electrons is the foundation of electricity.
- The flow of Electrons- The electrons flow from negative to positive ends as has been found by scientists. Because electrons are negatively charged, they are attracted to positively charged objects while repelling negatively charged objects.
- Current is defined as the flow of free electrons in the same general direction from one atom to the next, and it is measured in amperes (“amps” or “A”).
- The force that is supplied to a conductor to release electrons, causing Electrical Current to flow, is known as voltage. Volts, or “V”, are its units of measurement.
- Resistance is a unit of measurement for the restriction of electron flow via a conductor. It is measured in ohms.
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Sample Questions
Ques. What are the various types of current? (2 marks)
Ans. The two types of current in electricity are direct current (DC) and alternating current (AC).
Direct current only allows electrons to move in one direction. Batteries produce direct current. In alternating current, electrons move in both directions.
Ques. What is the difference between voltage and current? (2 marks)
Ans. Voltage is defined as the difference in charge between two places.
Current is the rate at which a charge travels.
Ques. Make a list of the factors that can influence the electric flux. (3 marks)
Ans. The following factors influence electric flux:
- Electric field surrounding the area.
- The surface area of the area element.
- The angle of inclination between the normal to the area element and field.
Ques. If an electric circuit's current and voltage are given as 3.5 amps and 16 volts, respectively. How do you calculate its electrical power? (3 marks)
Ans. According to question: V = 16V and I = 3.5A
As the formula for electric power is,
VI = P
So, P = 16 x 3.5
= 56 watts.
Hence, the calculated electric power is 56 watts.
Ques. Determine the voltage across an electrical circuit with an 8-amp current and a 150-ohm resistance. (3 marks)
Ans. The formula for calculating voltage is as follows:
As, V = I R
So, V = 8 x 150
= 1200.
Hence, the voltage of the electrical circuit is 1200V.
Ques. An electric bulb uses 0.8 amps and operates at 250 volts for an average of 8 hours per day.
a.Determine the amount of energy utilized by the bulb.
b.What is the monthly cost for 60 days if the electric distribution company changes Rs 5 for 6 KWH? (3 marks)
Ans. a) The electrical bulb's power is supplied by P = VI
P = 0.8 x 250
= 200W
= 0.2 KW
b) The bulb's total energy consumption over the course of 60 days (E) = P x t
= 0.2 x 8 x 60 E
= 96KWH.
As a result, the cost will be = (5 x 96) / 6
= Rs 80
Ques.
a.3 bulbs B1, B2, and B3 are connected in parallel with the voltage and the total current passed to them is 6A. Will the amount of current delivered to each bulb be the same and have a value of 2 A?
b.If Bulb B3 is blown out, the bulbs B1 and B2 will begin to shine brighter. Is it true or false?
c.If Bulb B1 is blown out, what will happen to all of the ammeter readings?
d.Even if any bulb fails, the current shown in Ammeter A stays constant. Is it true or false? (5 marks)
Ans.
- Because the bulbs are the same and are linked in parallel with the voltage. The amount of current delivered to each bulb will be the same. Because there is a total current of 6 A. 2 A will be the individual current.
- When Bulb B3 is blown out, the potential difference between the remaining bulbs remains constant, resulting in the same current flow and illumination. There has been no change.
- When Bulb B1 fails, the current in that section falls to zero.
As a result, the value on Ammeter A1 becomes zero.
Ammeter A2 reading will remain the same, i.e. 2 A
Reading of Ammeter A3 will remain the same, i.e. 2 A
Reading of Ammeter A will be = 2+2 = 4 A
- The reading of Ammeter A will vary, as seen above.
Ques. Calculate the potential difference of a 10-ohm resistance cable when a current of 20mA is sent through it. (3 marks)
Ans. First, use the formula to compute the potential difference, where R = 10 and I = 20mA.
As R x I = V
So, V = 10 x 20 x 10-3A
= 0.2 volts.
Hence potential difference V will be 0.2 volts.
Ques. When a 15 resistor wire is linked to a 60-volt battery, how much current will flow through it? (3 marks)
Ans. First of all, we need to find I, where R = 15 and V = 60V, so as to determine the current.
As R x I = V
So, V/R = I
Hence, I = 60/15
= 4A
The current will flow at a rate of 4A.
Ques. At any point, a force of 13 N acts on the charge at 9 μ C. Calculate the intensity of the electric field at that position. (3 marks)
Ans. Given F = 13 N and q = 9 μC
The formula for the electric field is as follows:
E = F / q
= 13 / (9 x 106)
= 1.45 x 10-6 N/C
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