The voltage drop identifies the amount of electric power produced or consumed when electric current flows throughout the voltage drop. Also, we can measure it in a circuit with a proper device, but also can be calculated in advance using suitable equations. Moreover, in this topic, we will learn about voltage drop, voltage drop formula, its derivation, and solved examples.
Voltage Drop
It refers to the decrease of electric potential along the path of a current flowing in an electrical circuit. Furthermore, it is a process similar to the electric circuit. Also, each point in the circuit can be assigned a voltage that’s proportional to its ‘electrical elevation’ to speak. In simple words, the voltage drop is the arithmetical difference between a higher voltage and a lower voltage.
In addition, the amount of energy per second (power) delivered to a component in a circuit is equal to the voltage drop from corner to corner of that component’s terminals multiplied by the current flow through the components.
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How to calculate voltage drop?
Voltmeters
One way to resolve the voltage drop across the circuit component is to build the circuit and measure the drop using the voltmeter (current measuring device). Also, they are designed in a way that they disturb the operation of the circuit they are connected to as little as possible. Moreover, they achieve this by minimizing the current that flows through the voltmeter to the smallest possible value.
KCL and KVL
These equations present knowledge of all voltage drops and all current flows in the circuit. Also, the engineers can adjust the various components values to obtain a final circuit that serves its principle in the best possible way.
KCL- Is the Kirchhoff’s Current Law that states that total current flow in and out of any junction of wires in the circuit is zero. Moreover, KCL equations are expressions of charge conservation.
KVL- Is the Kirchhoff’s Voltage Law that states that the sum of voltage drop around any closed path in a circuit is zero. Furthermore, its equations are an expression of energy conservation.
Voltage Drop Formula
The voltage drop formula points out how the supplied power from the voltage source is condensed as electric current flows throughout the elements that do not supply the voltage of the electrical circuit. Moreover, the voltage drops across the internal resistances and connectors of the source are unwanted since the supply energy is lost. Furthermore, the voltage drop across active circuit elements and loads are preferred, because supplied power executes competent work.
V = I × Z
V = IZ
Derivation of the Formula
I = refers to the current in amperes (A)
Z = refers to the impedance in omega (\(\Omega\))
V = refers to the voltage drop
Moreover, the single-phase voltage drop formula is given as
VD = \(\frac{2 LRI}{1000}\)
And three-phase voltage drop formula is given as
VD = \(\frac{2 LRI}{ 1000 \times 0.866}\)
Here,
L = refers to the length of the circuit
R = refers to the resistance in Omega (\(\Omega\))
I = refers to the load current in amperes
Solved Example on Voltage DropÂ
Example 1
Through a circuit, a current of 9A flows through that carries a resistance of 10 \(\Omega\). Find the voltage drop across the circuit.
Solution:
Given:
Current = 9A
Impedance Z = 10 \(\Omega\)
Putting values in the voltage drop formula we get
V = IZ
V = 9 × 10 = 90 V
So, the voltage drop is 90 V.
Example 2
Suppose a lamp has a 15 \(\Omega\) and 30 \(\Omega\) connected in a series. Also, a current of 4A is passing through it. So, calculate the voltage drop of the series?
Solution:
I = 4A
Resistance Z = (15 +30) \(\Omega\) = 45 \(\Omega\)
Putting values in voltage drop formula we get
V =IZ
V = 4 × 45 = 180 V.
So, the voltage drop is 180 V.
Typo Error>
Speed of Light, C = 299,792,458 m/s in vacuum
So U s/b C = 3 x 10^8 m/s
Not that C = 3 x 108 m/s
to imply C = 324 m/s
A bullet is faster than 324m/s
I have realy intrested to to this topic
m=f/a correct this
Interesting studies
It is already correct f= ma by second newton formula…