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Overview

A battery converts chemical energy into electrical energy by creating a voltage across its two terminals – that is, a difference in electrical potential. A resistor is a component that creates a certain resistance to electric current. When we connect the two terminals of the resistor to the two terminals of the battery, the charge carriers move through the circuit and we call it electric current.

Voltage conveys the ability to move charge from one point to another. For example, a 5V battery can do 5 joules of work with a single charge ball. Since current flows through a resistor, we can measure the amount of work (per unit charge) required to keep the current flowing through the resistor.

This is the essence of voltage drop: a battery (or voltage source) provides energy to drive a moving charge. When current flows, components such as resistors consume energy, and the work per unit charge associated with current passing through a given component is the component’s voltage drop.

The voltage drop of the component corresponds to part of the voltage produced by the battery. In other words, the work done by the battery is distributed among the components in the circuit.

We understand intuitively that it takes more work to move a given current through a larger resistor. Therefore, if two resistors are in series (ie, of the same current), the voltage drop across the resistor with the higher resistance will be greater. This is the basis of the operation of the voltage divider circuit.

A resistor always acts as a load, ie. a component that consumes energy. If we adopt the standard current model where the current flows from a higher voltage to a lower voltage, the voltage drop across the resistor is positive when it enters the resistor and negative when the current leaves the resistor:

Voltage drop

What is Voltage Drop?

Voltage drop is the decrease in electrical potential by current flowing in a circuit. Or, more simply, “voltage drop”. Voltage losses are caused by the internal resistance of the source, passive elements, the passage of wires, contacts and connectors is not desirable, because part of the supplied energy is dissipated.

The voltage drop across an electrical load is proportional to the power available at that load for conversion to another useful form of energy. The voltage drop is calculated according to Ohm’s law.

Ohm's Law

Ohm’s law and Kirchhoff’s circuit law are used to control the voltage drop, explained below.

Ohm’s Law is represented by

                                                      V= R x I

V → Voltage drop (V)

R → Electrical resistance (Ω)

I → Electrical current (A)

In closed DC circuits, we also use Kirchhoff’s circuit law to calculate the voltage drop It is as follows:

Live voltage = the sum of the voltage drops between each component in the circuit.

Calculating Voltage Drop on a DC Line

Here is an example of a 100 foot power line. So 2 x 100 feet for two rows Let the electrical resistance be 1.02 Ω/1000 ft and the current 10 A.

Voltage Drop in Alternating Current Circuits

In AC circuits, in addition to resistance (R), current flow has another opposite – Reactance (X), which consists of Xc and Xl. Both X and R also resist current. The sum of the two is called impedance (Z).

Xc → Capacitive reactance

Xl → Inductive reactance.

                                                     Z= R +jX

 

The magnitude of Z depends on factors such as magnetic permeability, electrical insulating elements and AC frequency.

Similar to Ohm’s Law in DC circuits, this is given as:

                                               E= Z x I

E-   Voltage drop (V)

Z –  Electrical impedance (Ω)

I –   Electrical current (A)

EXAMPLE 1:

A current of 9A flows through a circuit that carries a resistance of 10 Ω. Determine the voltage drop across the circuit.

Solution:

Given:

Current I = 9A,

Impedance Z = 10Ω

the voltage drop formula is given by

V = IZ

  = 9 × 10

   V= 90 v.

EXAMPLE 2:

A lamp of 15 Ω and 30 Ω are connected in series. A current of 4 A is made to flow through it. Determine the voltage drop.

Solution:

Given:

Resistance Z = (15 + 30) Ω

      Z = 45 Ω,

Current I = 4A

The voltage drop formula is given by,

V = IZ

   = 4 × 45

    V= 180 V

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