Physics Formulas

Voltage Divider Formula

A voltage divider is a fundamental circuit used in the field of electronics. It can produce a portion of its input voltage as an output. It is generally formed using two resistors and a voltage source. The resistors are connected in series form. And the voltage is acted across these two resistors. It is termed as a potential divider. In this article, we will discuss voltage divider formula with example. Let us learn the concept!

voltage divider formula

Voltage Divider Formula

What is the division of voltage in a circuit?

The input voltage is distributed among the resistors or components of the voltage divider circuit. Due to this the voltage division takes place. If we want help with the calculation for voltage division, we can use the voltage divider calculator.

Formally, a voltage divider is a simple circuit that turns a large voltage into the smaller one. Using just two series resistors and the input voltage, we can create an output voltage that is a fraction of the input. Voltage dividers are one of the very used circuits in electronics. Therefore we can say that learning Ohm’s law was like being introduced to the alphabets of a circuit then learning about voltage dividers can be considered as learning how to make words.

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The formula for Voltage Divider

The voltage divider is the series of resistors or capacitors which that can be tapped at any intermediate point to generate a specific fraction of the voltage applied between its ends.

It consists of an electric circuit composed of two resistors and one input voltage supply. The below figure shows a simple voltage divider. In this circuit, two resistors are connected in series. The output voltage of the voltage divider is a function of the input voltage. This circuit helps to determine how the input voltage divides among the components in the circuit.

A voltage divider is applying a voltage across a series of two resistors. We may draw in a few different ways, but they should always essentially be the same circuit.

Thus formula is given as follows:

\( V_{out} = \frac{R_b}{R_a+R_b} \times V_{in} \)

Where,

\( V_{out} \) Output Voltage
\( V_{in} \) Input Voltage
\( R_a \) Input Registor
\( R_b \) Output Registor

Voltage dividers are having numerous applications, and these are among the most common of circuits electrical engineers use. Many applications are there. Some areas Potentiometers, Reading Resistive Sensors, and Level Shifting, etc.

Solved Examples on Voltage Divider

Q.1: Find out the output voltage of the voltage divider circuit whose two registers are 6ω and 8ω  respectively and the input voltage is 20V. Where 8 ω is in parallel to the output voltage.

Solution:

\( R_a = 6 \omega \),

\( R_b = 8 \omega \)

\( V_{in}  = 20 V \)

Now, applying the Voltage divider formula,

\( V_{out} = \frac{R_b}{R_a+R_b} \times V_{in} \)

Substituting the known values,

\( V_{out} = \frac{8}{6+8} \times 20 \)

= \( \frac {8}{14} \times 20 \)

=  \( \frac{80}{7} \)

\( V_{out} = 11.43 V \)

Therefore output voltage will be 11.43 V.

Q.2: The value of the input voltage of a voltage divider circuit is 20V. The resistors are 5 ω and 7 ω Where 7 ω is in parallel to the output voltage. Compute the output voltage.

Solution: We are knowing,

\( R_a = 5 \omega \),

\( R_b = 7 \omega \)

\( V_{in}  = 20 V \)

Now, applying the Voltage divider formula,

\( V_{out} = \frac{R_b}{R_a+R_b} \times V_{in} \)

Substituting the known values,

\( V_{out} = \frac{7}{5+7} \times 20 \)

= \( \frac {7}{12} \times 20 \)

= \( \frac{35}{3} \)

\( V_{out} \) = 11.66 V

Therefore output voltage will be 11.66 V.

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5 responses to “Spring Potential Energy Formula”

  1. 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

  2. Malek safrin says:

    I have realy intrested to to this topic

  3. umer says:

    m=f/a correct this

  4. Kwame David says:

    Interesting studies

  5. Yashdeep tiwari says:

    It is already correct f= ma by second newton formula…

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