# Voltage divider Formula

A voltage divider is a configuration of an electrical circuit that produces an output voltage that is a fraction of its input voltage, dividing the source voltage between one or more impedances connected in series.

Suppose you have a source voltage (input voltage), connected in series with n impedances. To know the output voltage on the generic impedance, Ohm's law is used:

output voltage = (generic impedance / the sum from the first to the nth impedance of the circuit) * input voltage.

The equation is written

We have:

V_{out} = output voltage.

Z_{i} = generic impedance.

= the sum from the first to the nth impedance of the circuit.

V_{in} = input voltage.

Voltage divider formula Questions.

1)We have a circuit with two resistors, the first of 3 ohm and the second of 5 ohm connected in series, the voltage of the source has a value of 12 V, calculate the dividing voltage after the first resistance.

Answer:

The first thing to note is that this is a resistive divider circuit so the values of the impedances are the same as the values of the resistances. The impedance Z_{i} takes the value of 3 ohm. Now let's add the circuit impedances to obtain the voltage divider, this is

= 3 ohm + 5 ohm = 8 ohm.

Then, we divide the value of the second impedance by the sum of the impedances and multiply it by the value of the input voltage to obtain the voltage divider,

V_{out} = (5 ohm / 8 ohm) * 12 V = 0.625 * 12 V= 7.824 V

V_{out} = 7.824 V.

2) Calculate the voltage divider for two capacitors that are connected in series that have capacitance values of 10uF and 22uF respectively. Here the circuit voltage is 10V with a frequency of 40 Hz.

Answer:

The first thing is to obtain the value of the impedances of each capacitor.

Z = 1/(2 * π * f * C)

Z_{1} = 1/(2 * π * 40 Hz * 10 * 10^{-6} F)= 400 ohm.

Z_{2} = 1/(2 * π * 40 Hz * 22 * 10^{-6} F)= 180 ohm.

Then,

V_{out} = = 10 V = 0.31 * 10 V = 3.1 V

V_{out} = 3.1 V.

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