Problem with Voltage Dividers

ricardo7890

i know this is a very basic question, but there's something i don't understand about voltage dividers.V_out is supposed to be(v2/V2+v1)* V, or the voltage drop across a resistor, but i don't see how this works in practice. When you put a load across the output, this changes the voltage drop of all the resistors in the circuit, so how is a voltage divider useful if you can't have a constant out value.

gnurf

Quote by ricardo7890

When you put a load across the output, this changes the voltage drop of all the resistors in the circuit, so how is a voltage divider useful if you can't have a constant out value.

It's "constant" within limits. Quiz-Question: For V_in=10V and a desired V_out=5V, how would you design the voltage divider if you want to stay within, say, 5% your nominal value when you connect a load of 1kΩ?

Runei

Also, they are used when you need to give an input signal to a high impedance load. For example, if you have 5V and need a reference voltage of 3.3V to, say, an OP-AMP, you could use a voltage divider to get those 3.3V.

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Runei	Also, they are used when you need to give an input signal to a high impedance load. For example, if you have 5V and need a reference voltage of 3.3V to, say, an OP-AMP, you could use a voltage divider to get those 3.3V.