I?=I source(R equivalent / R?)
Ix = IT(Rp/Rx+Rp) where Ix is the current you are trying to find, IT is the total current, Rx is the resistor in question, Rp is/are the resistor(s) in parallel with the resistor in question.
The current that flows through an unloaded voltage divider is very small, close to zero. This is because there is no load connected to the output of the divider, so there is nowhere for the current to flow. The purpose of a voltage divider is to divide the input voltage between the two resistors, not to pass current.
Boss its a circuit not a device, you can also create one of yours..... just use simple logic of voltage divider and current divider rules...-satendra.svnit@gmail.com
A: that is true for less current a divider is OK it has to do with the series resistance and loading if the loading is forever fixed and the source is also fixed at a value then a divider can be used no matter what the current is.
The current should be high enough to maintain the voltage at each division. Generally, the current should be ten times the load current or the voltage will be across the voltage divider. If possible use regulators or zeners or regular diodes. I'm not saying dividers are bad but there are less current sensitive solutions.
beacause ammeter is use to measure the current not to measure the voltage
obstacle and divider
You get proportional (equal) amp pressure throughout.
Use a less powerful filter, and put one on each side.
a number that divide an other number
first we will put the divider in the starting of line then we will make points than we can measure it.
When a load is connected to the output of a voltage divider, the output voltage will typically decrease due to the loading effect. This occurs because the load draws current, which can change the voltage across the resistors in the divider. The extent of the voltage drop depends on the resistance of the load relative to the resistors in the voltage divider. If the load resistance is significantly lower than the divider resistances, the output voltage will drop more noticeably.