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.
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 Bleeder current.
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.
A voltage divider in which the base current is small compared to the current in R2 (resistor in other path to ground) is said to be a stiff voltage divider because the voltage is relatively independent of different transistors and temperature effects.
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.