what is the voltage across R252 & R253. Please do the needful asap. Thanks in advance!!

ericgibbs

Joined Jan 29, 2010
18,734
hi,
For two series resistors, Vjunc = Vin * Rtop/[Rtop +Rbot] ,assuming the junction is not loaded with another resistance.

Is this Homework.?
 

Thread Starter

veeresh540

Joined Dec 17, 2014
13
hi,
For two series resistors, Vjunc = Vin * Rtop/[Rtop +Rbot] ,assuming the junction is not loaded with another resistance.

Is this Homework.?
Hi,
Thanks for the quick response. Its required for my project. So,I need the voltage values across them in order to verify with mine. Please do the needful!!
 

ericgibbs

Joined Jan 29, 2010
18,734
Hi.
Vout = Vin * 100k/[100k+10k] , so Vo = 5v * [100k/110k], = 5v * 0.909' = 4.54' Volts.

This of course assumes the ON pin has a very high input impedance [ resistance]

Does this help.?
E
 

Thread Starter

veeresh540

Joined Dec 17, 2014
13
hi,
Voltage at the R252 is 5V only right. The circuit acts as voltage divider and we will get voltage divider output at the R253 and after the resistor R252.
So for calculating the power dissipation of R252 we should consider the voltage passing through the R252.
We will get potential divider output after R252.

Is my analysis right?? Please clarify me!!
 

ericgibbs

Joined Jan 29, 2010
18,734
hi,
The current thru the 10K and 100K in series is 5v/110k is 45 micro amps!
So the voltage drop across the 10K is, 0.000045 *10,000 = 0.45'v and the 100k is 0.000045*100,000 = 4.54'v

Dissipation 10k = 0.000045 * 0.45 = 20 micro Watt and the 100k= 0.000045 * 4.54 = 204 micro Watts

BTW:
the voltage passing is incorrect, the voltage is across a resistor, the current 'passes' , flows thru a resistor.
 
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