arduino coding

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1

This is my arduino code to measure current and voltage across a circuit on breadboard which is connected to the external mobile battery of 3.8V. I am trying to measure the power consumption.I dont understand why my raw values are not getting converted to proper voltage and current values with correct decimal places. It is rounding up to nearest integer . I am attaching the screenshot of the serial view. please help me out

#include "arduino.h"
const int x=A5;
const int y=A2;
int Rawvalue=0;
int Rawvalue1=0;
float voltage;
float voltage1;
float current;
float power;

void setup()
{
Serial.begin(9600);
}
void loop()
{
Rawvalue=analogRead(x);
voltage=(Rawvalue *5)/1023;
Rawvalue1=analogRead(y);
voltage1=(Rawvalue1*5)/1023;
current=(voltage1/0.20);
Serial.print("Raw value= ");
Serial.println(Rawvalue);
Serial.print("Raw value1=");
Serial.println(Rawvalue1);
Serial.print("voltage= ");
Serial.println(voltage,3);
Serial.print("voltage1= ");
Serial.println(voltage1,3);
Serial.print("current= ");
Serial.println(current,3);
power=(voltage*current);
Serial.print("power=");
Serial.println(power,3);
delay(60000);
}

Screenshot (45).png
Screenshot (45).png1920×1080 126 KB

2

You should divide by 1024, not 1023.

Calculations are be default done with integer math. You can force use of float math by making any number in the calculation a float.

voltage=(Rawvalue *5.0)/1024;

By the way, how is this a "website and forum" question?

3

@SRameswari, please do not cross-post and please stop posting in these old threads (2015) as your issues have often been different from the original thread.
Also, another reason to not revive old threads like this (which is 3 years old) is that MANY things have changed over this time period. The solution today is not necessarily the same as it was back then.
Finally read how to use this forum and insert attached images.
I did it for you this time.