From 262342dd7eb68d0c6cdd9ef1229b426398e0052c Mon Sep 17 00:00:00 2001
From: Lukas Schnelle <33157020+schnellecom@users.noreply.github.com>
Date: Sun, 22 Dec 2024 13:40:30 +0100
Subject: [PATCH 1/2] Correct mA to A conversion in
 Power-Monitoring-Calibration.md

Also rewrite some of the calculation formulas for clarity.
---
 docs/Power-Monitoring-Calibration.md | 13 +++++++++----
 1 file changed, 9 insertions(+), 4 deletions(-)

diff --git a/docs/Power-Monitoring-Calibration.md b/docs/Power-Monitoring-Calibration.md
index 00d5016fe5..becfcd7955 100644
--- a/docs/Power-Monitoring-Calibration.md
+++ b/docs/Power-Monitoring-Calibration.md
@@ -35,16 +35,21 @@
    [`VoltageSet <voltage>`](Commands.md#voltageset)  
    _Replace `<voltage>` with your standard voltage or with reading on your multi-meter if you have one. Your voltage will vary depending on the electrical standards and your electrical grid_  
 
-3. Verify the **Current** reading by calculating current value (amperage) using this formula: **P<sub>(W)</sub>/V<sub>(V)</sub>=I<sub>(A)</sub>**. Adjust the current offset if needed (in milliAmps (mA=A\*1000)):  
+3. Verify the **Current** reading by calculating current value (amperage) using this formula: **P<sub>(W)</sub>/V<sub>(V)</sub>=I<sub>(A)</sub>**. Adjust the current offset if needed (in milliAmps ($1000mA=1A$)):  
    [`CurrentSet <current>`](Commands.md#currentset)  
    _Replace `<current>` with your calculated value (in milliAmps)_  
 
    `CurrentSet` calculation:   
-   P/V=I
-   1000 * Watts/Volts = milliAmperes
+   $$\frac{Watts}{Volts}=Amps$$
+   
+   and thus
+   
+   $$1000 \* \frac{Watts}{Volts} = milliAmps$$
 
 !!! example
-     1000*(60.0/235.5) = 254.777
+     Consider a power usage of $60W$ at a voltage of $240V$. Then the current in milliAmps can be calculated as
+     
+     $$1000 \* \frac{60}{240} = 240.0$$
 
 4. Confirm the validity of your calibration process checking `Power Factor` from the web UI which should be as close as possible to `1.00`. In theory resistive loads will always provide a power factor of 1.00. If that is not the case, we recommend you repeat the calibration process and make sure everything was done correctly. 
    

From 472c145c7e7916f0207cc107ea6964dbc7bc6972 Mon Sep 17 00:00:00 2001
From: Lukas Schnelle <33157020+schnellecom@users.noreply.github.com>
Date: Mon, 23 Dec 2024 11:01:07 +0100
Subject: [PATCH 2/2] Update Power-Monitoring-Calibration.md

Fix the error I added in the example
---
 docs/Power-Monitoring-Calibration.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/docs/Power-Monitoring-Calibration.md b/docs/Power-Monitoring-Calibration.md
index becfcd7955..5b251745ca 100644
--- a/docs/Power-Monitoring-Calibration.md
+++ b/docs/Power-Monitoring-Calibration.md
@@ -49,7 +49,7 @@
 !!! example
      Consider a power usage of $60W$ at a voltage of $240V$. Then the current in milliAmps can be calculated as
      
-     $$1000 \* \frac{60}{240} = 240.0$$
+     $$1000 \* \frac{60}{240} = 250.0$$
 
 4. Confirm the validity of your calibration process checking `Power Factor` from the web UI which should be as close as possible to `1.00`. In theory resistive loads will always provide a power factor of 1.00. If that is not the case, we recommend you repeat the calibration process and make sure everything was done correctly.