Forums
New posts
Search forums
What's new
New posts
New media
New media comments
Latest activity
Media
New media
New comments
Search media
Members
Current visitors
Log in
Register
What's new
Search
Search
Search titles only
By:
New posts
Search forums
Menu
Log in
Register
Install the app
Install
Main Forums
Do-It-Yourself (DIY) Question & Answer Forum
Why doesn't RCD trip?
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Reply to thread
Help Support Talk Electrician Forum:
This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.
Message
<blockquote data-quote="SPECIAL LOCATION" data-source="post: 498018" data-attributes="member: 250"><p>Draw it out as a simple circuit.... (Ignore the Live side).</p><p></p><p>N coming back from the load through, MCB &amp; RCD (or RCBO) then through main switch then making its way back to supply to re-join Earth at the Star point.. </p><p></p><p>with a parallel path somewhere part way down this N wire, via a 2.7k fault, back to the installation main earth. (And if this is TN-C-S this is back onto the N wire incoming side of supply)..</p><p></p><p>So as ProDave said... the electrical resistance straight down the normal neutral path will be so low that hardly any current will flow down your 2.7k fault path...</p><p></p><p>or think of it as a voltage divider...</p><p></p><p>where your neutral current reaches a fault with one half down a copper conductor the other half down a 2.7k fault....</p><p></p><p>for the sake of arguments imagine the two parallel paths are 2700ohms and 0.35ohms..</p><p></p><p>calculate the voltage split and thus current you would expect down your neutral fault...</p><p></p><p>Whereas a fault on the live side would be splitting between 2.7k and the load resistance back to N..</p><p></p><p>Its one of those things that if you draw it out with the load and fault as simple resistors on a basic series/parallel circuit, and blocks indicating the supply/RCD/MCB etc..</p><p></p><p>It can make more sense than your brains logic saying Oh I have a low resistance on my IR tests this should trip the RCD!!!</p><p></p><p> Guinness</p></blockquote><p></p>
[QUOTE="SPECIAL LOCATION, post: 498018, member: 250"] Draw it out as a simple circuit.... (Ignore the Live side). N coming back from the load through, MCB & RCD (or RCBO) then through main switch then making its way back to supply to re-join Earth at the Star point.. with a parallel path somewhere part way down this N wire, via a 2.7k fault, back to the installation main earth. (And if this is TN-C-S this is back onto the N wire incoming side of supply).. So as ProDave said... the electrical resistance straight down the normal neutral path will be so low that hardly any current will flow down your 2.7k fault path... or think of it as a voltage divider... where your neutral current reaches a fault with one half down a copper conductor the other half down a 2.7k fault.... for the sake of arguments imagine the two parallel paths are 2700ohms and 0.35ohms.. calculate the voltage split and thus current you would expect down your neutral fault... Whereas a fault on the live side would be splitting between 2.7k and the load resistance back to N.. Its one of those things that if you draw it out with the load and fault as simple resistors on a basic series/parallel circuit, and blocks indicating the supply/RCD/MCB etc.. It can make more sense than your brains logic saying Oh I have a low resistance on my IR tests this should trip the RCD!!! Guinness [/QUOTE]
Insert quotes…
Verification
Post reply
Main Forums
Do-It-Yourself (DIY) Question & Answer Forum
Why doesn't RCD trip?
Top