Powershell open Ps1 if true - powershell

Question

What am I doing wrong here? I am trying to open a second PS1 file if True but an error is thrown (shown below).
#ABOUT
#GLOBALS
$userID = "admin"
$pswrd = "test"
$name = Read-Host 'What is your username?'
$pass = Read-Host 'And your password?' #-AsSecureString | ConvertFrom-SecureString
#$script = '.\sdsSysMain.ps1'
if($name -eq $userID -and $pass -eq $pswrd) #or blank?
{
#write-host "Well done! You're in. "
#Start-Sleep -s 5
Powershell -noexit ".\sdsSysMain.ps1"
}
elseif($name -ne $userID -or $pass -ne $pswrd)
{
write-host "Login Failed... :("
}
This is the error I'm getting:
powershell.exe : The term '.\sdsSysMain.ps1' is not recognized as the name of a cmdlet, function
At C:\Users\1234\Documents\Projects\sdsSys\sdsSysLogin.ps1:19 char:15
+ Powershell <<<< -noexit ".\sdsSysMain.ps1"
+ CategoryInfo : NotSpecified: (The term '.\sds...mdlet, function:String) [],
RemoteException
+ FullyQualifiedErrorId : NativeCommandError
, script file, or operable program. Check the spelling of the name, or if a path was included,
verify that the path is correct and try again.
At line:1 char:17
+ .\sdsSysMain.ps1 <<<<
+ CategoryInfo : ObjectNotFound: (.\sdsSysMain.ps1:String) [], CommandNotFoundException
+ FullyQualifiedErrorId : CommandNotFoundException

Answer 1

The error is rather self-explanatory. The script doesn't find sdsSysMain.ps1 in the current working directory. You can output the current working directory in your script like this:
(Get-Location).Path
If you want to run sdsSysMain.ps1 from the same directory as the script calling it, change this:
Powershell -noexit ".\sdsSysMain.ps1"
into this:
$scriptPath = Split-Path -Parent $script:MyInvocation.MyCommand.Path
Powershell -NoExit -File "$scriptPath\sdsSysMain.ps1"