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Python Sqlite OpenRead Error: OperationalError

I’m trying to open a sqlite table in Python 3.3 and I receive an error as follows:
OperationalError: cannot open file ‘test.sqlite’ for reading
I’ve checked the file location and it’s present. I’ve also tried to use a hard-coded path and still receive the same error.
import sqlite3
sqlite_filepath = r’C:\Users\rw\Desktop\test.sqlite’

db_handle = sqlite3.connect(sqlite_filepath)
cursor = db_handle.cursor()
cursor.execute(‘select * from test’)

Is there something special I need to do with a sqlite3 cursor to be able to read from a db?
EDIT
Added code that prints the error that I receive, including the path:
Traceback (most recent call last):
File “main.py”, line 7, in
cursor.execute(‘select * from test’)
sqlite3.OperationalError: cannot open file ‘C:\\Users\\rw\\Desktop\\test.sqlite’ for reading

A:

There is a typo in the path. Try r’C:\Users\rw\
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