If the resistance of a cathodic protection system doubles while the voltage output remains constant, what would the new current output be?

To determine how the current output changes when the resistance of a cathodic protection system doubles while keeping the voltage constant, we can utilize Ohm's Law, which states that voltage (V) is equal to the current (I) multiplied by resistance (R). This can be expressed with the formula:

[ V = I \times R ]

If the voltage remains constant and the resistance doubles, we can analyze the impact on current.

Let’s assume the original resistance is ( R ) and the original current is ( I ). According to Ohm’s Law, the initial conditions can be written as:

[ V = I \times R ]

When the resistance doubles, it becomes ( 2R ). Since the voltage doesn't change, we can express the new current (( I' )) with the modified resistance:

[ V = I' \times (2R) ]

Since V is the same, we set the two equations equal to each other:

[ I \times R = I' \times (2R) ]

By simplifying this equation, we can isolate the new current:

[ I' = \frac{I}{2} ]

This tells us that the new current output is half