How much current is "seen" by the circuit breaker when the insulation on the "hot" wire is damaged and the impedance of the Sealtite® is 500 ohms?

To determine how much current is "seen" by the circuit breaker when the insulation on the "hot" wire is damaged, we can use Ohm's law, which states that voltage (V) equals current (I) multiplied by resistance (R), expressed as V = I x R. In this scenario, we need to calculate the current where the impedance (in this case, the resistance due to the damaged insulation) is 500 ohms.

First, it's essential to note that the standard voltage of a typical circuit in many installations is often 120 volts. Assuming this voltage is present, we can rearrange Ohm's law to solve for current: I = V / R.

Plugging in the values we have:

• Voltage (V) = 120 volts (standard for many residential circuits)

• Resistance (R) = 500 ohms

Using the formula:

I = 120 V / 500 ohms = 0.24 A

However, if we consider a higher voltage common in certain applications, such as 240 volts, using the same formula would give us:

I = 240 V / 500 ohms = 0.48 A

When the options are assessed, it seems like an oversight