Question

How will the potential difference across the resistors in a parallel circuit change if ONE of the bulbs burns out, and the current cannot pass through that path?

Answer 1

In a parallel circuit, the potential difference between each bulb is the same, that is.

$V=V_{1}=V_{2}...V_{n}$

When a pump burns, the resistance of the circuit changes, the current passing through the other bulbs remains the same

Example.

Imagine two bulbs that are connected in parallel, both with a resistor R connected to a potential difference V.

$V_{1}=V_{2}=V$

$R_{1}=R_{2}=R$

The electric current in each bulb using Ohm's Law is equal to

$I_{1}=I_{2}=\frac{V}{R}$

if a bulb is damaged the potential difference in the others remains V, because it is a characteristic of parallel circuits

In summary

The potential difference remains constant.