Question

Bulbs A and B in FIGURE Q28.11 are identical, and both are glowing. Bulb B is removed from its socket. Does the potential difference $∆V_{12}$between points 1 and 2 increase, stay the same, decrease, or become zero? Explain.

Short answer

The potential difference is$\mathbf{∆}{\mathit{V}}_{\mathbf{12}\mathbf{}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}$

Step-by-step solution

Given information and formula used 

Given :

Theory used:

The potential difference is equal to the amount of current multiplied by the resistance. A potential difference of one Volt is equal to one Joule of energy being used by one Coulomb of charge when it flows between two points in a circuit.

Determining if the potential difference between points 1 and 2 increase, stay the same, decrease, or become zero 

Initially, the current flows between the two lights due to the battery's emf. Across bulb B, the voltage is greater than zero. When a bulb is removed from its socket, it takes on the appearance of a wire.

As a result, there is no possible difference between 1 and 2. That is,

$\mathbf{∆}{\mathit{V}}_{\mathbf{12}\mathbf{}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}$