Question

Let us assume that the old - fashioned serial lights were connected in a series across a $240V$ household line. If a string of these lights consists of twelve bulbs, calculate the potential difference across each bulb? When the bulbs were connected in parallel, calculate the potential difference across each bulb?

Answer 1

Hint: The potential difference of the source will get equally divided among the same resistance valued components in a circuit, if they are connected in series. And if they are in parallel connection, then the potential difference will be the same as in that of the source. This will help you in answering this question.

Complete step by step answer:
It has been mentioned in the question that the old - fashioned serial lights were connected in a series way across a $240V$ household line. From this we have to find the potential difference across each of the bulbs, if a string of these lights are included of twelve bulbs. And also the potential difference across each of the bulbs, if the bulbs were connected in parallel.
Let us assume that ${{V}_{1}}$​ be the potential difference across each of the bulb in series,
Therefore in accordance to the given situation, we can write that,
$\begin{align}
  & 12{{V}_{1}}=240V \\
 & \Rightarrow {{V}_{1}}=\dfrac{240}{12} \\
 & \therefore {{V}_{1}}=20V \\
\end{align}$
Therefore the potential difference across each of the bulbs also will be $20V$, if these twelve bulbs are connected in series. If the twelve bulbs are connected in parallel, then the potential difference across each of the bulbs will be equivalent to the total voltage.
Therefore the potential difference across each of the bulbs will also be $240V$.

Note: When the components with some specific resistances are connected in parallel in a circuit, then the current will get divided in that circuit in accordance with the value of each resistances. If the resistors with the same resistance are connected in a series way, then their value of the current in each of the components will be the same itself.