Question
Download Solution PDFThree Identical bulbs are connected in series and these together dissipate power ‘M’. If now the bulbs are connected in parallel, then the power dissipated will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
When two or more electrical devices of the same voltage are connected in parallel connection, their equivalent power rating can be calculated as
Peq = P1 + P2 + P3 + ……
When ‘n’ number of devices of the same voltage and same power rating (P) are connected in parallel connection, their equivalent power rating can be calculated as
Peq = n P
When two or more electrical devices of the same voltage rating are connected in a series connection, their equivalent power rating can be calculated as
\(\frac{1}{{{P_{eq}}}} = \frac{1}{{{P_1}}} + \frac{1}{{{P_2}}} + \frac{1}{{{P_3}}} + \; \ldots \ldots .\)
When ‘n’ number of devices of the same voltage and same power rating (P) are connected in series connection, their equivalent power rating can be calculated as
\({P_{eq}} = \frac{P}{n}\)
Calculation:
When three identical bulbs are connected in series, their equivalent power is ‘M’
\({P_{eq}} = \frac{P}{n}\)
M = P/3
P = 3M
∴ The power rating of each bulb, P = 3 M
When these three bulbs are connected in parallel, then the equivalent power dissipated is
Peq = 3 × P
Last updated on Jun 11, 2025
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