ICSE Class 10 Answered
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Asked by roshanisaywani | 04 Apr, 2022, 02:05: PM
Expert Answer
(a) Two resistors are in series , hence equivalent resistance = 2Ω + 2Ω = 4 Ω
(b) Two resistors are in parallel , hence equivalent resistance = ( 2 × 2 ) / ( 2+ 2 ) = 1Ω
(c) two 2 Ω resistors that are connected in series is parallel to 4Ω resistor , hence equivalenet resistance = ( 4 × 2 ) / ( 4+ 2 ) = 1.333 Ω
(d) paralel combination of two series resistors , hence equivalent resistance = ( 4 × 4 ) / ( 4 + 4 ) = 2 Ω
(e) The connection is similar to part (c) , hence equivalenet resistance = 1.333 Ω
(f) Equivalent resistance of Parallel combination of 4Ω and 2Ω is 1.333 Ω .
Hence given network of resistors is equivalent to series combination of 1.333 Ω, 2 Ω and 1.333 Ω .
Effective resistance between A and B is ( 1.333 Ω + 2 Ω + 1.333 Ω ) = 4.666 Ω
(g) Series combination five resistors , each of value 2Ω
Equivalent resistance = 2 Ω + 2 Ω + 2 Ω + 2 Ω +2 Ω = 10 Ω
Answered by Thiyagarajan K | 05 Apr, 2022, 09:20: PM
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