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Question
Calculate equivalent resistance in the following cases:
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Solution
(i) Resistance P and Q are in series = 3 Ω + 3 Ω = 6 Ω
Resistance of 6 Ω segment and 3 Ω are parallel.
∴ `1/"R" = 1/6 + 1/3 = (1 + 2)/6 = 3/6` or R = 2Ω
(ii) Resistance of the arm PS = 2Ω +2Ω = 4Ω (in series)
Resistance of the arm QR = 2Ω + 2Ω = 4Ω (in series)
Resistance of the arm PQ, QR and RS are in series
= (4 + 4 + 4)Ω = 12Ω
Now resistance of the arm RS and arms (PQ + QR + RS) are in parallel.
`1/"R" = 1/4 + 1/12`
`= (3 + 1)/12 = 4/12`
or R = 3Ω
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