Question
Calculate equivalent resistance in the following cases:

Answer

(i) Resistance P and Q are in series = 3 Ω + 3 Ω = 6 Ω
Resistance of 6 Ω segment and 3 Ω are parallel.
$\therefore \frac{1}{ R }=\frac{1}{6}+\frac{1}{3}=\frac{1+2}{6}=\frac{3}{6}$ or $R =2 \Omega$
(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.
$\frac{1}{ R }=\frac{1}{4}+\frac{1}{12}$
$=\frac{3+1}{12}=\frac{4}{12}$
or R = 3Ω

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