Physics
In the network shown in figure, calculate the equivalent resistance between the points (a) A and B (b) A and C

Current Electricity
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Answer
In the circuit, there are two parts. In the first part, three resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R's then
R's = (2 + 2 + 2) Ω = 6 Ω
In the second part, the resistance of first part (R's = 6 Ω ) and 2 Ω are connected in parallel. If the equivalent resistance of this part is Rp then
p} = \dfrac{1}{6} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1 + 3}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{4}{6} \\[0.5em] \Rightarrow Rp = 1.5 Ω \\[0.5em]
∴ Rp = 1.5 Ω
(b) In the circuit, there are three parts. In the first part, two resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R's then
R's = (2 + 2) Ω = 4 Ω
In the second part, two resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R''s then
R''s = (2 + 2) Ω = 4 Ω
In the third part, the two parts of resistance R's = 4 Ω and R''s = 4 Ω are connected in parallel. If the equivalent resistance between points A and C is Rp then
p} = \dfrac{1}{4} + \dfrac{1}{4} \\[0.5em] \dfrac{1}{Rp} = \dfrac{2}{4} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1}{2} \\[0.5em] \Rightarrow Rp = 2 Ω \\[0.5em]
∴ Equivalent Resistance between A and C = 2 Ω
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