Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in (i) series, and (ii) parallel. Draw the arrangements of…

Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in (i) series, and (ii) parallel. Draw the arrangements of the resistors and find the equivalent resistance of each arrangement.

Solution

(i) [circuit diagram: 2Ω, 3Ω, 6Ω resistors connected in series] In series, Rs = R1 + R2 + R3 = (2 + 3 + 6) Ω = 11 Ω (ii) [circuit diagram: 2Ω, 3Ω, 6Ω resistors connected in parallel] In parallel, $\dfrac{1}{Rp}$ = $\dfrac{1}{R1}$ + $\dfrac{1}{R2}$ + $\dfrac{1}{R3}$ = 1/2 + 1/3 +1/6 = (3+2+1)/6 Rp = 1.0 Ω

Asked in: CBSE

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