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How can three resistors of resistances 2 Ω , 3 Ω , and 6 Ω be connected to give a total resistance of (a) 4 Ω , (b) 1 Ω?

How can three resistors of resistances 2 Ω , 3 Ω , and 6 Ω be connected to give a total resistance of (a) 4 Ω , (b) 1 Ω?

Grade:12

1 Answers

Harshit Singh
askIITians Faculty 5963 Points
3 years ago
Dear Student

R1 = 2 ohm
R2 = 3 ohm
R3 = 6 ohm
(a)When R2 and R3 are connected in parallel with R1 in series we get
1/R = 1/R2 + 1/R3
g= 1/3 + 1/6
= 1/2
Thus, R = 2 ohm
Resistance in series = R + R1
= 2 + 2
= 4 ohm
(b)When R1,R2, R3 are connected in parallel we get
1/R = 1/R1 + 1/R2 + 1/R3
= 1/2 + 1/3 + 1/6
= 1 ohm.

Thanks

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