How can four resistors of resistances 4Ω, 8Ω, 12Ω and 24Ω be connected to get the highest resistance and lowest resistance respectively:

Engineering

Chemistry

Faraday Law

Question

All in series, all in parallel

All in parallel, all in series

$24Ω and 12Ω in series with the rest two in parallel, 4Ω and 8Ω in parallel with the rest two in series$

$4Ω and 8Ω in series with the rest two in parallel, 12Ω and 24Ω in parallel with the rest two in series$

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→ Given resistances:
Let (R₁= 4Ω, R₂= 8Ω.

R₃= 12Ω, R₄= 24Ω.
(A) For highest resistance, connect the given resistances in series combination:

R_Total = 4 + 8 + 12 + 24.
B] Lowest resistance is

∴ R_Total 48Ω.

(B) Lowest resistance is obtained when the given resistances are connected in parallel combination.

$\frac{1}{R_{Total}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}} .$
$∴ \frac{1}{R_{Total}} = \frac{1}{4} + \frac{1}{12} + \frac{1}{8} + \frac{1}{24}$

$∴ \frac{1}{R_{total}} = \frac{1}{2}$

∴ R_total = 2

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