Question

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω , 8 Ω , 12 Ω , 24 Ω?

Answer 1

(a) The highest resistance is when the resistances are connected in series:                 R1 = 4 ohm                R2 = 8 ohm                 R3 = 12 ohm                R4 = 24 ohm           Total resistance in series = R1 + R2 + R3 + R4                 = 4 + 8 + 12 + 24                 = 48 ohm           Thus, highest resistance is 48 ohm.

          (b)  The lowest resistance is when the resistances are connected in parallel           Total resistance in parallel           = 1/R1 + 1/R2 + 1/R3 + 1/R4           1/R = 1/2 + 1/8 + 1/12 + 1/24           = 12/24           1/R = 1/2 ohm, R = 2 ohm           Thus, lowest resistance is 2 ohm.

Answer 2

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• $\rm\:R_1=4\Omega$

• $\rm\:R_2=8\Omega$

• $\rm\:R_3=12\Omega$

• $\rm\:R_4=24\Omega$

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• $\rm\:The\:highest\:resistance=?$

• $\rm\:The\:lowest\:resistance=?$

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$\rm\:(a)$ The highest resistance can be secured by connecting,all four coil in series.

$\tt\:In\:this\:case:-$

$\rm\:R=R_1+R_2+R_3+R_4$

$\rm\:R=4+8+12+24$

$\rm\:R=48\Omega$

Thus,the highest resistance which can be secured is 48 Ω.

$\rm\:(b)$ The lowest resistance can be secured by connecting all four coils in parallel.

$\tt\:In\:this\:case:-$

$\rm\:\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}$

$\rm\:\dfrac{1}{R}=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}$

$\rm\:\dfrac{1}{R}=\dfrac{6+3+2+1}{24}$

$\rm\:\dfrac{1}{R}=\cancel\dfrac{12}{24}$

$\rm\:\dfrac{1}{R}=\dfrac{1}{2}$

$\rm\:R=2\Omega$

Thus,the lowest resistance which can be secured is 2 Ω.