What is, highest, and lowest, resistance which can be obtained by combining Com resistors having the following resistances?
$4 Ω, 8 Ω, 12 Ω, 24 Ω$
$4 Ω, 8 Ω, 12 Ω, 24 Ω$
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For obtaining the highest resistance by combining the given resistance, we must connect them in series.
We get,
R = 4 + 8 + 12 + 24 = 48 ohms
For obtaining the lowest resistance by combining the given resistance, we must connect them parallel.
We get,
$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24 }$
On solving we get, R = 2 ohms.
We get,
R = 4 + 8 + 12 + 24 = 48 ohms
For obtaining the lowest resistance by combining the given resistance, we must connect them parallel.
We get,
$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24 }$
On solving we get, R = 2 ohms.
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