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# Show how you would connect three resistors, each of resistance $6 \Omega$, so that the combination has a resistance of

$(i)$. $9 \Omega$,

$(ii)$. $4 \Omega$.

**Given: **Three resistors, each of $6\ Ω$ resistance.

**To do: **To connect the given three resistors in such a way so that the combination has a resistance of $(i).\ 9\ Ω$, $(ii).\ 4\ Ω$.

**Solution: **If we connect the three resistors in series, the equivalent resistance will be $=6\ Ω+6\ Ω+6\ Ω=18\ Ω$, that is wrong.

If we connect the three resistors in parallel, the equivalent resistance will be $=\frac{1}{\frac{1}{6\ Ω}+\frac{1}{6\ Ω}+\frac{1}{6\ Ω}}$

$=\frac{1}{\frac{3}{6\ Ω}}$

$=2\ Ω$, that is also not the desired result.

**$(i)$. **If we connect two resistors in parallel,

Then, equivalent resistance $=\frac{1}{\frac{1}{6\ Ω}+\frac{1}{6\ Ω}}$

$=\frac{1}{\frac{2}{6}}$

$=3\ Ω$

Let\'s connect the third resistor of $6\ Ω$ in series with the combination of two paralleled resistors$(6\ Ω\ each)$.

Then, $R_{eq}=6\ Ω+3\ Ω=9\ Ω$

**$(ii)$. **If we connect the two resistors$(6\ Ω\ each)$ in series

Then equivalent resistance $=6\ Ω+6\ Ω=12\ Ω$

Now, let\'s connect the third resistor$(6\ Ω)$ in parallel with the series combination of two resistors$(6\ Ω\ each)$

Resultant resistance $R_{eq}=\frac{1}{\frac{1}{12\ Ω}+\frac{1}{6\ Ω}}$

$=\frac{1}{\frac{1}{4}}$

$=4\ Ω$

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