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Question

Show how you would connect three resistors, each of resistance $$6\Omega$$, so that the combination has a resistance of
(i) $$9\Omega$$,
(ii) $$4\Omega$$. 


Solution

If resistor are connected n series $$6\Omega+6\Omega+6\Omega=18\Omega$$ This is not correct 
When they are connected in parallel
$$\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=3$$ This is also wrong
(i)When they are connected in parallel
Two $$6\Omega$$ resistors are connected in parallel.
$$Resistance=\frac{1}{\frac{1}{6}+\frac{1}{6}}$$
$$=\frac{6\times 6}{6+6}=3\Omega$$
If $$3rd$$ resistor of $$6\Omega$$ and $$3\Omega$$ are connected in series, it becomes $$6\Omega+3\Omega=9\Omega$$
(ii)When they are connected in series
$$Resistance=6\Omega+6\Omega$$
$$=12\Omega$$
If $$3rd$$ resistor $$6\Omega$$ and is connected to $$12\Omega$$ in parallel
Total resistance =$$4\Omega$$

Physics

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