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 parallelTwo $$6\Omega$$ 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 parallelTotal resistance =$$4\Omega$$Physics

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