Question

# Three resistances of one ohm each are connected in parallel. Such connection is again connected with $$2 / 3\Omega$$ resistor in series. The resultant resistance will be

A
53Ω
B
32Ω
C
1Ω
D
23Ω

Solution

## The correct option is D $$1\Omega$$For parallel combination of n equal resistance R, each  $$R_{eq}=\dfrac{R}{n}$$ Resistance of $$1$$ ohm group $$=\dfrac{R}{n}=\dfrac{1}{3}\Omega$$This is in series with $$\dfrac{2}{3}\Omega$$ resistor$$\therefore$$ Total resistance $$=\dfrac{2}{3}+\dfrac{1}{3}=\dfrac{3}{3}\Omega=1\Omega$$Physics

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