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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
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B
32Ω
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C
1Ω
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D
23Ω
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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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