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Is it true that the total resistance of a circuit is $9 Ω$ if two resistors of $6 Ω$ are in parallel and a

third resistor of $6 Ω$ is in series with them?

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True

False

answer is A.

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Detailed Solution

The above statement is True.
If the resistance are connected in parallel then, the equivalent resistance of the resistances can be calculate by the given equation,

\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}

……………. (i) (

R_{P}

= resistance in parallel)
Let

R_{1} = 6 Ω

and

R_{2} = 6 Ω

Now put the above values in equation (i) we get,

\frac{1}{R_{P}} = \frac{1}{6} + \frac{1}{6}

\Rightarrow \frac{1}{R_{P}} = \frac{1 + 1}{6}

\Rightarrow \frac{1}{R_{P}} = \frac{2}{6}

\Rightarrow

\frac{1}{R_{P}} = \frac{1}{3}

\Rightarrow

R_{P} = 3 Ω

…………….. (ii)
Since a

6 Ω

resistance is in series with the resistances

R_{P}

3 Ω

and

6 Ω

. Therefore, the total resistance (

R_{s}

) is:

= > R_{s} = R_{p} + R_{3}

= > R_{s} = 3 Ω + 6 Ω

= > R_{s} = 9 Ω

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