Three identical resistors are first connected in parallel and then in series. The resultant resistance of the first combination to the second will be

9 times
$$rac{1}{9}$$ times
$$rac{1}{3}$$ times
3 times

The correct answer is: C. $\frac{1}{3}$ times

When three identical resistors are connected in parallel, the equivalent resistance is given by:

$$R_{eq} = \frac{R}{3}$$

When three identical resistors are connected in series, the equivalent resistance is given by:

$$R_{eq} = 3R$$

Therefore, the resultant resistance of the first combination to the second is:

$$\frac{R}{3} \div 3R = \frac{1}{3}$$


Explanation of each option:

  • Option A: 9 times. This is incorrect because the resultant resistance of the first combination is $\frac{R}{3}$, which is 1/3 of the resistance of the second combination.
  • Option B: $\frac{1}{9}$ times. This is incorrect because the resultant resistance of the first combination is $\frac{R}{3}$, which is 3 times the resistance of the second combination.
  • Option C: $\frac{1}{3}$ times. This is the correct answer because the resultant resistance of the first combination is $\frac{R}{3}$, which is 1/3 of the resistance of the second combination.
  • Option D: 3 times. This is incorrect because the resultant resistance of the first combination is $\frac{R}{3}$, which is 1/3 of the resistance of the second combination.
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