<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.
The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)
The capacitance of a series combination is
1 / (1/A + 1/B + 1/C + 1/D + .....) .
If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is
(product of the 2 individuals) / (sum of the individuals) .
In this problem, we have a humongous one and a tiny one.
Let's call them 1000 and 1 .
Then the series combination is
(1000 x 1) / (1000 + 1)
= (1000) / (1001)
= 0.999 000 999 . . .
which is smaller than the smaller individual.
It'll always be that way. </span>
Clock wise idk i think you should double check my answer
Answer:
They arn't usually guesses, but they are well made theorys or explanations. Its a well-substantiated explanation of facts that have been confirmed in expirements.
Answer:
D
Explanation:
F = G m1 m2 / r^2 now double r
F = G m1m1/ (2r)^2
F = 1/4 G m1m2/r^2 <===== this is 1/4 of the original
