1.6 KiB
id | challengeType | title | forumTopicId |
---|---|---|---|
5900f4081000cf542c50ff1a | 5 | Problem 155: Counting Capacitor Circuits | 301786 |
Description
The capacitors can be connected in series or in parallel to form sub-units, which can then be connected in series or in parallel with other capacitors or other sub-units to form larger sub-units, and so on up to a final circuit. Using this simple procedure and up to n identical capacitors, we can make circuits having a range of different total capacitances. For example, using up to n=3 capacitors of 60 F each, we can obtain the following 7 distinct total capacitance values:
If we denote by D(n) the number of distinct total capacitance values we can obtain when using up to n equal-valued capacitors and the simple procedure described above, we have: D(1)=1, D(2)=3, D(3)=7 ... Find D(18). Reminder : When connecting capacitors C1, C2 etc in parallel, the total capacitance is CT = C1 + C2 +...,
whereas when connecting them in series, the overall capacitance is given by:
Instructions
Tests
tests:
- text: <code>euler155()</code> should return 3857447.
testString: assert.strictEqual(euler155(), 3857447);
Challenge Seed
function euler155() {
return true;
}
euler155();
Solution
// solution required