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* fix: clean-up Project Euler 141-160 * fix: corrections from review Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com> * fix: corrections from review Co-authored-by: Tom <20648924+moT01@users.noreply.github.com> * fix: use different notation for consistency * Update curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-144-investigating-multiple-reflections-of-a-laser-beam.md Co-authored-by: gikf <60067306+gikf@users.noreply.github.com> Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com> Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
51 lines
1.8 KiB
Markdown
51 lines
1.8 KiB
Markdown
---
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id: 5900f4081000cf542c50ff1a
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title: 'Problem 155: Counting Capacitor Circuits'
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challengeType: 5
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forumTopicId: 301786
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dashedName: problem-155-counting-capacitor-circuits
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---
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# --description--
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An electric circuit uses exclusively identical capacitors of the same value C.
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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.
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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 $60 μF$ each, we can obtain the following 7 distinct total capacitance values:
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<img class="img-responsive center-block" alt="example circuits having up to three capacitors, each of 60 μF" src="https://cdn.freecodecamp.org/curriculum/project-euler/counting-capacitor-circuits.gif" style="background-color: white; padding: 10px;">
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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: $D(1) = 1, D(2) = 3, D(3)=7, \ldots$
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Find $D(18)$.
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Reminder: When connecting capacitors $C_1$, $C_2$ etc in parallel, the total capacitance is $C_T = C_1 + C_2 + \cdots$, whereas when connecting them in series, the overall capacitance is given by: $\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$.
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# --hints--
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`capacitanceValues()` should return `3857447`.
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```js
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assert.strictEqual(capacitanceValues(), 3857447);
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```
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# --seed--
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## --seed-contents--
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```js
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function capacitanceValues() {
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return true;
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}
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capacitanceValues();
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```
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# --solutions--
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```js
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// solution required
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```
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