A Hypothetical RC Circuit

Electricity and Magnetism Level 3

R ( a ) = ∑ n = 1 ∞ 1 a n = 1 a − 1 C ( a ) = ∑ n = 0 ∞ 1 a n = a a − 1 R(a) = \sum_{n=1}^{\infty} \frac{1}{a^n} = \frac{1}{a-1} \qquad \qquad C(a) = \sum_{n=0}^{\infty} \frac{1}{a^n} = \frac{a}{a-1}

An RC ciruit consisting of a multitude of resistors ( R R ) in series followed by a multitude of capacitors ( C C ) in parallel is prepared such that the resistance and capacitance can be calculated with the series written above. At a = 3 a= 3 , calculate the voltage ( V V ) of the capacitor when it has charged for t = 2.5 seconds t = 2.5 \space \text{seconds} , where the circuit's power supply voltage ( V ∘ V_{\circ} ) is 14 volts 14 \space \text{volts} .

Round your answer to three significant figures.

Note : V ( t ) = V ∘ ( 1 − e N ) V(t) = V_{\circ}(1-e^{N}) where N = -t RC N = \frac{\text{-t}}{\text{RC}} .


The answer is 13.5.

David Hontz by David Hontz , 26, USA

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2 solutions

David Hontz
Jun 15, 2016

R ( 3 ) = 1 3 − 1 = 1 2 C ( 3 ) = 3 3 − 1 = 3 2 N = − 2.5 1 2 3 2 = − 2.5 3 4 = − 10 3 V ( 2.5 ) = 14 ( 1 − e − 10 3 ) ≈ 13.50056 ⇒ V ( 2.5 ) = 13.5 volts R(3) = \frac{1}{3-1} = \frac{1}{2} \\ C(3) = \frac{3}{3-1} = \frac{3}{2} \\ N = \frac{-2.5}{\frac{1}{2} \frac{3}{2}} = \frac{-2.5}{\frac{3}{4}} = \frac{-10}{3} \\ V(2.5) = 14(1 - e^{\frac{-10}{3}}) \approx 13.50056 \Rightarrow V(2.5) = \boxed{13.5 \space \text{volts}}

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Eva Tba
Jun 18, 2016

Hi all,

Could anyone please explain the function of the two summs in it? Why is the summ of 1/a^n for n=1-infinity 1/a-1? Wouldn't the summ asymptotically approach 0? an infinit amount of 1.. in the nominator and a quicker ascending a^1 a^2 a^3 a^4 a^5 ..a^infinity in the denomimator.

Thank you, Eva

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Let's look at the following sum: ∑ n = 0 ∞ 1 2 n \sum_{n=0}^{\infty} \frac{1}{2^n} If you treat the sum like a sequence, then a 0 , a 1 , a 2 = 1 2 0 , 1 2 1 , 1 2 2 = 1 , 1 2 , 1 4 a_0,a_1,a_2=\frac{1}{2^0},\frac{1}{2^1},\frac{1}{2^2} = 1, \frac{1}{2}, \frac{1}{4} It becomes clear that as the sequence continues on for higher values of n n , the sequence approaches 0 0 .

In the problem above, we are dealing with series and n o t not sequences. Let's treat the same sum as a series: ∑ n = 0 ∞ 1 2 n \sum_{n=0}^{\infty} \frac{1}{2^n} The first three terms of the series would be: S 0 = a 0 = 1 S 1 = a 0 + a 1 = 1.5 S 2 = a 0 + a 1 + a 2 = 1.75 S_0 = a_0 = 1 \\ S_1=a_0+a_1=1.5 \\ S_2=a_0+a_1+a_2=1.75 It becomes clear that as n n approaches ∞ \infty , smaller and smaller pieces (terms) are summed together and the series approaches the value of 2 2 . You should notice that when a = 2 a=2 : ∑ n = 0 ∞ 1 a n = ∑ n = 0 ∞ 1 2 n = 2 = 2 2 − 1 = a a − 1 \sum_{n=0}^{\infty} \frac{1}{a^n} = \sum_{n=0}^{\infty} \frac{1}{2^n} = 2 = \frac{2}{2-1} = \frac{a}{a-1} Let's also look at a similar series with a = 2 a=2 : ∑ n = 1 ∞ 1 a n = ∑ n = 1 ∞ 1 2 n = 0.5 + 0.25 + 0.125... = 1 = 1 2 − 1 = 1 a − 1 \sum_{n=1}^{\infty} \frac{1}{a^n} = \sum_{n=1}^{\infty} \frac{1}{2^n} = 0.5+0.25+0.125 ... = 1 = \frac{1}{2-1}=\frac{1}{a-1}

David Hontz - 4 years, 12 months ago

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Thank you for the in depth reply, how do I know from the question that it is a series and not a sequence?

eva tba - 4 years, 12 months ago

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Fair question. I'll add in the question that the summations are series; however, I have in the problem what each approaches therefore knowing whether the sums are sequences or series is not needed to answer the problem.

David Hontz - 4 years, 12 months ago

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@David Hontz – Sure, I got caught in these sums. But in fairness. If you had written R(a)= 1/a-1 for a=3 this would have hardly been a level 4 :)

I am sure you can help me with this too. How do I know that the series approaches 1, unless I put it in the calculator. Could you please prove for the Sum of 1/a^n for n=1-infinity is 1/a-1.

Thank you

eva tba - 4 years, 12 months ago

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@Eva Tba – When you have a fraction in the form of 1 a \frac{1}{a} , it is well-known that the following series approaches as follows: ∑ n = 0 ∞ ( 1 a ) n = 1 1 − 1 a = 1 a − 1 a = a a − 1 \sum_{n=0}^{\infty} \Big( \frac{1}{a} \Big)^n = \frac{1}{1 - \frac{1}{a}} = \frac{1}{\frac{a-1}{a}} = \frac{a}{a-1} Now to answer your question, we will need to use some properties of summations. ∑ n = 1 ∞ ( 1 a ) n = ∑ n = 0 ∞ ( 1 a ) n − ∑ n = 0 0 ( 1 a ) n = a a − 1 − 1 a 0 = a a − 1 − 1 1 = a − ( a − 1 ) a − 1 = a − a + 1 a − 1 = 1 a − 1 \sum_{n=1}^{\infty} \Big( \frac{1}{a} \Big)^n = \sum_{n=0}^{\infty} \Big( \frac{1}{a} \Big)^n - \sum_{n=0}^{0} \Big( \frac{1}{a} \Big)^n = \frac{a}{a-1} - \frac{1}{a^0} = \frac{a}{a-1} - \frac{1}{1} = \frac{a - (a-1)}{a-1} = \frac{a-a+1}{a-1} = \frac{1}{a-1} Also, the question is dealing with a circuit consisting of an infinite amount of resistors and capacitors, so series are appropriate for the situation; but that's only my opinion, of course.

David Hontz - 4 years, 12 months ago

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@David Hontz – I just wanted to write "thank you" for the explanation... however the app complains thats not complex enough.

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eva tba - 4 years, 12 months ago

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