Open Question: How to find current and power in resister that's connected parallel and in series?
(a) Find the current in the 121.0-O resistor.
(b) Find the power dissipated in the 121.0-O resistor.
Open Question: What is the resistance of the light bulb when wired in series with 144-Ω resistor?
lower value O
higher value OAnswer QuestionBe the first to answer this question.
Open Question: Can I use the integral test for the series cos n/n?
Open Question: Is voltage and resistance shared in series circuits?
series:
Total resistance = individual resistance of each component added
total voltage = individual voltage of each component added
current = same everywhere
parallel:
total resistance = ????????????
total voltage = same everywhere
total current = individual current of each component added
is this correct? and what is the total resistance for parallel circuits? thanks!!!!
Open Question: Maths problem- Geometric series?
ar^4(1+r) = 4375 .... [ II ]
[ II ] / [ I ]
r^3 = 4375/280 which yields
r = 2.5
substituting in [ I ] yields a = 32
ans: a = 32, r = 2.5
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Open Question: Determine whether the series is convergent or divergent?
k/k^2 reduces to 1/k
as the limit approaches infinity you get 1/2,1/3,1/4,1/5.....1/x so you see the number is getting smaller and smaller, thus the limit approaches 0. Since the series approaches a finite number the series converges. If it went to infinity it would diverge.
Open Question: Taylor Series Problem?
Evaluate this limit using Taylor series:
lim x-->0 ((2cos2x - 2 + 4x^2)/2x^4) . This is 0/0 and you can use l'Hopital's rule but I am not supposed to as I am supposed to use Taylor series. Can anyone help me with a step-by-step?
Open Question: How do you find the power series representation for the following function?
f(x)=x^2 / ( (1-2x)^2 )
Thank you for any help
Open Question: series n^1/n converge or diverge?
Open Question: Infinite series --> converging or diverging and if converges --> find sum.?
lim (n-->infinity) ln(n + 1)/ln(n^3 + 2)
= lim (n-->infinity) ln[n(1 + 1/n)]/ln[n^3(1 + 2/n^2)]
= lim (n-->infinity) [ln(n) + ln(1 + 1/n)]/[3ln(n) + ln(1 + 2/n^2)]
= lim (n-->infinity) [1 + ln(1 + 1/n)/ln(n)]/[3 + ln(1 + 2/n^2)/ln(n)]
= (1 + 0)/(3 + 0)
= 1/3.
Since this limit does not converge to zero, this series is divergent by the limit test.
I hope this helps!
