MHB When the capacitor have 94% of its final voltage.

AI Thread Summary
To determine the time required to charge a 140 pF capacitor through a 77.0 MΩ resistor to 94% of its final voltage, the exponential charging formula is applied. The final voltage is denoted as V∞, and the voltage at time t is expressed as V(t) = V∞(1-e^(-t/RC)). When V(t) equals 0.94V∞, the equation simplifies to (1-e^(-t/RC)) = 0.94. By using logarithms, the time t can be calculated based on the RC time constant, which is 0.000108 seconds. The discussion emphasizes the correct interpretation of resistance values and the application of logarithmic functions in solving the problem.
cbarker1
Gold Member
MHB
Messages
345
Reaction score
23
Dear Everybody,

Using the exact exponential treatment, find how much time (in s) is required to charge an initially uncharged 140 pF capacitor (C) through a 77.0 M[FONT=&quot]Ω resistor (R) to 94.0% of its final voltage ($\varepsilon$).

Work:
C=140 pF=1.4x10^-10
R=77.0 Mohm=7.7x10^5
$q(t)=C\varepsilon(1-e^{-\frac{t}{RC}})$ or $I(t)=\frac{\varepsilon}{R}e^{\frac{-t}{RC}}$
RC=.000108

What am I missing?
.94*C or what is the battery?

Thanks
Carter Barker
 
Mathematics news on Phys.org
Cbarker1 said:
Dear Everybody,

Using the exact exponential treatment, find how much time (in s) is required to charge an initially uncharged 140 pF capacitor (C) through a 77.0 M resistor (R) to 94.0% of its final voltage ($\varepsilon$).

Work:
C=140 pF=1.4x10^-10
R=77.0 Mohm=7.7x10^5
$q(t)=C\varepsilon(1-e^{-\frac{t}{RC}})$ or $I(t)=\frac{\varepsilon}{R}e^{\frac{-t}{RC}}$
RC=.000108

What am I missing?
.94*C or what is the battery?

Thanks
Carter Barker

Hi Carter,

Note first that $77\,\text{M$\Omega$}$ is $77\times10^6 = 7.7\times10^7$.

The voltage has the form $V(t) = V_\infty(1-e^{-\frac{t}{RC}})$, where $V_\infty$ is is the final voltage, corresponding to $t\to\infty$. $V(t)$ will be equal to $0.94V_\infty$ when $(1-e^{-\frac{t}{RC}}) = 0.94$.

You can easily compute the corresponding value of $t$ using logarithms.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top