Finding the Charging Equation of a Capacitor Using Laplace Transform

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SUMMARY

The discussion focuses on using the Laplace transform to derive the charging equation of a capacitor with an initial voltage. The participant attempts to apply the formula V_i = RC(dVc/dt) + V_c, leading to the expression V_c(s) = V/(s(sRC+1)) + λ/(sRC+1). However, a mistake is identified in the Laplace transform of dVc/dt, specifically in the application of the initial condition λ. The correct approach involves using L[RC V'] = RC(sV(s) - λ), which resolves the discrepancy in the derived formula.

PREREQUISITES
  • Understanding of Laplace transforms
  • Familiarity with capacitor charging equations
  • Knowledge of initial conditions in differential equations
  • Basic calculus, particularly differentiation and integration
NEXT STEPS
  • Study the properties of Laplace transforms in electrical circuits
  • Review the derivation of capacitor charging equations
  • Learn about initial conditions in differential equations
  • Explore common mistakes in applying Laplace transforms
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Electrical engineering students, circuit designers, and anyone studying differential equations in the context of electrical circuits will benefit from this discussion.

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Homework Statement


I'm trying to use the Laplace transform to find the charging equation of a capacitor with an initial voltage already on the capacitor.

Homework Equations



V_i = RC\frac{dVc}{dt} + V_c

The initial condition of Vc will be \lambda.

The Attempt at a Solution



V_i(s) = sRC*V_c(s) + V_c(s) - \lambda

Vi will be a step function of magnitude V, so

\frac{V_i}{s} + \lambda = (sRC + 1)V_c(s)

V_c(s) = \frac{V}{s(sRC+1)} + \frac{\lambda}{sRC+1}

So using a table of Laplace transforms I get:

V_c(t) = V_i(1-e^{\frac{-t}{RC}}) + \frac{\lambda}{RC}(e^{\frac{-t}{RC}})

The problem is that this doesn't seem to agree with the formula I've seen, it's close but lambda shouldn't be divided by RC. I'm wondering how I went wrong? Thanks for any advice.
 
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Your mistake when you took the Laplace transform of dVc/dt:

L[RC V']=RC L[V']=RC(sV(s)-\lambda)
 
Whoops! Thanks for the catch...:redface:
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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