Charge on Capacitor in RLC Circuit

In summary, the tutorial wants you to find an expression for the time by solving for t from the potential energy equations. Once you have t, you are supposed to take the natural log of both sides and isolate t to get ln(8)L/R.
  • #1
Drakkith
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Homework Statement


At t = 0, the charge stored on the capacitor plates is maximum in an oscillating series RLC circuit. At what time will the maximum possible energy that can be stored in the capacitor fall to one-eighth of its initial value if R = 7.20 Ω and L = 21.0 H?

The differential equation for an RLC circuit is Ld2q/dt2 + Rdq/dt + q/C = 0 and the solution to this equation is q = qmaxe−Rt/2L cos ωdt.
Assume that the damping is very weak (that is, assume the resistance R << sqrt(4L/C), so that the amplitude of the charge does not change by much during one oscillation).

Homework Equations


Given Above

The Attempt at a Solution



I've been trying to follow a "tutorial" on my online homework program, and after about an hour I finally managed to get most of the way through it. However, I'm now stuck on one particular part.

The tutorial wants me to determine a bunch of expressions in terms of different variables without putting any numbers in. The only I'm stuck on is finding an expression for the time.

Starting from the equations for potential energy, I eventually end up determining that e-Rt/2L = 1/√8. Solving for t by taking the natural log of both sides and then isolating t is supposed to give me ln(8)L/R. At least that's what the program tells me. But, where did the 2 in the denominator of the exponent go?! That's pretty much it. Once I figure that out I think I'll be able to solve the problem completely.

Thanks.
 
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  • #2
Remember that log(sqrt(x)) = log(x^(1/2)) = 1/2 log(x). So when you take the log of both sides, you get:
[tex] \frac{-Rt}{2L} = \frac{-log(8)}{2} [/tex]
[tex] t = \frac{log(8) L}{R} [/tex]
 
  • #3
I'm sorry, you've lost me. Where did the 2 on the right side, under the -log(8) come from?
 
  • #4
Hold on. ln(1/81/2) = ln(1)-ln(81/2) = 0-ln(8)/2
That look right?
 
  • #5
Yes.
 
  • #6
Okay. Somehow I thought the ln(1/√8) was just a ln(1/8)...

Thanks!
 

1. What is the formula for calculating the charge on a capacitor in an RLC circuit?

The formula for calculating the charge on a capacitor in an RLC circuit is Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor in volts.

2. How does the charge on a capacitor change over time in an RLC circuit?

The charge on a capacitor in an RLC circuit follows an exponential decay curve. It starts at its maximum value and decreases over time until it reaches zero, then it cycles back to its maximum value again.

3. What factors affect the charge on a capacitor in an RLC circuit?

The charge on a capacitor in an RLC circuit is affected by the capacitance of the capacitor, the voltage across the capacitor, and the resistance and inductance of the circuit. These factors determine how quickly the charge will change over time.

4. How does the charge on a capacitor in an RLC circuit affect the overall behavior of the circuit?

The charge on a capacitor in an RLC circuit plays a crucial role in the overall behavior of the circuit. It affects the voltage and current in the circuit, which in turn affects the frequency and amplitude of the oscillations in the circuit. It also determines the energy stored in the capacitor, which can be released during the discharge phase of the circuit.

5. How can the charge on a capacitor in an RLC circuit be controlled?

The charge on a capacitor in an RLC circuit can be controlled by adjusting the capacitance, voltage, and resistance values in the circuit. By changing these values, the rate of charge and discharge of the capacitor can be altered, affecting the overall behavior of the circuit. Additionally, external devices such as switches or diodes can be used to control the flow of charge in the circuit.

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