Charge on the Capacitor at a Specific Time

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SUMMARY

The discussion revolves around calculating the charge on a capacitor in an RC circuit at a specific time, t = 0.35 ms, using the formula q(t) = CV(1 - e^(-t/RC)). The user applied the formula with values C = 62 µF, V = 15 V, and R = 10.8 Ω, resulting in a charge of 3.78 x 10^-4 C. The user expressed concern about the answer format, as it was required in millicoulombs (mC), indicating a need for unit conversion.

PREREQUISITES
  • Understanding of RC circuit theory
  • Familiarity with the formula for capacitor charge q(t) = CV(1 - e^(-t/RC))
  • Basic knowledge of exponential functions and their applications in electrical engineering
  • Ability to convert between coulombs and millicoulombs
NEXT STEPS
  • Learn about unit conversions between coulombs and millicoulombs
  • Study the behavior of RC circuits over time, focusing on time constants
  • Explore the implications of capacitor charge in practical applications
  • Investigate the effects of varying resistance and capacitance on charge time
USEFUL FOR

Students studying electrical engineering, particularly those focusing on circuit analysis and capacitor behavior in RC circuits.

brittydagal
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Homework Statement



The switch in the RC circuit shown in the figure is closed at t=0.

What is the charge on the capacitor at the time t = .35 ms?

I'm 95% sure I'm using the right formula but it's not working so I was wondering if I was possibly confusing something?

Homework Equations



q(t)= CV(1-e^(-t/RC))

The Attempt at a Solution



q(t)= CV(1-e^(-t/RC))
= (62*10^-6)(15 V)(1 - e^(-.00035/(10.8*(62*10^-6)))
= 3.78*10^-4 C ----> this is definitely wrong because the answer is asked for in mC -> any ideas?
 
Physics news on Phys.org
Figure's missing.

You got an answer in C, but you can certainly express it in mC (millicoulomb), can't you? Are you saying that that answer is not accepted?
 

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