- #1
Kelsi_Jade
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Homework Statement
If so, how do I go about finding the total current?
Homework Equations
See photo above.
The Attempt at a Solution
See above.[/B]
After 14ms the current drops -- Find the capacitance at C
- Thread starter Kelsi_Jade
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In summary, this conversation is about a student seeking help with finding the total current in an RC circuit. They have already simplified the circuit and are looking for guidance on the general formula for current with respect to time in such a circuit. The student is directed to do a web search or refer to their class notes for more information.
If so, how do I go about finding the total current?
See photo above.
See above.[/B]
- #2
cnh1995
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Is that what you are asked?Kelsi_Jade said:
- #3
gneill
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You are given two values for the current at different times. You'll want to use them at some point...Kelsi_Jade said:
What kind of circuit is this? If you simplified the circuit, combining the capacitors and combining the resistors, what are you left with? What kind of behavior would you expect when the switch closed? What's the general formula for the current with respect to time for such a circuit?
- #4
Kelsi_Jade
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Well, since the resistors are in series you can combine them to equal one resistor with a resistance of 3+1= 4kOhm.
The capacitors in parallel would equate to one capacitor with 3+C= (3+C)microFerrads.
Current with respect to time would be = Q/Δt ?
The capacitors in parallel would equate to one capacitor with 3+C= (3+C)microFerrads.
Current with respect to time would be = Q/Δt ?
- #5
gneill
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Yes, so you have a basic RC circuit (resistor - capacitor circuit).Kelsi_Jade said:
Your class notes or textbook should show a general formula for the charge or voltage or current with respect to time for such a circuit. If not, do a web search on "RC circuit discharge". Here's an example hit:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html
FAQ: After 14ms the current drops -- Find the capacitance at C
1. What is the significance of the current dropping after 14ms?
The drop in current after 14ms indicates that the capacitor has reached its fully charged state. This is due to the fact that the capacitor's voltage has increased to match the voltage of the power supply, resulting in a decrease in current flow.
2. How is capacitance calculated at C?
Capacitance at C can be calculated using the formula C = Q/V, where Q is the charge stored on the capacitor and V is the voltage across the capacitor. In this case, the charge and voltage can be measured at the 14ms mark to determine the capacitance at C.
3. What factors affect the capacitance at C?
The capacitance at C is primarily affected by the distance between the capacitor's plates, the surface area of the plates, and the type of material used as the dielectric between the plates. Additionally, the amount of charge and voltage applied can also impact the capacitance.
4. How does the capacitance at C change over time?
The capacitance at C is constant as long as the physical properties and conditions of the capacitor remain unchanged. However, if there is a change in any of these factors, the capacitance may also change. For example, if the distance between the plates is altered, the capacitance will also be affected.
5. How can the capacitance at C be increased?
The capacitance at C can be increased by increasing the surface area of the capacitor's plates, decreasing the distance between the plates, or using a material with a higher dielectric constant. This allows for more charge to be stored on the plates, resulting in a higher capacitance value.