Need help finding current on RC circuit

• Ruskointhehizzy
In summary, the current in the circuit at the instant that the capacitors have lost 80.0% of their initial stored energy is 13.9A.
Ruskointhehizzy

Homework Statement

In the circuit shown in the figure each capacitor
initially has a charge of
magnitude 3.60nC on its plates. After the switch S is closed, what will be the
current in the circuit at the instant that the capacitors
have lost 80.0% of their initial stored energy?
the shown 3 capacitors are 10, 15, and 20 pF and the one resistor has 25ohms - all in series with one switch S.

Homework Equations

E = 1/2Q^2/c = QV/2
V_c(t)=Q/c * (1-e^(-t/RC))
i = Q/RC * e^(-t/RC)

I have since solved this but I would really appreciate it if I could get this checked for correctness.
Thank you.

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Last edited:
I think I have figured out my mistake but please can I still get a check to see if my solution is correct?
first I find what Q needs to be to get .2E
to do this I say:
E0 = ½Q2/C, E = ½ ( √.2 Q )2/C = .2(½ Q2/C) = .2E0
I then use this to find -t
-t = RCln(Q/Q0) = RCln(√.2Q0/Q0) = RCln(√.2)
then finally
i = - Q/RC * e-t/RC = Q/RC * eRCln(√.2)/RC = Q/RC * eln√.2 = = -(3.6nC/(25Ω*4.62pF)) * √.2
i = = -3.6e-9C * √.2 / ( 25Ω * 4.62e-12F) = 13.9A

We can't see your images, which is probably why nobody has responded.

Ruskointhehizzy
phyzguy said:
We can't see your images, which is probably why nobody has responded.
Yeah I tried to post a image with a link to my google drive but it didn't let me - then when I tried to remove it wouldn't let me.
I finally resorted to posting it in the comments lmao.
Did you have a chance to check over my solution? I hope I got this correct I've been stuck on it for awhile

I don't see the figure anywhere. Have you tried just dragging the figure into the comment entry window?

FactChecker said:
I don't see the figure anywhere. Have you tried just dragging the figure into the comment entry window?
oh that worked thank you

that was just my first wrong attempt - I will change the op

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1. How do I calculate the current on an RC circuit?

The current on an RC circuit can be calculated using Ohm's Law, which states that current (I) is equal to the voltage (V) divided by the resistance (R). In an RC circuit, there is both resistance from the resistor (R) and capacitance from the capacitor (C), so the total resistance is equal to the sum of these two components (Rtot = R + 1/ωC). Therefore, the current can be calculated using I = V/Rtot.

2. What is the significance of the time constant in an RC circuit?

The time constant (τ) in an RC circuit is a measure of how quickly the capacitor charges or discharges. It is equal to the product of the resistance (R) and capacitance (C) (τ = RC). The larger the time constant, the longer it takes for the capacitor to reach its maximum charge or discharge. It is an important parameter in understanding the behavior of RC circuits.

3. How do I find the current at a specific time in an RC circuit?

The current in an RC circuit changes over time as the capacitor charges or discharges. To find the current at a specific time, you can use the equation I = I0 * e^(-t/RC), where I0 is the initial current and t is the time. This equation is derived from the exponential charging or discharging curve of the capacitor in an RC circuit.

4. How does the value of the capacitor affect the current in an RC circuit?

The value of the capacitor (C) in an RC circuit affects the current in two ways. First, a larger capacitor will take longer to charge or discharge, resulting in a longer time constant (τ = RC). This means that the current will change more slowly. Second, a larger capacitor will have a larger capacitance (C) value in the denominator of the current equation (I = V/Rtot), resulting in a smaller current for the same voltage and resistance.

5. Can I measure the current on an RC circuit with a multimeter?

Yes, it is possible to measure the current on an RC circuit with a multimeter. However, since the current is constantly changing, the multimeter will only provide an average value. It is important to note that the multimeter must be connected in series with the circuit to measure the current, and the resistance of the multimeter must be taken into account when calculating the total resistance (Rtot) in the circuit.

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