Voltage Across 10, 20uF Capacitors in RC Circuit

In summary, two capacitors with values of 10 and 20 microfarads were connected in series and charged with a 100 volt battery. After being disconnected from the battery, the capacitors were connected to a 2500 ohm resistor. Using the formula V_C=V_0e^(-t/RC), the voltage across the capacitors after 1 second was found to be small due to the exponential decay of voltage over time. It was also noted that the minus sign was missing in one of the formulas provided.
  • #1
Logan Land
84
0
Two capacitors of value 10, 20 microfarads are
connected in series. Then charged with a 100 volt battery. If the capacitors is
disconnected from the battery, and then connected to a 2500
ohm resistor, what is the voltage across the capacitors after 1 second?
 
Last edited:
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  • #2
LLand314 said:
Two capacitors of value 10, 20 microfarads are
connected in series. Then charged with a 100 volt battery. If the capacitors is
disconnected from the battery, and then connected to a 2500
ohm resistor, what is the voltage across the capacitors after 1 second?

Hi LLand314!

Can you come up with a couple of formulas that are applicable?

One for the relationship between voltage, capacity, and charge?
Another one for the discharge of a charged capacitor versus time?
And perhaps one for the combined capacity of 2 capacitors in series? (Wondering)
 
  • #3
I like Serena said:
Hi LLand314!

Can you come up with a couple of formulas that are applicable?

One for the relationship between voltage, capacity, and charge?
Another one for the discharge of a charged capacitor versus time?
And perhaps one for the combined capacity of 2 capacitors in series? (Wondering)

these formulas?
View attachment 4022
 

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  • #4
LLand314 said:
these formulas?

Yep.

When the capacitors get charged, a charge Q flows from the first to the second.
Afterwards, they both hold the same charge Q.

From your formulas, you can get that:
$$Q=C_1 V_1$$
$$Q=C_2 V_2$$
$$V_{total} = V_1 + V_2$$

What will the charge Q and the voltages be?
 
  • #5
I like Serena said:
Yep.

When the capacitors get charged, a charge Q flows from the first to the second.
Afterwards, they both hold the same charge Q.

From your formulas, you can get that:
$$Q=C_1 V_1$$
$$Q=C_2 V_2$$
$$V_{total} = V_1 + V_2$$

What will the charge Q and the voltages be?

since the capacitors are in series can I just replace the 3 with a single using (1/(1/c1)+(1/c2))?
 
  • #6
LLand314 said:
since the capacitors are in series can I just replace the 3 with a single using (1/(1/c1)+(1/c2))?

Yep, you can.
Since you didn't give a formula for the combined capacity of 2 capacitors in series, I assumed you didn't have that formula and had to do without.

Then the only thing left is to apply $V_C=V_0 e^{-t/RC}$.
 
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  • #7
I like Serena said:
Yep, you can.
Since you didn't give a formula for the combined capacity of 2 capacitors in series, I assumed you didn't have that formula and had to do without.

Then the only thing left is to apply $V_C=V_0 e^{t/RC}$.
then 100e^(1/(2500*(6.666*10^-6)))?
 
  • #8
LLand314 said:
then 100e^(1/(2500*(6.666*10^-6)))?

Yup! (Happy)
 
  • #9
I like Serena said:
Yup! (Happy)

the voltage I get is a huge number how can it be so big?

edit: oh wait shouldn't it be Vc=V0e^(-t/rc)? a negative sign in the t/rc
 
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  • #10
LLand314 said:
the voltage I get is a huge number how can it be so big?

edit: oh wait shouldn't it be Vc=V0e^(-t/rc)? a negative sign in the t/rc

Quite right!

(Hmm. That minus sign also seems to be missing in your picture. (Worried))
 
  • #11
I like Serena said:
Quite right!

(Hmm. That minus sign also seems to be missing in your picture. (Worried))

Yes I see that, strange.
anyhow so my voltage should be small correct?
 
  • #12
LLand314 said:
Yes I see that, strange.
anyhow so my voltage should be small correct?

Correct.
The voltage really decays exponentionally instead of increasing.
And it decays with a characteristic time that is also called the RC-time.
 

FAQ: Voltage Across 10, 20uF Capacitors in RC Circuit

1. What is the purpose of a capacitor in an RC circuit?

A capacitor is used to store and release electrical energy in an RC circuit. It is able to do this by creating an electric field between two conductive plates, with one plate positively charged and the other negatively charged.

2. How do you calculate the voltage across a capacitor in an RC circuit?

The voltage across a capacitor in an RC circuit can be calculated using the formula V = Q/C, where V is the voltage, Q is the charge stored in the capacitor, and C is the capacitance. In an RC circuit, the voltage across the capacitor will decrease over time as the capacitor discharges.

3. What determines the capacitance of a capacitor in an RC circuit?

The capacitance of a capacitor in an RC circuit is determined by the physical characteristics of the capacitor, such as the distance between the two plates, the surface area of the plates, and the type of dielectric material used. The larger the capacitance, the more charge the capacitor can store.

4. How does the capacitance of a capacitor affect the voltage across it in an RC circuit?

The voltage across a capacitor in an RC circuit is inversely proportional to the capacitance. This means that a larger capacitance will result in a smaller voltage across the capacitor, while a smaller capacitance will result in a larger voltage. In other words, a capacitor with a higher capacitance will be able to store more charge and have a lower voltage, while a capacitor with a lower capacitance will store less charge and have a higher voltage.

5. What happens to the voltage across a capacitor as the RC circuit is turned on and off?

When the RC circuit is turned on, the voltage across the capacitor will increase until it reaches the same voltage as the power source. As the circuit is turned off, the capacitor will discharge and the voltage across it will decrease. The rate at which the capacitor discharges is determined by the resistance in the circuit and the capacitance of the capacitor.

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