Voltage Across 10, 20uF Capacitors in RC Circuit

[Logan Land]

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?

 

[I like Serena]

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)

 

[Logan Land]

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

 

[I like Serena]

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?

 

[Logan Land]

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))?

 

[I like Serena]

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 $V_C=V_0 e^{-t/RC}$.

 

[Logan Land]

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)))?

 

[I like Serena]

then 100e^(1/(2500*(6.666*10^-6)))?

Yup! (Happy)

 

[Logan Land]

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

 

[I like Serena]

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))

 

[Logan Land]

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?

 

[I like Serena]

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.