Archived Maximum Charge on a Capacitor, Help

[ryeager]

Hi, thanks in advance to all who help

a) Determine the time constant for charging the capacitor in the circuit given.
b) what is the maximum charge on the capacitor?
R1
______vvvv____________
| | |
| < |
E R2 > C
| | |
|______________|______|

E = emf and C is the capacitor, sorry for the poor illustration.

Mentor's edit: Here's a better illustration:

For a, I've determined the equivalent resistance of parallel resistors to be (1/R1 + 1/R2)^-1 = R1R2/(R1+R2), thus the time constant would be tau = CReq = CR1R2/(R1+R2)

For b) I have a feeling Qmax for the capacitor will be Qmax = CV where V is the voltage across R2 since R2 and C are parallel, but I can't figure out V across R2 in terms of E, R1, R2, C. Can anyone help? Thanks again.

 

[cnh1995]

For a, I've determined the equivalent resistance of parallel resistors to be (1/R1 + 1/R2)^-1 = R1R2/(R1+R2), thus the time constant would be tau = CReq = CR1R2/(R1+R2)

I got the same result after solving with differential equations.
Let the time constant be T.
Current supplied by the source is,
I(t)=ET/(R₁R₂C)*(1-e^-t/T)+(E*e^-t/T)/R₁
From this equation, it can be verified that at t=0, I=E/R₁(capacitor acts as a short at t=0) and at t=∞, I=E/(R₁+R₂)..(capacitor acts as open circuit at t=∞).

For b) I have a feeling Qmax for the capacitor will be Qmax = CV where V is the voltage across R2 since R2 and C are parallel, but I can't figure out V across R2 in terms of E, R1, R2, C.

In steady state, the capacitor will act as an open circuit. Hence, current will flow though the series combination of R₁ and R₂.
So, Voltage across C=Voltage across R₂ in steady state= E*R₂/(R₁+R₂)
∴Final charge on the capacitor=Q=CV= ECR₂/(R₁+R₂).

 

[cnh1995]

The above approach involves some mathematical calculations and a generalized expression for current I(t). Instead, I believe the best and simplest approach to solve these two questions a and b is Thevenin's theorem.

 

[Kenygreen]

Hi, thanks in advance to all who help

a) Determine the time constant for charging the capacitor in the circuit given.
b) what is the maximum charge on the capacitor?
R1
______vvvv____________
| | |
| < |
E R2 > C
| | |
|______________|______|

E = emf and C is the capacitor, sorry for the poor illustration.

Mentor's edit: Here's a better illustration:
View attachment 95530

For a, I've determined the equivalent resistance of parallel resistors to be (1/R1 + 1/R2)^-1 = R1R2/(R1+R2), thus the time constant would be tau = CReq = CR1R2/(R1+R2)

For b) I have a feeling Qmax for the capacitor will be Qmax = CV where V is the voltage across R2 since R2 and C are parallel, but I can't figure out V across R2 in terms of E, R1, R2, C. Can anyone help? Thanks again.

 

[Kenygreen]

Why do you regard R1 and R2 as being in parallel?
When C is fully charged then the current through the capacitor is ?

Which makes the fial voltage across C ? ?Hence the max charge will be ? ? ?

 

[cnh1995]

Why do you regard R1 and R2 as being in parallel?
When C is fully charged then the current through the capacitor is ?

Which makes the final voltage across C ? ?Hence the max charge will be ? ? ?

Are those your own queries or are you trying to give hints to the OP?

 

[Ronie Bayron]

Hi, thanks in advance to all who help

a) Determine the time constant for charging the capacitor in the circuit given.
b) what is the maximum charge on the capacitor?
R1
______vvvv____________
| | |
| < |
E R2 > C
| | |
|______________|______|

E = emf and C is the capacitor, sorry for the poor illustration.

Mentor's edit: Here's a better illustration:
View attachment 95530

For a, I've determined the equivalent resistance of parallel resistors to be (1/R1 + 1/R2)^-1 = R1R2/(R1+R2), thus the time constant would be tau = CReq = CR1R2/(R1+R2)

For b) I have a feeling Qmax for the capacitor will be Qmax = CV where V is the voltage across R2 since R2 and C are parallel, but I can't figure out V across R2 in terms of E, R1, R2, C. Can anyone help? Thanks again.

V_c=V_R2 where R1 & R2 are voltage divider.
I = E/(R1+R2) and V_R2=IR2, so V_c =E[R2/(R1+R2)]

 

[Ronie Bayron]

V_c=V_R2 where R1 & R2 are voltage divider.
I = E/(R1+R2) and V_R2=IR2, so V_c =E[R2/(R1+R2)]

Since C=Q/V_c, Q =C V_c, so the charge Q depends on the value of capacitance you put on your circuit and magnitude of V_c