# RC Circuit and 120 V battery

1. Feb 12, 2005

### Jacob87411

A current is released from a 120 V battery. The current passes through a resistor of 1200 ohms and then the current splits into two different capacitors. C1=2.0 uF C2=3.0 uF. If there is no charge on the capacitators before the current is released, determine the charges of Q1 and Q2 on capacitators C1 and C2 respectively after the current is released.

I should equal .1

V=IR
120 = I(1200)

Confused on what happens when the capacitators come into play. Any help is appreciated

2. Feb 12, 2005

### vsage

From my understanding, as a circuit is charged up, less potential will drop through the resistor and more through the capacitors as they are charged up, so at very large time the voltage drop across the capacitors is approximately equal to the emf applied to the system by the battery.

Using $$C=\frac{q}{V}$$ should be all that's required for this problem.

3. Feb 13, 2005

### Jacob87411

But isnt all the voltage sort of 'eaten up' by the 1200 ohm resistor? How much voltage is left after it passes through it, i thought it was 0.

4. Feb 13, 2005

### xanthym

The resistor R does not exactly "consume" voltage -- it more "responds" to it. The resistor will conduct current "I" thru itself whenever a voltage E occurs across it. That current will equal:
I = E/R

In this case, the voltage across R will equal 120 volts MINUS the voltage from the capacitor's accumulated charge:
{Voltage Across Resistor} = {120 - (Voltage From Accumulated Charge On C)}

The total rate of charge accumulation (dQ/dt) onto both capacitors C will be the current through the resistor:
(dQ/dt) = {Current Thru Resistor} = {Voltage Across Resistor}/R
= {120 - (Voltage Accumulation On C)}/R

Since the two capacitors are in parallel, we can treat them like one large capacitance C for discussion purposes for now. Thus, if a total charge Q has accumulated on C:
{Voltage From Accumulated Charge On C} = Q/C

Hence, we have for the charge accumulation rate (dQ/dt) on C:
(dQ/dt) = {Voltage Across Resistor}/R
(dQ/dt) = {120 - (Voltage Accumulation On C)}/R = {120 - (Q/C)}/R
(dQ/dt) = {120 - (Q/C)}/R

This last equation is the "BASIC" differential equation for this curcuit. (You would still need to account for C=(C1 + C2), but that is fairly easy). By solving this equation for Q(t) (which is a function of time "t"), you will determine the charge build-up on the capacitors with time. It can be shown the solution for battery voltage V is:

$$: \ \ \ \color {red} Q(t) = VC(1 - exp( \frac {-t} {RC} ) )$$

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Last edited: Feb 13, 2005
5. Feb 13, 2005

### Jacob87411

$$: \ \ \ \color {red} Q(t) = VC(1 - exp( \frac {-t} {RC} ) )$$

So do I end up with two seperate equations since it wants it for each capacitator? In one I plug in C1 (2uF) and the other C2 (3uF)? With V=120 and R=1200 in each of them?

Last edited: Feb 13, 2005
6. Feb 13, 2005

### vsage

The voltage drop across each is the same. You'll have two numbers in the end: one for each capacitor. Keep in mind that equation converges to Q = VC at large t.

7. Feb 13, 2005

### Jacob87411

Well I was told prior to the problem that you wont come out with answers just equations

8. Feb 13, 2005

### vsage

Alright then xanthym's answers are right. Mine were just limits (yeah you plug in the capacitance for each equation, don't try and sum them unless you want to combine them into one capacitor for some reason)

9. Feb 13, 2005

### Jacob87411

Alright thanks a lot both of you, makes a lot more sense now

10. Feb 17, 2005

### learningphysics

Only one thing I'd like to point out... in the exponent the C=C1+C2... in other words:

$$Q_1(t) = VC_1(1 - exp( \frac {-t} {R(C_1+C_2)} ) )$$

and

$$Q_2(t) = VC_2(1 - exp( \frac {-t} {R(C_1+C_2)} ) )$$