Solving RC Circuit Problem Homework

In summary, the circuit contains a 85V battery, a 40Ω resistor, a 60Ω resistor, and a capacitor. The switch has been closed for a long time, causing the capacitor to be fully charged with a potential difference of 85V. To find the charge on the capacitor, the equation Q = CV can be used, yielding a value of 170 μC. To determine the time at which the charge on the capacitor has decreased to 10% of its initial value, Kirchoff's Law and Loop Rule can be used, or alternatively, the steady-state potential can be found by removing the capacitor from the circuit and finding the open-circuit voltage at the open terminals.
  • #1
majormaaz
62
1

Homework Statement


Consider the circuit in the figure below, in which V = 85 V, R = 40 , and the switch has been closed for a very long time.

p31-77alt.gif


(a) What's the charge on the capacitor?
(b) The switch is opened at t = 0 s. At what time has the charge on the capacitor decreased to 10% of its initial value?

[Here's the link for the picture if it doesn't come up]
http://www.webassign.net/knight/p31-77alt.gif

Homework Equations


C = Q/V
V = IR
Q = C*EMF*e^(-t/RC)

The Attempt at a Solution



For part (a), I thought that since the circuit's been connected like that for a long time, the voltage across the capacitor would be the same as the battery's voltage - i.e. 85 Volts. From there, a simple Q = CV would yield 1.7 e -4 C = 170 μC, but that's not right.

For part (b), I was thinking that since it's a parallel circuit, I'd have to use Kirchoff's Law and the Loop Rule to get those 2 loop equations and the Junction equation, but I'm not sure as to if there's another method to get to the answer.

I would appreciate any help!
 
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  • #2
majormaaz said:

Homework Statement


Consider the circuit in the figure below, in which V = 85 V, R = 40 , and the switch has been closed for a very long time.

p31-77alt.gif


(a) What's the charge on the capacitor?
(b) The switch is opened at t = 0 s. At what time has the charge on the capacitor decreased to 10% of its initial value?

[Here's the link for the picture if it doesn't come up]
http://www.webassign.net/knight/p31-77alt.gif

Homework Equations


C = Q/V
V = IR
Q = C*EMF*e^(-t/RC)

The Attempt at a Solution



For part (a), I thought that since the circuit's been connected like that for a long time, the voltage across the capacitor would be the same as the battery's voltage - i.e. 85 Volts. From there, a simple Q = CV would yield 1.7 e -4 C = 170 μC, but that's not right.

For part (b), I was thinking that since it's a parallel circuit, I'd have to use Kirchoff's Law and the Loop Rule to get those 2 loop equations and the Junction equation, but I'm not sure as to if there's another method to get to the answer.

I would appreciate any help!
(The image shows up just fine!)

For part (a):
After a long time, the capacitor is fully charged, so no current flows through it or the 10Ω resistor. So there is 0 voltage drop across the 10Ω resistor.

However, current does flow through the 60Ω and 40Ω (R) resistors. Thus they act as a voltage splitter, and C is in parallel with R .​
 
  • #3
SammyS said:
However, current does flow through the 60Ω and 40Ω (R) resistors. Thus they act as a voltage splitter, and C is in parallel with R .

I think you're saying that the voltage across C would be the same potential difference as the voltage drop across resistor R. But I don't get how that would work if the capacitor was fully charged - I mean, after a long time, I can see why there'd be no current in that branch, but wouldn't the capacitor have charged up to the voltage of the battery by then (and not the resistor)?
 
  • #4
Thanks for the help! I did get both answers with your help, but I'd still like to understand why the capacitor would charge up to (I)*40 Ω instead of the full battery voltage.
 
  • #5
majormaaz said:
I think you're saying that the voltage across C would be the same potential difference as the voltage drop across resistor R. But I don't get how that would work if the capacitor was fully charged - I mean, after a long time, I can see why there'd be no current in that branch, but wouldn't the capacitor have charged up to the voltage of the battery by then (and not the resistor)?

The potential difference across the capacitor cannot go higher than the potential difference available at the nodes where it connects.

To find the steady-state potential that a capacitor will achieve, remove the capacitor from the circuit and find the open-circuit voltage at the open terminals where it was connected.
 
  • #6
gneill said:
The potential difference across the capacitor cannot go higher than the potential difference available at the nodes where it connects.

To find the steady-state potential that a capacitor will achieve, remove the capacitor from the circuit and find the open-circuit voltage at the open terminals where it was connected.

I will keep this in mind. Thanks!
 

FAQ: Solving RC Circuit Problem Homework

1. How do I approach solving an RC circuit problem in my homework?

First, identify the type of RC circuit (series or parallel) and the given values for resistance, capacitance, and voltage. Then, use the appropriate formulas (such as Ohm's Law and the time constant equation) to solve for the unknown variables.

2. What is the time constant and how is it calculated in an RC circuit problem?

The time constant (represented by the symbol tau, τ) is the amount of time it takes for a capacitor to charge or discharge to 63.2% of its maximum voltage. It is calculated by multiplying the resistance value (in ohms) by the capacitance value (in farads).

3. How do I solve for the voltage or current at a specific time in an RC circuit problem?

To solve for the voltage or current at a specific time, use the formula V(t) = V0(1-e^(-t/tau)) for a charging capacitor, or V(t) = V0(e^(-t/tau)) for a discharging capacitor. Plug in the given values for time, voltage at time 0 (V0), and the time constant (tau) to solve for the unknown variable.

4. Can I use Kirchhoff's Laws to solve an RC circuit problem?

Yes, you can use Kirchhoff's Laws (specifically Kirchhoff's Voltage Law for series circuits and Kirchhoff's Current Law for parallel circuits) to solve for the unknown variables in an RC circuit problem. However, it may be more efficient to use the appropriate formulas for RC circuits instead.

5. What are some common mistakes to avoid when solving an RC circuit problem?

Some common mistakes to avoid include forgetting to convert units (such as resistance values from kilohms to ohms), mixing up the formulas for charging and discharging capacitors, and not using the correct values for the time constant (such as using the resistance value instead of the time constant). It is also important to double check your calculations and make sure they make sense in the context of the problem.

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