Calculating Charge, Current, and Energy in a Simple RC Circuit

In summary: I cannot help you with part d) since it asks for the power dissipated by a resistor in terms of current and voltage.
  • #1
Apprentice123
92
0
The values of components in a simple RC circuit in series with a key are: C = 1x10^{-6}F, R=2x10^{6} Ohm and epsilon = 10 V. 10s at the moment after the key is closed, calculate:

a) the charge in capacitor
b) the current in the resistor
c) the rate at which energy is stored in capacitor
d) the rate at which energy is supplied by the battery


My solution:

a) Qmax = C * epsilon = 1x10^5 c

b) I = - dQ/dt
Rule of Kirchhoff
I*R = Q/C
I = Q / R*C = 5x10^{-6} A

c) How to calculate?
d) How to calculate?
 
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  • #2
Apprentice123 said:
The values of components in a simple RC circuit in series with a key are: C = 1x10^{-6}F, R=2x10^{6} Ohm and epsilon = 10 V. 10s at the moment after the key is closed, calculate:

a) the charge in capacitor
b) the current in the resistor
c) the rate at which energy is stored in capacitor
d) the rate at which energy is supplied by the battery


My solution:

a) Qmax = C * epsilon = 1x10^5 c

b) I = - dQ/dt
Rule of Kirchhoff
I*R = Q/C
I = Q / R*C = 5x10^{-6} A

c) How to calculate?
d) How to calculate?

Maybe consider:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/filter.html#c2
 
  • #4
Apprentice123 said:
thanks, but I could not resolve the (c) and (d)

Well, what is the "rate of energy build up"?
 
  • #5
Apprentice123 said:
thanks, but I could not resolve the (c) and (d)

Nevermind (c) and (d), your answers for (a) and (b) are not correct yet.

LP is reminding you that there is a time constant associated with the RC circuit, and that the charging is exponential with time. You need to write the equation for the capacitor voltage as a function of time, and use that equation to solve for a-d.
 
  • #6
berkeman said:
Nevermind (c) and (d), your answers for (a) and (b) are not correct yet.

LP is reminding you that there is a time constant associated with the RC circuit, and that the charging is exponential with time. You need to write the equation for the capacitor voltage as a function of time, and use that equation to solve for a-d.

Yes. New:

a) Q = C * E * (1 - e^{(-t)/R*C}
Q = 9,932x10^{-6} c


b) I = E/R * e^{(-t)/R*C}
I = 3,37x10^{-8} A

(c) and (d) I not find equation
 
  • #7
Apprentice123 said:
Yes. New:

a) Q = C * E * (1 - e^{(-t)/R*C}
Q = 9,932x10^{-6} c


b) I = E/R * e^{(-t)/R*C}
I = 3,37x10^{-8} A

(c) and (d) I not find equation

c) the rate at which energy is stored in capacitor
d) the rate at which energy is supplied by the battery

What is the definition of power, in terms of energy? What is the power dissipated in a resistor in terms of I and V? How is the energy stored on a capacitor defined?
 
  • #8
berkeman said:
What is the definition of power, in terms of energy? What is the power dissipated in a resistor in terms of I and V? How is the energy stored on a capacitor defined?

Energy in a capacitor charged:

1/2 * C * V^2
 
  • #9
Apprentice123 said:
Energy in a capacitor charged:

1/2 * C * V^2

Since you already found a relationship between charge and time and another between charge flow rate (i.e. current) and time, it's wise to express energy in terms of charge. Then you can use calculus to find the equation for power.
 
  • #10
ideasrule said:
Since you already found a relationship between charge and time and another between charge flow rate (i.e. current) and time, it's wise to express energy in terms of charge. Then you can use calculus to find the equation for power.

c)
U = 1/2 * E^2 * C
U = 5x10^{-5}J

d)
E = (Q*E)/(t)
E = 1x10^{-5}J


It is correct ?
 
  • #11
Apprentice123 said:
c)
U = 1/2 * E^2 * C
U = 5x10^{-5}J

d)
E = (Q*E)/(t)
E = 1x10^{-5}J


It is correct ?

Part (c) is not correct. What you have calculated is the total energy stored in the capacitor (after it has charged fully). The question is asking for the *rate* at which energy is stored (during the charging process)

For part (d), I would follow berkeman's advice in his latest post and investigate power.
 
  • #12
I meant that you should express, for part c), the energy stored on the capacitor in terms of the charge on the capacitor.
 

1. What is a simple RC circuit in series?

A simple RC circuit in series is a circuit that contains a resistor (R) and a capacitor (C) connected in series with each other. This means that the current flows through both components in a single path.

2. How does a simple RC circuit in series work?

In a simple RC circuit, the resistor and capacitor work together to control the flow of electric current. The resistor limits the flow of current, while the capacitor stores and releases electrical energy in the form of an electric field. As the current flows through the circuit, the capacitor charges and discharges, creating a time-varying voltage across the circuit.

3. What is the purpose of a simple RC circuit in series?

A simple RC circuit in series has many practical applications, such as filtering signals in electronic circuits, smoothing out power supply voltages, and creating time delays. It can also be used in timing circuits and as a simple low-pass filter.

4. How do you calculate the time constant of a simple RC circuit in series?

The time constant (τ) of a simple RC circuit in series is calculated by multiplying the resistance (R) in ohms by the capacitance (C) in farads. It is represented by the equation τ = RC. The time constant is a measure of how quickly the capacitor charges or discharges in the circuit.

5. What is the difference between a series RC circuit and a parallel RC circuit?

In a series RC circuit, the resistor and capacitor are connected in series, meaning that the current flows through both components in a single path. In a parallel RC circuit, the resistor and capacitor are connected in parallel, meaning that the current splits between the two components. Additionally, the time constant (τ) is different for each type of circuit, with the series circuit having a longer time constant than the parallel circuit.

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