Another DC Circuit (But more complicated with a SWITCH and Capacitor)

In summary, the circuit shown has initially uncharged capacitors and a closed switch. After a very long time, the capacitors are fully charged and behave as an open circuit, with a current of 20A passing through both resistors. When the switch is reopened, the initial current through the 100 ohm resistor can be determined using Kirchoff's loop rules, and the current through the resistor can be found as a function of time using the equation for current through a capacitor. The charge on the capacitor falls to 10% of its fully charged state after a certain amount of time, which can be calculated using the voltage across the capacitor and the resistance of the 100 ohm resistor.
  • #1
sweetdion
64
0

Homework Statement


For the circuit shown below, the capacitors are initially uncharged. At t=0, the switch S is closed.

a) Determine the current in each resistor immediately after the switch is closed.
b) Determine the current in each resistor a very long time after the switch is closed.
c) Determine the voltage across the capacitor a very long time after the switch is closed.

After the switch has been closed for a very long time, it is reopened.

d) Determine the initial current I0 through the 100 ohm resistor, as a function of time after the switch is reopened.

e) Determine I(t), the current through the 100 ohm resistor, as a function of time after the switch is reopened.

f) How long after the switch is reopened does the charge on the capacitor fall to 10% of its fully charged state?

100_0323.png




Homework Equations


V=IR
I through a capacitor is I = Cdv/dt = Q/C
Kirchoff's Loops Rules

The Attempt at a Solution



a) Immediately after the switch is closed, the capacitor behaves ls a short circuit and hence no current flows through the 100 ohm resistor.

I in the 200 ohm resistor = V/R=15V/200ohms=.075A

b) long after the switch is closed, the capacitor is fully charged and behanves as an open circuit. All of the current passes through both resistors.

Left Loop Clockwise:
15V-200ΩI1-100ΩI1=0
Right Loop Counter clockwise: Q/C-100ΩI1-0

I1 is the only current passing through both the resistors

I= 20A

as for the rest of the parts, I want to make sure I get A and B right before I move on.
 
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  • #2
sweetdion said:
1.
b) Determine the current in each resistor a very long time after the switch is closed.


b) long after the switch is closed, the capacitor is fully charged and behanves as an open circuit. All of the current passes through both resistors.

Left Loop Clockwise:
15V-200ΩI1-100ΩI1=0

Right Loop Counter clockwise: Q/C-100ΩI1-0

I1 is the only current passing through both the resistors

I= 20A

I do not understand. What is that I=20 A? A few hundred ohm and 15 V emf, how can be the current 20 A?

ehild
 
  • #3
sweetdion said:
I through a capacitor is I = Cdv/dt = Q/C

I = C dv/dt is correct.

But I = Q/C is not. Don't use it. V = Q/C is the correct equation.
 
  • #4
ehild said:
I do not understand. What is that I=20 A? A few hundred ohm and 15 V emf, how can be the current 20 A?

ehild

I solved the top equation and that's what I got. Did I label my currents right?
 
  • #5
Do you mean 15V-200ΩI1-100ΩI1=0? That is 15 V = 300Ω * I1. Do you mean I1= 20 A?

ehild
 
  • #6
Yeah, that's what I meant.
 
  • #7
Amazing. Tell me please, how much is 300*20?

ehild
 
  • #8
A lot, 6000. Which, is wrong.

I'm asking for your help because I don't understand what I'm doing wrong, ehild.
 
  • #9
You have an equation. 300 I = 15. This equation is correct.
Solve it for I. Is not it 15/300 A?

ehild
 
  • #10
ehild said:
You have an equation. 300 I = 15. This equation is correct.
Solve it for I. Is not it 15/300 A?

ehild

Oh. Now I see what I was doing wrong. Simple Math error. I = 0.05A
 
  • #11
Moving on to part C now.

When the switch is closed for a very long time, the capacitor starts discharging through the resistors. After a long time, the voltage across the capacitor becomes equal to the voltage across the 100 Ω resistor.

V= IR=(0.05A)(100Ω)=5V
 
  • #12
sweetdion said:
Moving on to part C now.

When the switch is closed for a very long time, the capacitor starts discharging through the resistors. After a long time, the voltage across the capacitor becomes equal to the voltage across the 100 Ω resistor.

V= IR=(0.05A)(100Ω)=5V

That's right. After a very long time, the capacitor becomes essentially an open circuit.
 
  • #13
Thanks Melawrghk. :)
 

1. What is the purpose of a switch in a DC circuit with a capacitor?

The switch is used to control the flow of current in the circuit. When the switch is closed, the capacitor will charge, and when the switch is open, the capacitor will discharge. This allows for the storage and release of electrical energy in the circuit.

2. How does a capacitor affect the behavior of a DC circuit with a switch?

A capacitor is a passive electronic component that stores electrical energy in the form of an electric field. When a capacitor is placed in a DC circuit with a switch, it will charge when the switch is closed and discharge when the switch is open. This affects the behavior of the circuit by introducing a time delay and smoothing out any fluctuations in the voltage.

3. What is the role of a resistor in a DC circuit with a switch and capacitor?

A resistor is used to limit the flow of current in a circuit. In a DC circuit with a switch and capacitor, a resistor is often placed in series with the capacitor to control the rate of charging and discharging of the capacitor. This helps to prevent any damage to the components and ensures a more controlled flow of current.

4. How is the time constant calculated in a DC circuit with a switch and capacitor?

The time constant in a DC circuit with a switch and capacitor is calculated by multiplying the resistance and capacitance values in the circuit. This is represented by the equation τ = RC, where τ is the time constant in seconds, R is the resistance in ohms, and C is the capacitance in farads.

5. What are some common applications of a DC circuit with a switch and capacitor?

A DC circuit with a switch and capacitor has many practical applications, such as in timing circuits, power supplies, audio amplifiers, and electronic filters. It is also commonly used in electronic devices to reduce power supply noise and improve overall performance.

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