Find the transient current in this circuit

[Helly123]

Homework Statement

Homework Equations

The Attempt at a Solution

I often do circuit. But still there is always stuff i don't get..

I was asked to find
1) I1 I2 I3 immidiately after S closed
2) I1 I2 I3 after sufficient of time
3) I1 I2 I3 immidiately after S opened

Just give me clue please? Where should i start?
I find the current I1 is 1 mA because V/R = 6V/6kohm is 1 mA
Is I3 is 0 mA?

 

[kuruman]

You start by realizing that (a) as soon as the switch is closed, the capacitor acts as if it were a short (straight wire); (b) after sufficient time, the capacitor acts as an open circuit (cut wire); (c) immediately after the switch is opened again, the capacitor acts as a battery of voltage equal to the voltage it had before the switch is opened. Draw 3 separate circuits and analyze them.

 

[Helly123]

You start by realizing that (a) as soon as the switch is closed, the capacitor acts as if it were a short (straight wire); (b) after sufficient time, the capacitor acts as an open circuit (cut wire); (c) immediately after the switch is opened again, the capacitor acts as a battery of voltage equal to the voltage it had before the switch is opened. Draw 3 separate circuits and analyze them.

What kind of theories i should look up? Is there websites you suggest pls?

 

[kuruman]

There are no theories other than those you must have already seen. What I wrote in #2 is described by the following two equations
1. Charging capacitor
$$V(t)=V_{\infty}\left( 1-e^{-t/\tau} \right)$$Here $V_{\infty}$ is the voltage across the capacitor after sufficiently long time has elapsed and $\tau$ is the time constant. It is assumed that at t = 0 some kind of switch is moved to get the capacitor to start charging.
2. Discharging capacitor
$$V(t)=V_0e^{-t/\tau}$$ Here $V_0$ is the initial voltage at time t = 0. It is assumed that at t = 0 some kind of switch is moved to get the capacitor to start to discharge.

As always, the charge on the capacitor is related to the voltage across its plates by $Q=CV$. To see where (a) comes from in post #2 start with an uncharged capacitor connected to a battery and a resistor through an open switch. At time t = 0, the switch is closed. At that instant the voltage across the capacitor is zero because $Q=0$ and $V=Q/C$. The equation for the charging capacitor reflects that because $V_{\infty}(1-e^0)=0$ Well, the voltage is also zero across any straight segment of wire in the circuit. Hence statement (a) in post #2. It will be instructive to you to see how statements (b) and (c) in #2 follow from the equations.

 

[Helly123]

There are no theories other than those you must have already seen. What I wrote in #2 is described by the following two equations
1. Charging capacitor
$$V(t)=V_{\infty}\left( 1-e^{-t/\tau} \right)$$Here $V_{\infty}$ is the voltage across the capacitor after sufficiently long time has elapsed and $\tau$ is the time constant. It is assumed that at t = 0 some kind of switch is moved to get the capacitor to start charging.
2. Discharging capacitor
$$V(t)=V_0e^{-t/\tau}$$ Here $V_0$ is the initial voltage at time t = 0. It is assumed that at t = 0 some kind of switch is moved to get the capacitor to start to discharge.

As always, the charge on the capacitor is related to the voltage across its plates by $Q=CV$. To see where (a) comes from in post #2 start with an uncharged capacitor connected to a battery and a resistor through an open switch. At time t = 0, the switch is closed. At that instant the voltage across the capacitor is zero because $Q=0$ and $V=Q/C$. The equation for the charging capacitor reflects that because $V_{\infty}(1-e^0)=0$ Well, the voltage is also zero across any straight segment of wire in the circuit. Hence statement (a) in post #2. It will be instructive to you to see how statements (b) and (c) in #2 follow from the equations.

V at t = 0 is zero. But why the Capacitors act like a straight wire? Why the voltage of Resistor 2 that become 0 instead..

 

[kuruman]

V at t = 0 is zero. But why the Capacitors act like a straight wire?

Because when the voltage or potential difference between any points A and B on a straight wire with no circuit elements in between is zero. Straight wires are equipotentials and so are the plates of a capacitor that have zero voltage across them. If you take a capacitor that is uncharged and connect the two plates with a wire, nothing will happen. Remember that capacitors act like straight wires instantaneously. As soon as any charge collects on the plates, the uncharged capacitor will no longer act as a straight wire. The voltage across R₂ will also be instantaneously zero and will match the voltage across the capacitor at any time because the two are in parallel.

 

[Helly123]

Because when the voltage or potential difference between any points A and B on a straight wire with no circuit elements in between is zero. Straight wires are equipotentials and so are the plates of a capacitor that have zero voltage across them. If you take a capacitor that is uncharged and connect the two plates with a wire, nothing will happen. Remember that capacitors act like straight wires instantaneously. As soon as any charge collects on the plates, the uncharged capacitor will no longer act as a straight wire. The voltage across R₂ will also be instantaneously zero and will match the voltage across the capacitor at any time because the two are in parallel.

So the Voltage for Capacitor is 0 at first before charged and act like a straight wire?

 

[Helly123]

Because when the voltage or potential difference between any points A and B on a straight wire with no circuit elements in between is zero. Straight wires are equipotentials and so are the plates of a capacitor that have zero voltage across them. If you take a capacitor that is uncharged and connect the two plates with a wire, nothing will happen. Remember that capacitors act like straight wires instantaneously. As soon as any charge collects on the plates, the uncharged capacitor will no longer act as a straight wire. The voltage across R₂ will also be instantaneously zero and will match the voltage across the capacitor at any time because the two are in parallel.

At first the Capacitor is zero (not charged). Then, it gets charged. Do you mean Capacitor stop act as straight wire instantaneously? (From not charged to get charged.)

 

[kuruman]

So the Voltage for Capacitor is 0 at first before charged and act like a straight wire?

Yes, this is the case immediately after (at t = 0) switch S is closed. It's an instantaneous situation.

At first the Capacitor is zero (not charged). Then, it gets charged. Do you mean Capacitor stop act as straight wire instantaneously? (From not charged to get charged.)

Yes. It's like releasing a ball from rest in free fall. Immediately after you let go, the velocity of the ball is zero but it's not zero at any later time. You say that the ball has zero velocity at time t = 0. Likewise, you can say that the capacitor acts as a straight wire at time t = 0.

 

[Helly123]

Yes, this is the case immediately after (at t = 0) switch S is closed. It's an instantaneous situation.

Yes. It's like releasing a ball from rest in free fall. Immediately after you let go, the velocity of the ball is zero but it's not zero at any later time. You say that the ball has zero velocity at time t = 0. Likewise, you can say that the capacitor acts as a straight wire at time t = 0.

So, Capacitor and Resistor 2 have same voltage that is zero?
If Capacitor act like a straight wire, what does this imply to I2? What i get is the Capacitor is just like a wire without circuit elements, so I3 = I1, and since R2 is parallel to Capacitor, and I3 + I2 = I1. Will it make I2 zero?
So the voltage C = R2 = 0. I3 = I1. I2 = 0

 

[kuruman]

So, Capacitor and Resistor 2 have same voltage that is zero?

Yes, they always have the same voltage and that voltage is instantaneously zero at t = 0.

So the voltage C = R2 = 0. I3 = I1. I2 = 0

Yes. What is the value in Amperes for I1 (or I3)?

 

[Helly123]

Yes, they always have the same voltage and that voltage is instantaneously zero at t = 0.

Yes. What is the value in Amperes for I1 (or I3)?

That is. I am still confused. Should we take into account the R2? When R2 has no current? I think no right? So, I1 = 6V/2kohm = 3mA

 

[kuruman]

Should we take into account the R2? When R2 has no current? I think no right?

Right. When R2 has no current it's like a straight wire the same as the capacitor and there is zero voltage across it.

So, I1 = 6V/2ohm = 3A

Not exactly. R1 = 2 kΩ, not 2 Ω.

 

[Helly123]

Right. When R2 has no current it's like a straight wire the same as the capacitor and there is zero voltage across it.

Not exactly. R1 = 2 kΩ, not 2 Ω.

I have edited. Btw, after sufficient of time. Capacitor gets charged. So, I3 = 0? No direct current across capacitor?

 

[kuruman]

Btw, after sufficient of time. Capacitor get chared. So, I3 = 0? No direct current across capacitor?

Correct. This gives you a new circuit to analyze in which the capacitor is like an open switch. What does this say about the currents I1, I2 and I3?

 

[Helly123]

Correct. This gives you a new circuit to analyze in which the capacitor is like an open switch. What does this say about the currents I1, I2 and I3?

I3 = 0. I2 = I1. But.. if we take into account R2 to find I3, R2 + R1 = 4/3 kohm. Should we take into account R2? Capacitor will have voltage. Will the voltage affect the 6V ?

 

[kuruman]

R2 + R1 = 4/3 kohm.

How do you figure? If R1 = 2 kΩ and R2 = 4 kΩ, what is R1 + R2?

 

[Helly123]

How do you figure? If R1 = 2 kΩ and R2 = 4 kΩ, what is R1 + R2?

Ok. R1 and R2 are series. So total R = 6kohm
I1 = I2 = 1mA
And I3 = 0mA

 

[kuruman]

That is correct. So you have answered parts (a) and (b). Can you figure out part (c)?

 

[Helly123]

That is correct. So you have answered parts (a) and (b). Can you figure out part (c)?

The charged Capacitor will act like a battery now, with V = 6V? I1 = 0 mA. I2 = 6V/4kohm ?

 

[kuruman]

The charged Capacitor will act like a battery now, with V = 6V?

No. Remember that the voltage across the capacitor is always the same as the voltage across R2. What is that voltage after sufficient time has elapsed?

 

[Helly123]

No. Remember that the voltage across the capacitor is always the same as the voltage across R2. What is that voltage after sufficient time has elapsed?

4V. Then I2 = 1mA
How about I3? Is I3 = I2 = 1mA?

 

[kuruman]

4V. Then I2 = 1mA
How about I3? Is I3 = I2 = 1mA?

Yes I2 is 1 mA because you know that the voltage across it is 4 V. What is the argument that I2 = I3? The sure way to find I3, is to find I1 first because you know that I2 + I3 = I1.

 

[Helly123]

Yes I2 is 1 mA because you know that the voltage across it is 4 V. What is the argument that I2 = I3? The sure way to find I3, is to find I1 first because you know that I2 + I3 = I1.

I3 = -1mA? Does it mean opposite direction?

 

[kuruman]

I3 = -1mA?

No. Can you find I1? What is the voltage across R1? Look at the circuit. What must the voltage across R1 plus the voltage across the capacitor be equal to?

 

[Helly123]

No. Can you find I1? What is the voltage across R1? Look at the circuit. What must the voltage across R1 plus the voltage across the capacitor be equal to?

S is opened. R1 not connected? I1 =0 mA

 

[kuruman]

Correct. So now you know I1 and I2. To complete the picture, is the current in branch with the capacitor flowing clockwise or counterclockwise?

 

[Helly123]

Correct. So now you know I1 and I2. To complete the picture, is the current in branch with the capacitor flowing clockwise or counterclockwise?

Since the result is negative . The direction is counter clockwise. Battery is clockwise

 

[kuruman]

Correct. Also note that the upper plate of the capacitor is positively charged and current flows from the positive to the negative plate.

 

[Helly123]

Correct. Also note that the upper plate of the capacitor is positively charged and current flows from the positive to the negative plate.

Maybe you mean the bottow is positive, so current go from bottom (+) to upper (-) , counterclockwise

 

[kuruman]

Maybe you mean the bottow is positive, so current go from bottom (+) to upper (-) , counterclockwise

No, the top plate of the capacitor is connected to the positive terminal of the battery according to your diagram. So the top plate is at higher potential than the bottom plate. Current flows in the direction of I2 in your drawing and opposite to I3.

 

[Helly123]

No, the top plate of the capacitor is connected to the positive terminal of the battery according to your diagram. So the top plate is at higher potential than the bottom plate. Current flows in the direction of I2 in your drawing and opposite to I3.

Ok. Thanks sir. You save me