RLC Circuit DC Homework Solution

[Arman777]

Homework Statement

Homework Equations

The Attempt at a Solution

I used the loop rule where $ΣΔV=0$ and junction rule.
From here I get 4 equations
$i_3=i_1+i_2$
$ε-i_3R_3-i_1R_1-L(di\dt)=0$
$ε-i_3R_3-i_2R_2-Q\C=0$
$i_1R_1-L(di\dt)+Q\C+i_2R_2=0$

1-I said Q\C will disappear since Q=0 at t=0. But what about (di\dt) ? Is (di\dt)=0 ? I think it should be, but If its zero I didnt understand why it asked again since we need to claim its zero to solve the question.If its not zero then I couldn't solve the upper 4 equations.İts like something is missing.

 

[cnh1995]

You do not need any equations here.
Just apply what you know about the behaviour of L and C at the moment of energization i.e. t=0.

What is the current through the inductor "just before" the switch is closed?

 

[Arman777]

You do not need any equations here.
Just apply what you know about the behaviour of L and C at the moment of energization i.e. t=0.

What is the current through the inductor "just before" switch is closed?

Isnt it a time depented value ? Function like I(t)=Awcos(wt)+Bwsin(wt) ? I don't know...

 

[cnh1995]

Isnt it a time depented value ?

Yes.
But you only need to know what happens at t=0.

What is the current through the inductor "just before" the switch is closed?

 

[Arman777]

Yes.
But you only need to know what happens at t=0.

Can you help a bit more..I couldn't see the solution

 

[cnh1995]

There was no source in the circuit before the switch is closed.
So what can you say about the inductor current "before" the switch is closed?

 

[Arman777]

Zero ?

 

[cnh1995]

Zero ?

Yes. So what is the inductor current just after closing the switch?

 

[Arman777]

Yes. So what is the inductor current just after closing the switch?

zero ?

 

[cnh1995]

zero ?

Yes.

 

[Arman777]

1-I said Q\C will disappear since Q=0 at t=0. But what about (di\dt) ? Is (di\dt)=0 ? I think it should be, but If its zero I didnt understand

You could have just say yes.. ? Are my equations wrong ? Cause I need to find currents.

 

[cnh1995]

You could have just say yes.. ?

I just answered your question. You need to be more specific about your problems.

Cause I need to find currents.

Only at t=0. As I said, you do not need any of the equations you've written.

What can you say about the current through the capacitor at t=0 i.e. just after the switch is closed?

 

[cnh1995]

Ok. I think you do need to solve the equations for solving b and c because of presence of R1. The current through R1 is governed by both L and C values. My apologies.
But you do not need them to know i1 and i2 at t=0.

 

[Arman777]

you do not need them to know i1 and i2 at t=0.

Question asks that,I need to know

 

[cnh1995]

Question asks that,I need to know

Yes. So what is the current through the capacitor at t=0?
How does a capacitor behave when energized at t=0?

 

[ehild]

Homework Statement

View attachment 203932

Homework Equations

The Attempt at a Solution

I used the loop rule where $ΣΔV=0$ and junction rule.
From here I get 4 equations
$i_3=i_1+i_2$
$ε-i_3R_3-i_1R_1-L(di\dt)=0$
$ε-i_3R_3-i_2R_2-Q\C=0$
$i_1R_1-L(di\dt)+Q\C+i_2R_2=0$

The fourth equation is not needed. And di/dt should be di₁/dt.
So you have the system of equations
$i_3=i_1+i_2$
$ε-i_3R_3-i_1R_1-L(di_1/dt)=0$
$ε-i_3R_3-i_2R_2-Q/C=0$
Substitute $i_3=i_1+i_2$ for i_3. You might need the relation between Q and i₂. Then you have a system of equations for two currents and their derivatives.
The problem asks the initial values. You found that i₁(0)=0. What about i₂(0)? Answer @cnh1995's question.

 

[Arman777]

Yes. So what is the current through the capacitor at t=0?
How does a capacitor behave when energized at t=0?

Zero ? At the beginning there should be no current ? Seems wrong but also makes sense.

The fourth equation is not needed. And di/dt should be di₁/dt.
So you have the system of equations
$i_3=i_1+i_2$
$ε-i_3R_3-i_1R_1-L(di_1/dt)=0$
$ε-i_3R_3-i_2R_2-Q/C=0$
Substitute $i_3=i_1+i_2$ for i_3. You might need the relation between Q and i₂. Then you have a system of equations for two currents and their derivatives.
The problem asks the initial values. You found that i₁(0)=0. What about i₂(0)? Answer @cnh1995's question.

I see well I find $-i_2R_2=Q\C$ but in any case at t=0,Q=0

 

[BvU]

Yes but for t>0 you should get something else

 

[Arman777]

Yes but for t>0 you should get something else

Rest of the question.Well at $t=∞$ there would be no current on the capacitor brench.

In the brench where inductor exist, there would be current.I can write an equation like, $ε-iR_3-iR_1-L(di\dt)=0$.Do I need to solve this ? I couldn't directly see the answer.
For (e) probably its $εC$.
For (f), I need to find that current passes through inductor,I know the equations and ıf my answer to e is true then I can find it easily.

 

[ehild]

Zero ? At the beginning there should be no current ? Seems wrong but also makes sense.

I see well I find $-i_2R_2=Q\C$ but in any case at t=0,Q=0

No, it is not true . You ignored i3, and it is not zero at t=0.

 

[BvU]

can write an equation like, $\varepsilon-iR_3-iR_1-L(di/dt)=0$. Do I need to solve this ? I couldn't directly see the answer

Yes, no and Oh? What do you expect of the $dI/dt$ in the last term ?
(e) No. draw the circuit a little differently: $\varepsilon$ on the left and all the R etc on the right
(f) What you say is true...

 

[BvU]

No, it is not true . You ignored i3, and it is not zero at t=0.

Shame on me . Wake up first and then check PF. Over to you, Ersbeth

 

[Arman777]

I didnt understand...I ll ask my proffesor I guess...

 

[ehild]

I didnt understand...I ll ask my proffesor I guess...

Substitute I3=I1+I2 into your second and third equations. What do you get?
Use that i1=0 and Q=0 at t=0 in the second equation. What is i2 at t=0?
Knowing i2(0) and i1(0) you get d(i1)/dt from the first equation.

 

[Arman777]

Substitute I3=I1+I2 into your second and third equations. What do you get?
Use that i1=0 and Q=0 at t=0 in the second equation. What is i2 at t=0?
Knowing i2(0) and i1(0) you get d(i1)/dt from the first equation.

I find $i_2=2A$ and $di_1\dt=6A\s$

 

[ehild]

Correct, but use / in fractions, instead of \.
an oblique stroke (/) in print or writing, used between alternatives (e.g., and/or ), in fractions (e.g., 3/4 ), in ratios (e.g., miles/day ), or between separate elements of a text.
So $di_1/dt=6A/s$.
Now take the derivative of the second equation, and use that i₂=dQ/dt. Calculate di₂/dt.

 

[Arman777]

Calculate di2/dt.

Why do we need this ? .I ll do that in a moment

 

[Arman777]

In Part c it asks $q_{upper}/dt$ which that's equal to $i_2=2A$ as you said.

 

[ehild]

Why do we need this ? .I ll do that in a moment

Oh, well, it was not asked.
Then go to the questions about infinite time. What are i1 and i2?

 

[Arman777]

Oh, well, it was not asked.
Then go to the questions about infinite time. What are i1 and i2?

Well $i_2=0$ since capacitor is full.There can't be current.I couldn't find $i_1$...I think its 1A but I am ot sure...I thought $ε=i_1(R_1+R_3)+L(di_1/dt)$
Part (e) I guess Q=εC
Part (f) If its one ampere and I find charge correctly then $U_L=1/2Li^2$ and $U_C=Q^2/2C$

 

[ehild]

Well $i_2=0$ since capacitor is full.There can't be current.I couldn't find $i_1$...I think its 1A but I am ot sure...I thought $ε=i_1(R_1+R_3)+L(di_1/dt)$

After very long time, the circuit is in stationary state. Nothing changes, the time derivative of the currents. i ₁=1 A and i₂=0 are correct.

Part (e) I guess Q=εC
Part (f) If its one ampere and I find charge correctly then $U_L=1/2Li^2$ and $U_C=Q^2/2C$

The capacitor is not connected directly to the battery, there are two resistors in between. Is some potential drop across those resistors?

 

[Arman777]

The capacitor is not connected directly to the battery, there are two resistors in between. Is some potential drop across those resistors?

Oh you are right yes, then $Q/C=ε-i_3R_3-i_2R_2$ $i_2=0$ and $i_3=1A$ ?

 

[ehild]

Oh you are right yes, then $Q/C=ε-i_3R_3-i_2R_2$ $i_2=0$ and $i_3=1A$ ?

Yes.

 

[Arman777]

Yes.

Thanks a Lot.

 

[ehild]

Well done!