Finding expression for the current

In summary, the conversation discusses finding an expression for the current i(t) as a function of C and V_out (t) after t = 0 in a circuit. The equation V(t) = Ri(t) + V_out (t) is deemed incorrect and the concept of Kirchoff's Voltage Law is introduced to solve the problem. The final equation is written as +Vs - Vc - Vout = 0 and the process of substituting equations and solving is explained.
  • #1
Ahmed Hassona
3
1
< Mentor Note -- thread moved to HH from the technical forums, so no HH Template is shown >

Hi all,
I'm trying to find an expression for the current i(t) as a function of C and V_out (t) after t = 0. I have an answer but still not sure at all if its correct or not.
Here is my answer:
I know that
i(t) = C*dV(t) / dt
V(t) = Ri(t) + V_out (t)

combining them together

i(t) = C*dV_out(t)/dt - dV_out(t) / R

Here is the circuit.
ff.jpg


Thanks
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Ahmed Hassona said:
< Mentor Note -- thread moved to HH from the technical forums, so no HH Template is shown >

Hi all,
I'm trying to find an expression for the current i(t) as a function of C and V_out (t) after t = 0. I have an answer but still not sure at all if its correct or not.
Here is my answer:
I know that
i(t) = C*dV(t) / dt
V(t) = Ri(t) + V_out (t)

combining them together

i(t) = C*dV_out(t)/dt - dV_out(t) / R

Here is the circuit.
View attachment 114129

Thanks
Welcome to the PF. I've moved your thread to the schoolwork forums, where it fits better than in the EE forum.

This equation "V(t) = Ri(t) + V_out (t)" looks wrong to me. There is the voltage source, the voltage across the capacitor, and Vout which is I*R. Do you want to take another cut at the equations? Or am I just misreading them?
 
  • #3
berkeman said:
Welcome to the PF. I've moved your thread to the schoolwork forums, where it fits better than in the EE forum.

This equation "V(t) = Ri(t) + V_out (t)" looks wrong to me. There is the voltage source, the voltage across the capacitor, and Vout which is I*R. Do you want to take another cut at the equations? Or am I just misreading them?

Thanks for moving it and for your reply.
It's the first time I work with electrecity so I am sure that you are right about this eqn "V(t) = Ri(t) + V_out (t)" is wrong. But then how can I substitute V(t) with V_out (t) in the i(t) eqn. ?
 
  • #4
Ahmed Hassona said:
Thanks for moving it and for your reply.
It's the first time I work with electrecity so I am sure that you are right about this eqn "V(t) = Ri(t) + V_out (t)" is wrong. But then how can I substitute V(t) with V_out (t) in the i(t) eqn. ?
So you are basically writing a "loop" equation for the circuit. Have you learned Kirchoff's Voltage Law yet? It says that the sum of all voltages around a closed loop circuit will be zero. So call the voltage source on the left Vs, the voltage across the capacitor Vc, and the voltage across the resistor Vout. Add the voltages as you go around the loop clockwise, from the bottom node, all the way back around to the bottom node.

Since you see the - terminal of the voltage source first, and you are traveling from - to + on the part, you call that voltage a positive gain. Since you go across the capacitor and resistor from + to -, you call that a negative voltage gain (or a voltage drop).

So you would write an equation like this:

+Vs - Vc - Vout = 0

Then substitute the equations you wrote for the Vc and Vout, and solve the resulting equation. Does that help?
 
Last edited:
  • #5
berkeman said:
So you are basically writing a "loop" equation for the circuit. Have you learned Kirchoff's Voltage Law yet? That the sum of all voltages around a closed loo-p circuit will be zero? So call the voltage source on the left Vs, the voltage across the capacitor Vc, and the voltage across the resistor Vout. Add the voltages as you go around the loop clockwise, from the bottom node, all the way back around to the bottom node.

Since you see the - terminal of the voltage source first, and you are traveling from - to + on the part, you call that voltage a positive gain. Since you go across the capacitor and resistor from + to -, you call that a negative voltage gain (or a voltage drop).

So you would write an equation like this:

+Vs - Vc - Vout = 0

Then substitute the equations you wrote for the Vc and Vout, and solve the resulting equation. Does that help?

Yeah it helped me alot. Thank you so much, really appreciate it :smile:
 
  • Like
Likes berkeman

Related to Finding expression for the current

1. What is the formula for finding the expression for current?

The formula for finding the expression for current is I = Q/t, where I represents the current, Q represents the charge, and t represents the time.

2. How is the expression for current different from the formula for Ohm's law?

The expression for current is different from Ohm's law because it takes into account the charge and time, whereas Ohm's law only considers the voltage and resistance in a circuit.

3. Can the expression for current be used for both direct and alternating current?

Yes, the expression for current can be used for both direct and alternating current, as long as the value for charge and time are appropriately calculated for the specific type of current.

4. How is the expression for current affected by the presence of a resistor in a circuit?

The expression for current is affected by the presence of a resistor in a circuit because the resistance will limit the flow of current and result in a lower current value.

5. Is the expression for current affected by the number of branches in a circuit?

Yes, the expression for current is affected by the number of branches in a circuit because the current will divide among the different branches, resulting in a different current value for each branch.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
9
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
28
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
16
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
16
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
19
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
21
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
20
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
Back
Top