Finding expression for the current

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The discussion focuses on finding an expression for the current i(t) in relation to capacitance C and output voltage V_out(t) after time t = 0. The initial attempt at deriving the equation included an incorrect relationship between V(t), i(t), and V_out(t). A suggestion was made to apply Kirchhoff's Voltage Law to create a loop equation for the circuit, emphasizing the need to correctly account for voltage gains and drops. The user expressed gratitude for the clarification, indicating that the explanation was helpful in understanding the problem better. Overall, the conversation highlights the importance of accurate circuit analysis in electrical engineering.
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< 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
 
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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?
 
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. ?
 
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?
 
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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:
 
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