Differentiating Kirchoff's voltage law expression

In summary, V0 - iR - q/C = 0 to prove that di/dt = -i/RC. The Attempt at a Solution states that V0 - (dq/dt)R - q/C = 0 to prove that di/dt = -i/RC.
  • #1
And123
2
0

Homework Statement



Differentiate V0 - iR - q/C = 0 to prove that di/dt = -i/RC.

Homework Equations



V0 - iR - q/C = 0
^ derived from previous question for a circuit that had one battery with emf V0, a resistor of resistance R and a capacitor of capacitance C (all in series).
di/dt = -i/RC

The Attempt at a Solution



V0 - iR - q/C = 0
V0 - (dq/dt)R - q/C = 0
V0C - (dq/dt)RC - q = 0
(dq/dt)RC = V0C - q
d/dt((dq/dt)RC) = d/dt(V0C - q)
RC(d/dt(dq/dt) = -dq/dt
RC(di/dt) = -i
di/dt = -i/RC

^ not sure if my maths makes sense (not very good at doing dif/integration when there are multiple variables). I did get the "answer" (which is easy, since its given), but obviously the method's what matters.

Thanks a lot!
 
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  • #2
It's right but you may have gotten there quicker if you just differentiated this:

V0 - iR - q/C = 0

with respect to time right away. You'll notice the answer is in terms of di/dt so you don't really want to substitute i=dq/dt into that equation.
 
  • #3
Thanks for the reply.

Do you mean like this?:

V0 - iR - q/C = 0
d/dt(V0 - iR - q/C) = d/dt(0)
d/dt(V0) + d/dt(-iR) + d/dt(-q/C) = 0
0 - R(di/dt) - (1/C)(dq/dt) = 0
-R(di/dt) - (i/C) = 0
di/dt = -i/(RC)

^ I pull out the R and C like they're constants (which I think is correct), and got rid of V0 like you would when you dif a constant.

Edit: I realize the 3rd step isn't required, but just to step it out for myself, I included it.
 
  • #4
Yes that's right too. I wouldn't have shown so many steps but it's not too different anyway.

V0 - iR - q/C = 0
d/dt (V0 - iR - q/C) = d/dt (0)
-R di/dt - i/C = 0 ;; maybe they want you to say i = dq/dt here
di/dt = - i/(RC)
 
  • #5




Your solution appears to be correct. To differentiate the given expression, you first need to use the chain rule to differentiate the term (dq/dt)R. Then, using the product rule, you can differentiate the term (dq/dt)(RC) and equate it to the derivative of V0C - q. From there, you can solve for di/dt and arrive at the given expression of -i/RC. Your method is correct and shows a good understanding of differentiation. Keep up the good work!
 

Related to Differentiating Kirchoff's voltage law expression

1. What is Kirchoff's voltage law?

Kirchoff's voltage law is a fundamental principle in circuit analysis that states that the sum of all voltages in a closed loop in a circuit must equal zero. This law is based on the conservation of energy and is used to determine the voltage drops and current flows in a circuit.

2. How is Kirchoff's voltage law applied?

Kirchoff's voltage law is applied by writing a loop equation for a closed loop in a circuit. This equation includes the voltage sources, resistors, and any other circuit elements in the loop. The sum of all voltage drops in the loop must equal the sum of all voltage sources in the loop.

3. What is the difference between Kirchoff's voltage law and Kirchoff's current law?

Kirchoff's voltage law deals with the conservation of energy in a closed loop in a circuit, while Kirchoff's current law deals with the conservation of charge at a node in a circuit. In other words, Kirchoff's voltage law focuses on voltages while Kirchoff's current law focuses on currents.

4. Can Kirchoff's voltage law be applied to any circuit?

Yes, Kirchoff's voltage law can be applied to any circuit, as long as the circuit is a closed loop. This means that the starting and ending points of the loop must be the same, and there should be no open branches in the loop.

5. What are some practical applications of Kirchoff's voltage law?

Kirchoff's voltage law is essential in circuit analysis and is used to design and troubleshoot electronic circuits. It is also used in various fields such as power systems, telecommunications, and signal processing. Additionally, it is a foundational concept in understanding the behavior of batteries and other power sources.

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