contestada

We would like to use the relation V(t)=I(t)RV(t)=I(t)R to find the voltage and current in the circuit as functions of time. To do so, we use the fact that current can be expressed in terms of the voltage. This will produce a differential equation relating the voltage V(t)V(t)V(t) to its derivative. Rewrite the right-hand side of this relation, replacing I(t)I(t)I(t) with an expression involving the time derivative of the voltage.

Respuesta :

Answer:

V = -RC (dV/dt)

Solving the differential equation,

V(t) = V₀ e⁻ᵏᵗ

where k = RC

Explanation:

V(t) = I(t) × R

The Current through the capacitor is given as the time rate of change of charge on the capacitor.

I(t) = -dQ/dt

But, the charge on a capacitor is given as

Q = CV

(dQ/dt) = (d/dt) (CV)

Since C is constant,

(dQ/dt) = (CdV/dt)

V(t) = I(t) × R

V(t) = -(CdV/dt) × R

V = -RC (dV/dt)

(dV/dt) = -(RC/V)

(dV/V) = -RC dt

∫ (dV/V) = ∫ -RC dt

Let k = RC

∫ (dV/V) = ∫ -k dt

Integrating the the left hand side from V₀ (the initial voltage of the capacitor) to V (the voltage of the resistor at any time) and the right hand side from 0 to t.

In V - In V₀ = -kt

In(V/V₀) = - kt

(V/V₀) = e⁻ᵏᵗ

V = V₀ e⁻ᵏᵗ

V(t) = V₀ e⁻ᵏᵗ

Hope this Helps!!!

ajeigbeibraheem

Using the relation V(t) = I(t)R to find the voltage and the circuit as a

function of time. The voltage V(t) can be expressed as [tex]\mathbf{V(t) = - R C \dfrac{dV(t)}{dt}}[/tex]

For a given capacitor with capacitance C and a charge (q); At any instant, the charge stored in the capacitor can be expressed as:

  • q(t) = CV(t)

where;

  • q = charge
  • C = capacitance
  • V = voltage

Taking the differentiation with respect to time (t) from the above equation:

[tex]\mathbf{\dfrac{dq}{dt}= C \dfrac{dV}{dt}}[/tex]

By rewriting the right-hand side of the above relation and replacing it with I(t) we have:

[tex]\mathbf{I(t) = - C \dfrac{dV}{dt}}[/tex]

where;

  • the negative sign from above indicates that the capacitor is discharging.

Recall from Ohm's Law that:

the Voltage (V) in a circuit varies directly proportional to the current (I) and resistance (R)

i.e.

  • Voltage (V)  = current (I) × Resistance (R)

So, making the current (I) the subject of the formula, we have:

[tex]\mathbf{I = \dfrac{V}{R}}[/tex]

∴

[tex]\mathbf{I(t) = - C \dfrac{dV}{dt}}[/tex] can be rewritten as:  [tex]\mathbf{\dfrac{V(t)}{R}= - C \dfrac{dV(t)}{dt}}[/tex]

Making V(t) the as the subject of the equation, we get:

[tex]\mathbf{V(t) = - RC \dfrac{dV(t)}{dt}}[/tex]

Therefore, the voltage V(t) can be expressed as   [tex]\mathbf{V(t) = - RC \dfrac{dV(t)}{dt}}[/tex]

Learn more about voltage here:

https://brainly.com/question/16475751?referrer=searchResults

Otras preguntas