Respuesta :
Answer:
V(t) = (q0/C) * e^(−t/RC )
Explanation:
If there were a battery in the circuit with EMF E , the equation for V(t) would be V(t)=E−(RC)(dV(t)/dt) . This differential equation is no longer homogeneous in V(t) (homogeneous means that if you multiply any solution by a constant it is still a solution). However, it can be solved simply by the substitution Vb(t)=V(t)−E . The effect of this substitution is to eliminate the E term and yield an equation for Vb(t) that is identical to the equation you solved for V(t) . If a battery is added, the initial condition is usually that the capacitor has zero charge at time t=0 . The solution under these conditions will look like V(t)=E(1−e−t/(RC)) . This solution implies that the voltage across the capacitor is zero at time t=0 (since the capacitor was uncharged then) and rises asymptotically to E (with the result that current essentially stops flowing through the circuit).
The solution to the differential equation V(t)=−CRdV(t)dt for the initial conditions given in the problem introduction in order to find the voltage as a function of time for any time t is [tex]\mathbf{V(t) = \dfrac{q_o}{C} \ e ^{-\dfrac{1}{RC}t }}[/tex]
From the given differential equation:
[tex]\mathbf{V(t) = -CR \dfrac{dV(t)}{dt} }[/tex]
The best way to solve this differential equation is to use the separation of the variable method.
Separation of the variable method enables us to produce equivalence between two integrals that we can evaluate by rewriting their differential equation.
∴
[tex]\mathbf{\dfrac{dV(t)}{V(t)} = -\dfrac{1}{RC}t}[/tex]
By applying integration;
[tex]\mathbf{\int{ \dfrac{dV(t)}{V(t)} = - \int \dfrac{1}{RC} \ dt}}[/tex]
[tex]\mathbf{ In(V(t)) = - \dfrac{1}{RC} \ t+ In \ A}}[/tex]
[tex]\mathbf{In \dfrac{V(t)}{A} = e^{-\dfrac{1}{RC}t}}[/tex]
[tex]\mathbf{V(t) = Ae ^{-\dfrac{1}{RC}t} }[/tex]
where;
- In A = integration constant
From above, the initial conditions are not given. As such;
- when t = 0; V(t) = V₀
we can infer that: V₀ = A
∴
[tex]\mathbf{V(t) = V_oe ^{-\dfrac{1}{RC}t} }[/tex]
Recall from the amount of charge of a voltage in capacitance is:
[tex]\mathbf{V_o = \dfrac{q_o}{C}}[/tex]
Hence,
[tex]\mathbf{V(t) = \dfrac{q_o}{C}e ^{-\dfrac{1}{RC}t} }[/tex]
Therefore, we can conclude that the answer in terms of q0, C, R, and t is:
[tex]\mathbf{V(t) = \dfrac{q_o}{C}e ^{-\dfrac{1}{RC}t} }[/tex]
Learn more about differential equations here:
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