Question

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.

Answer 1

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!!!