# RC DIFFERENTIAL EQUATION

Cuthbert Nyack
Equating the voltages across the resistor and capacitor to the applied voltage gives the following equation. Which can be rearranged to give a first order differential equation for q(t). The differential equation above can also be deduced from conservation of energy as shown below. If an interval of time dt is considered during which time an amount of charge dq is transferred from the supply to the capacitor, then the work done by the supply must equal the energy dissipated in the resistor plus the increase in energy stored in the capacitor. Rearranging this equation for conservation of energy produces the same first order inhomogeneous differential equation as above.
The differential equation can be solved in more than one way. The solutions using integrating factors and Laplace Transform are illustrated below.

## Solving diff eqn with the integrating factor

The differential equation cannot be integrated directly because of the term on the right hand side. The general idea is to change q(t) into a(t)q(t) with a(t) chosen so the left hand side of the equation can be written as the derivative of a(t)q(t). Multiplying by a(t) gives the following eqn. The aim is to choose a(t) so the left hand side of the diff eqn can be written as:- Equating the second term on the left hand side of the diff eqn with the second term on the right hand side of the above eqn and solving for a(t) gives:- The differential equation can now be written as:- The above equation can now be integrated directly to give give the following general solution. Assuming V(t) is a constant V, then the above eqn simplifies to:- and rearranging gives an expression for the capacitor voltage after the supply is switched on. ## Solving diff eqn with the Laplace Transform

Taking the Laplace Transform of both sides of the differential equation gives the following, where Q(s) is the LT of q(t) and V(s) is the LT of V(t). Solving for Q(s) gives Assuming the supply voltage is a step function with the following LT Then Q(s) simplifies to:- with inverse q(t) given by:- As above the voltage across the capacitor can be written as:- When the supply is removed and the capacitor discharges through the resistor, the differential equation becomes. This can be integrated directly to give:- With the capacitor and resistor voltages given by:- 