- #1

U.Renko

- 57

- 1

## Homework Statement

A resistor with resistance R and a capacitor with capacitance C are connected in series to a direct current battery ε.

find the current and charge on the circuit as function of time.

it looks more like a review of differential equations, so I'm not really sure if I should post here or in the calculus forum...feel free to move it if you think it's better.

## Homework Equations

potential at resistor: [itex] V_r = Ri [/itex]

potential at capacitor:[itex] V_c = \frac{q}{C} [/itex]

## The Attempt at a Solution

Applying one of the Kirchhoff's law:

[itex] \epsilon - Ri - \frac{q}{C} = 0 [/itex]

[itex] \epsilon = Ri +\frac{q}{C} [/itex]

since [itex] i = \frac{dq}{dt} [/itex] we can rewrite the equation as

[itex] \frac{\epsilon}{R} = \frac{dq}{dt} + \frac{q}{RC} [/itex] and solve with an integration factor [itex] e^{\frac{t}{RC}} [/itex]

so we have:

[itex] \frac{\epsilon}{R} e^{\frac{t}{RC}} = \frac{d}{dt}(q e^{\frac{t}{RC}}) [/itex]

and then:

[itex] \int \frac{\epsilon}{R} e^{\frac{t}{RC}} dt = q e^{\frac{t}{RC}} [/itex]

here I'm kinda stuck:

because the problem did not give any initial conditions, should I just solve an indefinite integral or integrate from zero to an arbitrary t??

also: [itex] q e^{\frac{t}{RC}} [/itex] comes from integrating [itex] \int_{a}^{b} \frac{d}{dt}(q e^{\frac{t}{RC}})dt [/itex] and applying the fundamental theorem of calculus.

However, don't I need the values a and b to properly use it?