# Hi please explain it step by steps. ThanksA constant voltage is applied to an electrical circuit containing a capacitor of capacitance C in series with a resistor R. By Kirchoff’s law the charge...

Hi please explain it step by steps.

Thanks

A constant voltage is applied to an electrical circuit containing a capacitor of capacitance *C *in series with a resistor *R*. By Kirchoff’s law the charge *q *in the circuit satisfies the equation R=dq/dt +1/c*q =V0 .Derive the formula for *q *given that *q=0 *at time *t=0*. Derive the formula for the current, *i*, in the circuit where .i = dq/dt . Derive the time it takes for the current to reach i =V0 /2R

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### 1 Answer

We can solve for dq/dt

`dq/dt = R-1/c*q`

Separating variables we get

`(dq)/(Rc-q) = dt/c`

Integrating we get

`-ln(Rc-q) = t/c + C`

`ln(Rc-q) = -(t/c + C)`

`Rc - q = e^(-(t/c+C))`

`q = Rc - e^(-C)(e^(-t/c))`

`q = Rc - C_1e^(-t/c)`

Give us `q = Rc - Ce^(-t/c)`

We can determine C by initial conditions of q=0 when t = 0. Substituting we get

`0 = cR - Ce^(-0/c) = cR - C`

So C = cR so our equation of charge as a function of time is

`q = cR(1 - e^(-t/c))`

We already have `(dq)/(dt) = R - 1/c*q`

We can substitute our equation for q to get I as a function of time

`(dq)/(dt) = I = R - 1/c(cR(1-e^(-t/c)) = R - R(1-e^(-t/c)) = R - R + Re^(-t/c) = Re^(-t/c)`

So `I = Re^(-t/c)`

We are finally asked when is `I = V_0/(2R)` So when is `V_0/(2R) = Re^(-t/c)`

This gives `e^(-t/c) = V_0/(2R^2)` , take the ln of both sides we get

`-t/c = ln(V_0/(2R^2))` solving for t we get

`t = c ln((2R^2)/V_0)`

So to sumarize we get

`q = Rc(1-e^(-t/c))`

`(dq)/(dt) = I = Re^(-t/c)`

and `I = V_0/(2R)` when `t = cln((2R^2)/V_0)`