Deriving Capacitor Voltage for a Discharging Capacitor

[Potatochip911]

Homework Statement

I'm trying to derive the voltage waveform across the capacitor for a discharging capacitor that has been fully charged by a DC power supply $v_0$, i.e. $v_c(t=0)=v_0$ and then at $t=0$ the switch is flipped and the capacitor starts to discharge.

Homework Equations

The Attempt at a Solution

From KVL we obtain
$$v_c=\frac{q}{c}=iR$$
Taking the time derivative of this and since capacitor is discharging $-\frac{dq}{dt}=i$... $$-\frac{i}{RC}=\frac{di}{dt}\\ -\int_0^t \frac{dt}{RC}=\int_{i_0}^{i}\frac{di}{i}\\ -\frac{t}{RC}=\ln (\frac{i}{i_0})\\ i=i_0e^{-\frac{t}{RC}}$$
Now to solve for $i_0$ at t=0 $\frac{q}{c}=i_0Re^{-\frac{t}{RC}}$ becomes $$\frac{v_0}{c}=i_0R\\ i_0=\frac{v_0}{RC}\Longrightarrow i=\frac{v_0}{RC}e^{-\frac{t}{RC}}$$
Finally from KVL $$v_c=iR=\frac{v_0}{C}e^{-\frac{t}{RC}}$$

which is not the correct answer.

 

[cnh1995]

vc=iR=v0Ce−tRC

which is not the correct answer

Of course it isn't. You can see it is dimensionally incorrect.(Vo/C doesn't give voltage).

v0c=i0R

Where does this come from?

 

[Potatochip911]

Of course it isn't. You can see it is dimensionally incorrect.(Vo/C doesn't give voltage).

Where does this come from?

I got that from the KVL equation (q/c=ir) and at t=0 so $\frac{q(0)}{c}=i(0)R$

 

[ehild]

Q/C=Vo, not Vo/C .

 

[Potatochip911]

Q/C=Vo, not Vo/C .

Whoops, after making the proper substitution then we get $i_0=\frac{v_0}{R}\Longrightarrow i=\frac{v_0}{R}e^{-\frac{t}{RC}}\Longrightarrow V_c=iR=v_0e^{-\frac{t}{RC}}$ which is the correct expression :)

 

[ehild]

It is useful to check the dimensions during a derivation.