- #1

Lambda96

- 158

- 59

- Homework Statement
- Calculate the current as a function of time when discharging the capacitor

- Relevant Equations
- none

Hi,

I am not sure if I have calculated the task b correctly.

I always interpret an open switch as an infinitely large resistor, which is why no current is flowing through this "resistor". So there is no current in the red circle, as it was the case in task part a, but only in the blue circle.

Since the resistors are connected in series, I combined them into one, so ##R_g=R_1+R_3## Then I set up Kirchhoff's rules ##\frac{q}{C}=-R_g I## after that I rewrote the current as follows ##\frac{dq}{dt}## and put it into the equation ##\frac{q}{C}=-R_g \frac{dq}{dt}## and rewrote it as follows and integrated it afterwards.

$$-\frac{1}{R_g C}dt=\frac{1}{q}dq$$

$$ \int_{0}^{t} -\frac{1}{R_g C}dt= \int_{Q_0}^{Q} \frac{1}{q} dq $$

$$-\frac{t}{R_g C}=\ln(\frac{Q}{Q_0})$$

Solve the equation for Q: ##Q=Q_0e^{-\frac{t}{R_g C}}##

To get the current now, I simply derived the equation for the charge with respect to time.

$$I(s)=\frac{dQ}{dt}=-\frac{Q_0}{R_g C}e^{-\frac{t}{R_g C}}$$

I am not sure if I have calculated the task b correctly.

I always interpret an open switch as an infinitely large resistor, which is why no current is flowing through this "resistor". So there is no current in the red circle, as it was the case in task part a, but only in the blue circle.

Since the resistors are connected in series, I combined them into one, so ##R_g=R_1+R_3## Then I set up Kirchhoff's rules ##\frac{q}{C}=-R_g I## after that I rewrote the current as follows ##\frac{dq}{dt}## and put it into the equation ##\frac{q}{C}=-R_g \frac{dq}{dt}## and rewrote it as follows and integrated it afterwards.

$$-\frac{1}{R_g C}dt=\frac{1}{q}dq$$

$$ \int_{0}^{t} -\frac{1}{R_g C}dt= \int_{Q_0}^{Q} \frac{1}{q} dq $$

$$-\frac{t}{R_g C}=\ln(\frac{Q}{Q_0})$$

Solve the equation for Q: ##Q=Q_0e^{-\frac{t}{R_g C}}##

To get the current now, I simply derived the equation for the charge with respect to time.

$$I(s)=\frac{dQ}{dt}=-\frac{Q_0}{R_g C}e^{-\frac{t}{R_g C}}$$