Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Current through a particular branch of RC circuits

1 vote

I'm done with part A, B and C, please help me to find the current through the resistor if I can calculate the current through $1.2M$ ohm then using Kirchoff's laws we can do part D, but I don't know how, please help.

asked May 18, 2019 in Physics Problems by n3 (508 points)

1 Answer

2 votes
Best answer

Initially no current flows through the $600k\Omega$ resistor. It all flows onto the uncharged capacitor. After a long time the capacitor will be fully charged and the current through the $600k\Omega$ resistor is the same as that through the battery, which is $\frac{80}{3}\mu A$.

So the current through the resistor increases exponentially from $0$ to a limit of $$I_{\infty} = \frac{80}{3} \mu A$$ The rate of increase is set by the time constant of the parallel combination of resistor and capacitor, which is $\tau=CR$, where $R=600k\Omega$ and $C=2.5nF$. Such an increase can be written as $$I(t)=I_{\infty}(1-e^{-t/ \tau})$$

answered May 22, 2019 by sammy gerbil (28,876 points)
selected May 22, 2019 by n3
From "the rate of increase is set...", please explain more. why such an expression of $I(t)$ can be written?
The expression for $I(t)$ can be written from experience. A current which decreases exponentially from an initial value $I_0$ to a limiting value of $0$ can be written $I_0 e^{-t/\tau}$. A current which increases from $0$ to a final limiting value of $I_{\infty}$ can be written $I_{\infty}(1-e^{-t/\tau})$.

The rate at which any exponential grows or decays depends on the factor in the exponent. So $e^{2at}$ grows faster than $e^{at}$. One way of writing the exponential factor is as a fraction such as $e^{-t/\tau}$. Here $\tau$ (aka "time constant") is a characteristic time with the meaning that the current grows by a factor $e$ or decays by a factor of $1/e$ after a time interval of $\tau$.

If you differentiate the last equation in the answer box and substitute $t=0$ you will see that although the current through the resistor is initially zero it increases linearly at first : $$\frac{dI}{dt}|_0=\frac{I_{\infty}}{\tau}$$