2
\$\begingroup\$

I'm a beginner with RC circuits, I would appreciate any hint or help

The problem asks to find the voltage across the current source in the below circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

while the graph of i_s(t) is given like this:

enter image description here

desmos.com

I know that from KVL we have $$v_s=V_R+V_C$$ while V_R and V_C are the Voltages across the Resistor and the Capacitor

we also know that: $$V_R(t)=R.i_s(t)$$ and $$V_C(t)=\frac{1}{C}\int _{t_0}^ti_s\left(t\right)dt\:+\:V_C\left(0\right)$$ But I have problem wiht determining the bounds of integrals or how to devide time intervals in order to calculate the integral

I also found the i_s based on singularity functions: $$i_s(t)=2r(t+1)-3r(t)+2r(t-2)-r(t-3)-r(t-4)+r(t-5)$$ but I don't know if it is helpful

sorry for my weak grammar

\$\endgroup\$
1
\$\begingroup\$

You need to be given or make an assumption about the initial capacitor voltage (at \$t=-1\$).

I would suggest that you perform a separate integral over each time period where the current has a constant slope, such as from \$t=-1\$ to \$t=0\$. The initial condition for each subsequent integral would be the final capacitor voltage for the integral over the previous time interval.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks for your help, the initial capacitor voltage was not given in the problem but I assumed it is 0 \$\endgroup\$ – Emily May 7 '20 at 0:08
1
\$\begingroup\$

Start with the capacitor voltage. You can write equations for the current as a function of time for each interval. You need to incorporate the initial capacitor voltage.

Integrate over each section to find the capacitor voltage as a function of time. The constant in the integration will be such that the initial capacitor voltage equals that at the end of the previous section. All the currents are >= 0 so you know the capacitor voltage will either increase or remain steady during that section. If the current is constant the capacitor voltage will obviously increase linearly. And if the current is ramped up or down- you can work it out from the integral.

The current source voltage is then just the capacitor voltage summed with the resistor voltage. You know the current source voltage will be greater than or equal to the capacitor voltage.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks a lot for your help, I think my fault was to write i_s with singularity functions, rather than a piecewise form \$\endgroup\$ – Emily May 7 '20 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.