0
\$\begingroup\$

Suppose there is a circuit with two capacitors connected in parallel with a current source \$i_L\$, each capacitor with initial voltage \$v_i\$ and capacitance \$C\$. Is it correct to say voltage with respect to time \$v(t) = 2v_i + \frac{i_L}{2C}\$?

I got this by saying that \$v(t) = V_i + v_{current}\$, and \$v_{current} = \frac{i_L(t)}{C_t}\$.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ Two ideal capacitors in ideal parallel connection will always have the same voltage across them. They cannot have differing voltages. The total ideal capacitance will be the same as the sum of the individual ideal capacitances. Is there a reason to avoid lumping them together for simplicity's sake? Perhaps I misunderstand the question? \$\endgroup\$
    – jonk
    Nov 1 '19 at 5:30
  • \$\begingroup\$ What do you mean? I think I accounted for the fact that they have the same voltage across them... that's why Ceq = 2C, which I put in the equation \$\endgroup\$ Nov 1 '19 at 5:33
  • \$\begingroup\$ I wasn't sure. My apologies. This leaves me only with the fact that I must be misunderstanding the essence of your question. Which is quite possible. So I'll leave it to my betters who can read your intent better than I'm able. \$\endgroup\$
    – jonk
    Nov 1 '19 at 5:37
  • \$\begingroup\$ The RHS of your equation for V(t) misses the independent variable t. \$\endgroup\$
    – Bart
    Nov 1 '19 at 9:18
  • \$\begingroup\$ @melanie Sanders, I think jonk means to say that the fact that two capacitors are in parallel is an irrelevant detail to your question. Just use C for total capacitance. \$\endgroup\$
    – Bart
    Nov 1 '19 at 13:16
1
\$\begingroup\$

If you have two capacitors connected in parallel, each with initial voltage Vi, then the initial voltage of the pair will also be Vi. Two elements in parallel will always have the same voltage across them.

As the current is going to charge the total capacity of 2C, the voltage with respect to time will be \$V(t) = V_i + \frac{1}{2C}\int{I(t)}dt\$, not 2Vi + ... as you've written.

\$\endgroup\$

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.