2
\$\begingroup\$

While solving the simple RC circuit below in a particular way, I came across an interesting mathematical quirk.

schematic

simulate this circuit – Schematic created using CircuitLab

Using nodal analysis, we can readily get the expected solution \$v = V_0e^{\frac{-2t}{RC}}\$, where \$V_0\$ is the initial condition.

Since the resistors are in parallel, they can be replaced by a single resistor valued \$R/2\$ and, by KVL, we can arrive at the same result uneventfully. However, it gets at little more interesting if we keep both loops and try to use mesh analysis to solve for one of the currents.

The right loop equation is $$ \frac{1}{C}\int{i_1+i_2 \mathrm{d}t} + Ri_2 = 0\\ RC\frac{\mathrm{d}i_2}{\mathrm{d}t} + i_2 = -i_1 $$ and the left loop equation is, similarly $$ RC\frac{\mathrm{d}i_1}{\mathrm{d}t} + i_1 = -i_2. $$

Solving this system for \$i_2\$, for instance, gives the second-order ODE $$ \frac{\mathrm{d}^2i_2}{\mathrm{d}t^2} + \frac{2}{RC}\frac{\mathrm{d}i_2}{\mathrm{d}t} = 0, $$ which has the solution $$ i_2 = A + Be^{\frac{-2t}{RC}}. $$ Appropriate initial conditions \$i_2(0)\$ and \$\mathrm{d}i_2/\mathrm{d}t(0)\$ can be found, giving \$A=0\$ and \$B=V_0/R\$ as expected. Alternatively, since \$i_2(\infty)\$ has to be 0, we can deduce that \$A=0\$ and only one initial condition is needed to find \$B\$.

While in this case this "phantom" order can be easily avoided, it can be a pitfall for young players and may not be so obvious in more complex circuits. It can also make solving a simpler circuit more difficult. That being said, is there any strategy or best practice that can be used to avoid this apparent extra order creeping up in your analysis?

\$\endgroup\$
5
  • \$\begingroup\$ Yes, the mathematical blunder that turned a first-order ODE into a second-order one was failing to realise that i1 and i2 are proportional (identical, in this particular case) and not independent variables. My question regards the best way to approach a problem so as to avoid walking into those kinds of traps \$\endgroup\$ Jan 26, 2022 at 7:48
  • \$\begingroup\$ Why differentiate an equation for solving? One can solve by adding the two first-order equations and have the solution? \$\endgroup\$
    – Antonio51
    Jan 26, 2022 at 13:34
  • \$\begingroup\$ I don't see how adding them would eliminate i1 or i2 \$\endgroup\$ Jan 26, 2022 at 19:39
  • 1
    \$\begingroup\$ Don't eliminate i1 or i2, but (i1+i2) appear, with d(i1+i2)/dt. So equation is first order (?). \$\endgroup\$
    – Antonio51
    Jan 26, 2022 at 20:49
  • \$\begingroup\$ Yep, you're in the right here. \$\endgroup\$ Jan 26, 2022 at 21:04

1 Answer 1

2
\$\begingroup\$

Using :

$$ \ RC\frac{\mathrm{d}i_2}{\mathrm{d}t} + i_2 = -i_1 $$ $$ RC\frac{\mathrm{d}i_1}{\mathrm{d}t} + i_1 = -i_2. $$

Add equations

$$ RC\frac{\mathrm{d}(i_1+i_2)}{\mathrm{d}t} +(i_1+i_2) = -(i_1+i_2). $$

Or

$$ RC\frac{\mathrm{d}(i_1+i_2)}{\mathrm{d}t} +2*(i_1+i_2) = 0. $$

One can solve vs (i1+i2) ... and then obtain separately i_1 and 1_2.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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