EECE 2112 Module 55: The Natural Response of RC Circuits

This is a series of lectures from the Circuits I class taught at Vanderbilt University.

[@bertank]

Amazing teaching, thank you

[@Festus2022]

At 13:50 why do you have I-C and I-R going in opposite directions. Would seem there is one current flowing clockwise from the capacitor.

[@wtimothyholman]

Physically there is one current, but analytically it makes sense to define the currents according to the passive sign convention for both the capacitor and resistor. Note that the capacitor will have negative power (supplying energy) as it discharges through the resistor, so the physical current will flow in the opposite direction of Ic. But from an analytic point of view, you'll get the correct answer no matter what directions you choose for currents. That's a fundamental principle of nodal analysis.

[@Festus2022]

@@wtimothyholman Thanks so much for the follow-up!

[@SaltyStargazer]

@@wtimothyholman if we assume the opposite direction for I-C, the equation will turn out to be e^(t/rc) right ? how will it still correctly predict the graph if we plug in the values?

[@wtimothyholman]

@@SaltyStargazer, if you reverse the direction of I-sub-C, then you must also flip the polarity of V-sub-C to satisfy the passive sign convention for the capacitor, and in turn flip the direction of i-sub-R to satisfy the passive sign convention for the resistor. Solving this modified KCL equation will still result in an equivalent solution.
Your choice of current direction or voltage polarity is usually arbitrary in circuit analysis, but once you choose one, then the other will be constrained by the passive sign convention for passive elements.