RC / RL circuit “cheat sheet”
Resistor Properties:
Cylindrical resistor: R = ρ L/A ( ρ = resistivity, L = length, A = cross-sectional area)
Basic resistor equation (Ohm’s Law): V = IR
Series resistors:
Parallel resistors:
R eq
1/R
= R eq
1
+ R
= 1/R
2
1
+ …
+ 1/R
2
+ … --or-- G eq
= G
1
+ G
2
+ … where G = conductance = 1/R
Capacitor Properties:
Parallel-plate cap: C = κε
0
A/d ( ε
0
= 8.85 x 10 -12 F/m, κ = dielectric constant, A,d = plate area, separation)
Basic capacitor equations: Q = CV ; I = C dV / dt series capacitors: 1/C eq
= 1/C
1
+ 1/C
2
+ … parallel capacitors: C eq
= C
1
+ C
2
+ …
Inductor Properties:
Solenoid inductor: L = ( µ
0
N 2 A/ l ) ( µ
0
= 4 π x 10 -7 H/m, N = # of turns, A = cross-sectional area, l = length)
Basic inductor equation: V = L dI/dt (where the “Lenz” minus sign has been omitted)
Series inductors: L eq
= L
1
+ L
2
+ …
Parallel inductors: 1/L eq
= 1/L
1
+ 1/L
2
+ …
Simple RC Circuit
I(t) = ( ε /R) e -t/ τ
τ = RC
(t=0: I = I
0
= ε /R; t → ∞ : I = 0)
V
R
(t) = R I(t) = ε e -t/ τ
V
C
(t) = ε -V
R
(t) = ε (1-e
Q(t) = CV
C
(t) = C ε (1-e
-t/ τ )
-t/ τ )
Uncharged capacitor has Q = 0 → V
C
= 0, so it acts like a short:
Can solve any problem with uncharged capacitors (i.e. t=0 for a “charging” circuit, t → ∞ for a
“discharging” circuit) by replacing them with shorts and applying KVL!
A fully charged capacitor (in a charging circuit, t → ∞ ) will have a voltage which sets I = 0. Thus you can use KVL to determine all the voltages by setting all currents through capacitors to zero.
To solve for behavior at intermediate times, try to use series/parallel decomposition rules to reduce your circuit to the “simple RC circuit” above. If a circuit cannot be decomposed, use Kirchhoff methods to set up simultaneous differential equations, and look for an “easy” solution. Or use one of your linear network methods (e.g. Thevenin) to replace the rest of the circuit (i.e. the part that has no capacitors) with something simpler! Finally, you can check that your solution matches the above “end behaviors!”

Simple RL Circuit τ = L/R
I(t) = (
V
R
ε /R) (1 - e -t/ τ ) (t=0: I=0; t → ∞ : I = ε /R)
(t) = R I(t) = ε (1 - e -t/ τ )
V
L
(t) = ε - V dI/dt = V
L
R
(t) = ε e
/L = ( ε /L) e
-t/ τ
-t/ τ (t=0: dI/dt = ε /L; t → ∞ : dI/dt = 0)
Inductor attempts to maintain constant current, so:
Can solve any problem in which a circuit suddenly changes (i.e. t=0 when the switch is closed) by ensuring that the current through inductors does not suddenly change! Usually this means I(t=0) = 0 for when a circuit is first connected, in which case inductors can be replaced by open circuits, and the voltages across them found using KVL.
Further, whenever an inductor is inducing a voltage, the rate of change of current through it can be found as dI/dt = V
L
/L.
Finally, if you attempt to force a sudden change of current through an inductor (i.e. by abruptly opening the switch in the above circuit, then dI/dt becomes huge, which induces a huge voltage: the inductor will “fire!”
In the steady state configuration (t → ∞ ) ideal inductors will have zero voltage drop since dI/dt=0:
V
L
(t → ∞ ) = 0. Thus you can use KVL to determine all the voltages by setting the voltage drops across all inductors to zero.
To solve for behavior at intermediate times, try to use series/parallel decomposition rules to reduce your circuit to the “simple RL circuit” above. If a circuit cannot be decomposed, you can use your
Kirchhoff methods to set up simultaneous differential equations, and look for an “easy” solution. Or you can use one of your linear network methods (e.g. Thevenin) to replace the rest of the circuit (i.e. the part that has no inductors) with something simpler! Finally, you can check that your solution matches the above “end behaviors!”