ECE 221 Review
I. Basics
i = dq/dt (A) {q = charge in C}, v = dw/dq (V) {w = work in J},
p = dw/dt = vi (W). For i flowing + to -, passive sign convention, p = power absorbed.
For i flowing - to +, active sign convention,
p = power delivered. Power absorbed = -power delivered.
II. Resistive Circuits
Kirchoff’s Laws:
KCL ientering = ileaving (or ileaving = 0, or ientering = 0)
KVL vloop = 0 {+ for voltage drop, - for voltage rise}
Ohm’s Law for resistors: v = iR (passive signs)
Equivalent Resistance:
Series (same i through) Rs = R1 + R2 + …
Voltage division vj = “Vsource”Rj/Rs
Parallel (same v across) Rp = [(1/R1) + (1/R2) + … ]-1
if only 2, Rp = R1R2/(R1 + R2)
Current division ij = “Isource”Rp/Rj
if only 2, i1 = “Isource”R2/(R1 + R2)
Dependent Sources: write KCL, KVL equations, then substitute for dependent source.
III. Analysis Techniques
Node Voltage Analysis:
(i) Identify essential nodes and choose one as reference.
(ii) Use node voltages as set of independent variables.
(iii) Write node (KCL) equations at each essential, non-ref., non-Vs node:
[ (conductances connected to node x)]Vx –  [(conductances between x and y)Vy] =  (Isources into x)
(iv) For voltage source between node and ground, use Vnode = +Vsource.
(v) For voltage source between nodes, use “supernode” equation.
(vi) # node equations = # non-reference nodes - # Vsources.
Mesh Current Analysis:
(i) Draw mesh currents in each mesh (same direction).
(ii) Use mesh currents as set of independent variables.
(iii) Write mesh (KVL) equations around each mesh without current sources:
[ (resistances in mesh x)]ix –  [(resistances between x and y)iy] =  (Vsources in x)
(Vsource positive if i leaves + terminal)
(iv) For current source in one mesh only, use imesh = +Isource.
(v) For current source between meshes, write KVL around loop without current sources (supermesh).
(vi) # mesh equations = # meshes - # Isources.
IV. Circuit Theorems
Superposition: the response to multiple sources =  (responses to each source alone). Find response to
each source by removing others (open Is, short Vs), then add.
Source Transformations: Vs + R in series ↔ Is||R using Vs = IsR
Thevenin/Norton Equivalent Circuits
VT = VOC, IN = ISC, RT = RN = VOC/ISC or = VTest/ITest or = Req with sources removed (for independent
RT
sources only)
a
Maximum power transferred when RLOAD = RT
VT
a
IN
RN
b
Thevenin
Norton
V. Op Amps
Ideal approximation only (assumes A = ∞, Rin = ∞, Rout = 0)
i+ = i- = 0
v+ = vStandard circuits (inverting amp, non-inverting amp, follower, summing amp, difference amp)
– use known results.
Unknown circuits – write KCL on input side of op amp and use ideal approximation.
VI. Inductors and Capacitors – summary:
C
L
Symbol, Units, Polarity
dvc
dt
di
vL  L L
dt
I-V Equation
iC  C
DC Behavior
Open Circuit
Continuous Variable
vC
iL
Stored Energy
1 Cv 2
C
2
1 Li 2
2 L
Series/Parallel
Opposite Resistors
Like Resistors
Short Circuit
b
Review Problems
1. The power absorbed in a 2 k resistor is 50 mW. Find the voltage across the resistor. [10 V]
2. State if the following KVL equations are T or F for the circuit below. If F, explain why.
(i) –10 + 6i1 + 12 – 10i1 = 0
6
4
(ii) 12 + 6i2 + 3 = 0
i1
i2
10 V
12 V
10
V3
[both F]
2
3V
3. q(t) = 4e-2t C and v = -20 V. Find i(t), p(t) and the work done between 0 and 1 s.
[-8e-2t A, 160e-2t W, 69.17 J]
4. Find v1 [10 V]
5. Find i1 [8 A]
6. State the number of simultaneous equations needed to solve this circuit by (i) node analysis and
(ii) mesh analysis.
[2, 3]
7. a) vx = ?
b) I2 = ?
[i, ii]
i) v1 – v2
ii) v2 - v1
iii) v1 + v2
iv) v2/3
i) i2 + i3
ii) i2 - i3
iii) i3 - i2
iv) i2
8. a) Find Req, I1, I2, I3
[10 , 5A, 2.5 A, 1 A]
9. Find I using node analysis.
[7 A]
10. Repeat problem 9 using mesh analysis.
11. Repeat problem 9 using superposition.
12. Repeat problem 9 using source transformations.
13. Find the Thevenin and Norton equivalent circuits.
a
2
[32 V, 16/3 A, 6 ]
48 V
6
4A
12
b
14. Find Vout. [-6 V]
+
+
5 V
-
4
20
OUT
6
-
10
-
OUT
+
20
40
+
6
3
Vout
-
15. iC = 0.2sin(½t) mA and C = 5 F. If vC(0) = 0, find vC(t) and wC(t) for t > 0.
[80(1 – cos(½t)) V, 16(1 – cos(½t))2 mJ]