E UCLA LECTRICAL ENGINEERING DEPARTMENT: EE 10: CIRCUIT ANALYSIS 1 Midterm Review
EE10 Midterm Review Example Problems: Supplement
P RO B L E M 1
Write down all the equations that are required to solve this circuit problem. All equations should be in terms of Mesh Current variables. Mesh A) Mesh B) 150 (iA iD)35 vs = 0 vs - (iB iD)10 - (iB iC)40 = 0
With iA iB = 3iy Or a Super Mesh equation: Super Mesh A-B) 150 (iA iD)35 - (iB iD)10 - (iB iC)40 = 0 With iA iB = 3iy Mesh C) iC = - 5i Mesh D) -(iD iA)35 +15ix (iD iB)10 = 0 And constraint equations iy = iB iC ix = - iA
EE 10 M IDTE RM S UPP LEME N T PAGE 1 OF 7
UCLA E LECTRICAL ENGINEERING DEPARTMENT: EE 10: CIRCUIT ANALYSIS 1 Midterm Review
P RO B L E M 2
+
30!
v1 -
182!
30!
i2
10!
60!
27!
44!
5A
vS +
Use a Delta-to-T circuit transformation to solve for the indicated circuit variables We will start by replacing the 30, 60, and 10 ohm resistances with a T construct
a b R1 R2
a
Rc
b
Rb
Ra c
=
R3 c
We identify elements and obtain Ra = 10, Rb = 30, and Rc = 60, Then, the corresponding T structure resistances are computed with this transformation
EE 10 M IDTE RM S UPP LEME N T PAGE 2 OF 7
UCLA E LECTRICAL ENGINEERING DEPARTMENT: EE 10: CIRCUIT ANALYSIS 1 Midterm Review
Expression for the Y structure resistance value in terms of the Delta structure resistance values
Rb Rc Ra + Rb + Rc Ra Rc R2 = Ra + Rb + Rc Ra Rb R3 = Ra + Rb + Rc R1 =
And, R1 = 18, R2 = 6, and R3 = 3,
Now, we can first find vS We combine the resistors appearing in series and parallel into a single resistance producing a 100 value o (The 18 and 182 series in parallel with the 6 and 44 series pair produces a 40 resistance. This lies in series with the 27, 3, and 30 resistances
EE 10 M IDTE RM S UPP LEME N T PAGE 3 OF 7
UCLA E LECTRICAL ENGINEERING DEPARTMENT: EE 10: CIRCUIT ANALYSIS 1 Midterm Review
Then vS = -500V (note polarity !) Now, we can use the information regarding the current divider and the 5A current flow to determine the current flowing through the 182 and 44 resistors. The current flow through the 182 resistor will be 5A[50/(200+50)] = 1A while the remaining 4A flows through the 44 resistor. Then, the voltage drop across the 60 resistor that would induce a current of polarity as indicated by i2 must be determined by the difference in voltage drops across the 44 and 182 resistors. And v44 v182 = 44(4A) 182(1A) = -6V and i2 = -6V/60 = -0.1A Finally, to compute v1 we can compute the voltage drops across the 30, 182 and 27 resistors combine and this with the net 500V drop. o v1 = 500V 5A(30) 5A (27) 1A(182) = 33V
EE 10 M IDTE RM S UPP LEME N T PAGE 4 OF 7
UCLA E LECTRICAL ENGINEERING DEPARTMENT: EE 10: CIRCUIT ANALYSIS 1 Midterm Review
P RO B L E M 3
NOTE: Operational Amplifiers will not be included on the Midterm. However, we will use the following problem to illustrate source equivalent methods (and reinforce operational amplifier principles.
The non-ideal amplifier in this circuit has an open loop voltage gain of AV , infinite input resistance, but finite output resistance, RO. Find an expression for the Thevinin resistance between the reference and the output terminal, above. First, we introduce the operational amplifier model.
v+ v-
RO + - A (v - v ) V + RF1
VO
+ -
Vi RF2
EE 10 M IDTE RM S UPP LEME N T PAGE 5 OF 7
UCLA E LECTRICAL ENGINEERING DEPARTMENT: EE 10: CIRCUIT ANALYSIS 1 Midterm Review
Then, we recognize that since a dependent source appears, we must use a Test Voltage method to compute RTH. We proceed to replace independent sources by their zero equivalents. This provides the new circuit,
Now, we can write a node voltage equation at the output node:
IT VT/(RF1 + RF2) + [AV(v+ - v-) - VT ]/RO = 0 Now, we see that v+ = 0 Also, since a voltage divider exists at the inverting input,
v- = VTRF2/(RF1 + RF2) Then, the node voltage equation becomes:
IT VT/(RF1 + RF2) + [-AVVTRF2/(RF1 + RF2) - VT ]/RO = 0 Or,
EE 10 M IDTE RM S UPP LEME N T PAGE 6 OF 7
UCLA E LECTRICAL ENGINEERING DEPARTMENT: EE 10: CIRCUIT ANALYSIS 1 Midterm Review
IT = (VT/RO ) [1 + RO/(RF1 + RF2) + AVRF2/(RF1 + RF2)] Finally, the Thevenin Resistance expression produces the following interesting result:
RTH VT/IT = RO / [1 + ( RO + AV RF2) /(RF1 + RF2)] Now, consider the limit of AV large, then
RTH = RO [(RF1 + RF2)/AVRF2] This demonstrates that the output resistance of this amplifier is reduced relative to the open loop output resistance, RO , by a factor depending on both open loop gain and the gain of the non-inverting voltage amplifier. Recall that we had found that the gain for this amplifier (for an ideal operational amplifier device) was (RF1 + RF2)/RF2 Note that in the limit where RF2 is large compared to RF1 then the amplifier gain falls to unity and the output resistance reaches RO/AV Note also that in the limit of AV = 0, then
RTH = RO / [1 + ( RO /(RF1 + RF2)] = 1 / [1/RO + 1/(RF1 + RF2)] This is just the resistance of RO in parallel with the resistance (RF1 + RF2)
EE 10 M IDTE RM S UPP LEME N T PAGE 7 OF 7