University of California, Berkeley Spring 2012EE 42/100 Prof. A. Niknejad

Problem Set 2Solutions

Please note that these are merely suggested solutions. Many of these problems can beapproached in different ways.

1. In problems like this, you may find it helpful to redraw the circuit to ignore all theirrelevant terminals and such that everything is rectangular. Note that since we arefinding Req from terminals a and b in each of these, we must simplify the circuitstoward these terminals.

(a) Terminals c and d are connected together, while e and f are irrelevant. If youredraw the circuit as below, you can see that the two resistors in the firstcolumn are in parallel as, they are connected together at both pairs of terminals(due to the short). This is also true for the second column of resistors. So eachof the pairs can be redrawn as one resistor with 1

2R.

a b

c

d

e f a b

c

dR

R

R

R

R

R

R

R

R

R

RR a

R R/2

RR/2 b

It is easy to see that all the resistors are in series, so Req = 3R.

(b) Terminals e and f are shorted. Intuitively, Req should only include the two innerresistors, since the other four are shorted out. That is, any current flowing froma to b would only go through 2R of resistance and proceed between e and f viathe short.

a b

c

d

e f a bR

R

R

R

RRe

fa b

R

R

2R

2R

a bR

R

R R

RR

R R

(c) The short between c and e gets rid of the one resistor between the two nodes.After that, everything else simplifies as before.

a b

c

d

e f

R R

RR

R R

a bR

R

RRRe

fa b

R

R

2R

Rc

d

a b8R/3

2. The key to this problem is that the ladder is infinite. Using this fact, we can concludethat the equivalent resistance Rin should not change if we remove one “rung” of theladder. So we can replace the ladder, save for the first rung, with one resistor Rin.

R

RR in

R

R in

Now we can write an equation in terms of R and Rin and solve.

Rin = R + (R||Rin) +R = 2R +RRin

R +Rin

R2in − 2RRin − 2R2 = 0

Since this is a quadratic equation, there are two solutions. We know that resistancecannot be negative, so we discard the negative solution. The positive solution is

Rin = (1 +√

3)R

3. We have two unknown nodes, so we’ll have two nodal equations, one at node 1 andthe other at node 2.

V1 − V2

200+V1

30− 8 = 0

V2 − V1

200+V2

20− 3 = 0

Solving this system gives us V1 = 218.4 V and V2 = 74.4 V.

We can also do this using superposition. Here we split the problem into twosubcircuits, each one with a different current source zeroed out. Recall that thezeroed out current source becomes an open circuit.

30Ω

200Ω

20Ω8A

1 2

2

V1 − V2

200+V1

30− 8 = 0

V2 − V1

200+V2

20= 0

This first circuit gives us V1 = 211.2 V and V2 = 19.2 V.

30Ω

200Ω

20Ω -3A

1 2

V1 − V2

200+V1

30= 0

V2 − V1

200+V2

20− 3 = 0

The solutions are V1 = 7.2 V and V2 = 55.2 V. Clearly, the sum of these with thesolutions for the first subcircuit yields the same answers as the original circuit.

4. There are five nodes in this circuit, labeled below. Placing the reference node at vEreduces the number of unknowns to three, since now vE = 0 V and vD = 5 V.

50Ω 30Ω1.2A

18Ω 75Ω

50Ω

+

5V

A B C

D E

The three nodal equations are the following:

VA − VC50

+VA − VB

18+VA − 5

50= 0→ 43VA − 25VB − 9VC = 45

VB − VA18

+VB − VC

75− 1.2 = 0→ −25VA + 31VB − 6VC = 540

VC − VA50

+VC − VB

75+VC30

= 0→ −3VA − 2VB + 10VC = 0

The solutions are VA = 32.8 V, VB = 47.6 V, and VC = 19.3 V. The branch currentscan be found using Ohm’s law and KCL: IAC = 0.27 A, IAB = −0.82 A, IBC = 0.38 A,IAD = 0.56 A, ICE = 0.64 A, IDE = 0.56 A.

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5. The circuit has four unknown nodes, labeled below.

+

+

100Ω 500Ω

505Ω

90Ω

+

1V

−

10 Vx+5V− Vx

80Ω

VA VB VC

Vx

In addition, we must also write a supernode equation at the 1 V source. Thedependent current source does not present a problem, since we treat Vx as one of theunknowns. The nodal equations at VA and Vx are

VA − 5

100+VA80

+VA − VB

500− 10Vx = 0

Vx − VC505

+Vx90

= 0

The two equations at the supernode are

VB − VA500

+VC − Vx

505= 0

VB − VC = 1

The solution to these equations is VA = 0.966 V, VB = 0.981 V, VC = −0.019 V,Vx = −0.003 V.

6. Note that the left part of the circuit is independent of the right part. Since there is nocomplete loop connecting the two parts, no current can flow between them; the wireconnecting them only serves as a common ground. The equivalent resistance on theleft is 3 + 12||12 = 9 Ω. So the total current from Vs is just Vs

9. Also, we have that Ix

is just half of that, since the two parallel branches have equal resistance. So Ix = Vs18

.

On the right half, the two resistors are in parallel, so the equivalent resistance is just2 Ω. Since the current going through them is given by AIx = AVs

18, the voltage across

them is Vout = IReq = AVs9

. Solving for A and plugging in VoutVin

= 9, we have thatA = 81.

7. The steps to finding the equivalent circuits are to find Voc, Isc, and/or Req. Sincethere is a dependent source in this circuit, it is easier to find the first two. Voc issimply the voltage difference between a and b. By inspection, this is just

Voc = −(100 kΩ)βIx = −(100 kΩ)βVs

10 kΩ= −10βVs

Isc is found by shorting the terminals and solving for Iab. Note that shorting out the100 kΩ resistor will leave us with Iab = −βIx = −β Vs

10 kΩ(note the sign!).

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Thus, our desired values are VTh = −10βVs and IN = − βVs10 kΩ

. Also, the Thevenin and

Norton resistance is just given by RTh = RN = VTh

IN= 100 kΩ.

8. (a) The maximum power transfer theorem tells us that the load resistance should beequal to the Thevenin resistance of the circuit. In this case, it is trivial to seethat that RL = RTh = Rs. If you want to review the derivation of this result,please consult the lecture notes or textbook.

(b) We want the power across the load, PL, to be equal to 80% of the powerdelivered by the voltage source, Ps. The former can be found using PL = I2RL,whereas the latter is Ps = IVs. In this circuit,

I =Vs

Rs +RL

So putting everything together, we have

V 2s RL

(Rs +RL)2= 0.8

V 2s

Rs +RL

Solving for RL gives us RL = 4Rs.

(c) Power is dependent on both voltage and current. While the answer to part (b)gives us a higher voltage across RL and a greater fraction of the total power, thetotal power is actually much smaller in this case. This is because the current is 5times smaller due to the greater resistance, while the voltage across RL is only1.6 times larger.

9. Intuitively we know that closing the switch puts R2 and R3 in parallel, lowering theeffective resistance. So it must be in the open case that vo takes the greater voltagevalue, 26 V. In this scenario, the circuit looks like the following:

++39−

V

50 Ω

+vo−

R = 26V2

Solving for this circuit, using voltage divider, nodal analysis, or otherwise, gives usR2 = 100 Ω.

Now if we close the switch, vo = 24 V. The value of R2 remains unchanged, so allthat’s remaining is to solve for R3.

++39−

V

50 Ω

+vo−

R100 Ω 3

5

We can write a nodal equation at vo = 24 V:

vo − 39

50+

vo100

+voR3

= 0

The solution is R3 = 400 Ω.

10. (a) When the bridge is balanced, no current flows across the galvanometer. By KCL,the current across R1 and R2 are the same, as well as the current across R3 andR4. In addition, the voltages va and vb must be equal due to the 0 current.

R 1

R 2

R 3

R x

GVs

+I1 I2a b

By KVL, the voltage drops across R1 and R3 must then be the same. The samereasoning goes for R2 and Rx.

I1R1 = I2R3

I1R2 = I2Rx

Substituting for the currents and solving for Rx, we have Rx = R2R3

R1.

(b) Given our values for R1 and R3, we can measure Rx = 2R2. Since the highestvalue that R2 can take is 1000 Ω, the maximum resistance that can be measuredfor Rx is just 2000 Ω. As R2 comes in increments of 10 Ω, the measurement canbe made with an accuracy to within 20 Ω.

(c) The Thevenin equivalent is found by finding the voltage and equivalentresistance across the open terminals a and b. Note that here we cannot assumethe bridge is balanced, as we are looking at the general case. voc is simply thedifference in potential between nodes a and b, both of which can be found via avoltage divider:

va =R2

R1 +R2

Vs

vb =Rx

R3 +Rx

Vs

VTh = va − vb =

(R2

R1 +R2

− Rx

R3 +Rx

)Vs

The equivalent resistance can be found by zeroing out the voltage source,making it a short, and finding Req.

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R 1

R 2

R 3

R x

a b

We have R1 and R2 in parallel with each other, and R3 and Rx in parallel witheach other. These two pairs are in series, so the equivalent resistance isRTh = (R1||R2) + (R3||Rx).

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