ENGG1015 Tutorial
„
„
„
Circuits (IV)
(Class A) 3 Oct and (Class B) 4 Oct
Learning Objectives
‰
Analysis power of circuits through circuit laws (Ohm’s Law,
KCL and KVL)
„
News
„
HW 1 (Deadline: 14 Oct)
Class A lecture arrangement
‰ Project Group and Project Stage
Ack.: MIT OCU 16.01, Circuit analysis I: with MATLAB
‰
‰
applications (S.T. Karris)
1
Analyzing Circuits
„
All circuits can be analyzed by systematically
applying
Ohm’s Law
KVL/KCL
‰ Parallel/Series combination
‰ Voltage/Current divider
and then solve the resulting equations
‰
‰
2
Question 1: Computation of power
„
Compute the power supplied or absorbed by
each device
3
1
Solution 1
„
We assign +ve and -ve polarities and current
directions
„
We observe that iA=iB and iE=iD.
Also, by KCL at Node X, ic = iB+iD = 24+6 = 30A
„
4
Solution 1
„
„
„
PA= vAiA = 12 x 24 = 288 w (absorbed)
PE= vEiE = 36 x 6 = 216 w (absorbed)
PC= vC(-iC) = 60 x (-30) = -1800 w (supplied)
5
Solution 1
„
By KVL, vA+vB=vC
vB = vC-VA = 60-12 = 48V
PB = vBiB = 48 x 24 = 1152 w (absorbed)
„
By KVL, vD+vE=vC
vD = vC-VE = 60-36 = 24V
PD =vDiD = 24 x 6 = 144 w (absorbed)
Power supplied = Power absorbed
„
6
2
Question 2: Circuit Analysis I
„
„
„
„
Assume all resistors have the same resistance, R.
Determine the voltage vAB.
Denote the power dissipated at each resistor i as Pi,
determine the values of P1, P2, P3, P4 in terms of R.
The power generated
from the 5V voltage
source is 5 × i1. Is this
value equal to the sum
of power dissipated at
each resistor?
7
Solution 2a
„
„
Determine VAB
We assign VG=0
VA = 5 ⋅
R2
= 2.5V
R1 + R2
VB = −3 ⋅
R4
= −1.5V
R3 + R4
VAB = VA − VB = 2.5 − ( −1.5 ) = 4V
8
Solution 2b
„
„
„
Positive power as power dissipation
Negative power as power production
No current flows
through node G
9
3
Solution 2b
„
Power produced by the 5V power supply:
5
25
P5V = −5 ⋅ iA = −5 ⋅ R = − R
2
2
„
Power produced by the 3V power supply:
3
9
P3V = −3 ⋅ iB = −3 ⋅ R = − R
2
2
10
Solution 2b
„
Power dissipated at R1 and R2:
5 5 25
⋅ =
2R 2 4R
5 5 25
P2 = iAV2 =
⋅ =
2R 2 4R
P1 = iAV1 =
„
Power dissipated at R3 and R4:
⎛ 3 ⎞ ⎛ 3⎞ 9
P3 = −iBV3 = − ⎜
⎟⋅⎜ − ⎟ =
⎝ 2R ⎠ ⎝ 2 ⎠ 4R
⎛ 3 ⎞ ⎛ 3⎞ 9
P4 = −iBV4 = − ⎜
⎟⋅⎜ − ⎟ =
⎝ 2R ⎠ ⎝ 2 ⎠ 4R
Caution: Directions of
voltage, current and
power
11
Solution 2c
„
Loop 1: P5V + P1 + P2 = −
25 25 25
+
+
=0
2R 4R 4R
„
Loop 2: P3V + P1 + P2 = −
9
9
9
+
+
=0
2R 4R 4R
„
All the power produced by
the 2 power supplies are
dissipated by the 4 resistors
12
4
Question 3: Circuit Analysis II
„
„
Find the branch voltages and branch currents using
the node method.
Verify that the net power dissipation of the circuit is
zero, i.e., that
7
∑i v
n =1
n n
=0
R1 = 3Ω, R2 = 2Ω, R3 = 3Ω,
R4 = 3Ω, R5 = 1Ω,
I 8 = 5 A, V7 = 3V
13
Solution 3
„
„
Step 1a: Define the node voltage
Step 1b: Define the current direction
14
Solution 3
„
Step 2: Write down KCL equations
⎧ ⎛ 1
1 ⎞
1
1
⎪ ⎜ + ⎟ e1 − e2 = V7 + I 6
R2
R1
⎪ ⎝ R1 R2 ⎠
⎪
⎛ 1
1
1 ⎞
1
⎪ 1
⎨− e1 + ⎜ + + ⎟ e2 − e3 = 0
R4
⎝ R2 R3 R4 ⎠
⎪ R2
⎪
⎛ 1
1
1 ⎞
⎪
− e2 + ⎜ + ⎟ e3 = − I 6
⎪⎩
R4
⎝ R4 R5 ⎠
15
5
Solution 3
„
„
„
„
„
Therefore, e1 = 9V, e2 = 3V,e3 = -3V
v1 = -6V, v2= 6V, v3 = 3V, v4 = 6V,
v5 = -3V, v6 = -12V, v7 = 3v
i1 = -2A, i2= 3A, i3 = 1A,
i4 = 2A, i5 = -3A, i6 = 5A
v7 = V7 for all i7
Applying KCL at V7 node,
i7+i1 = 0, therefore i7=2A
16
Solution 3
„
The net power dissipated by the circuit
Power = ∑ in vn
n
= ( −2 )( −6 ) + ( 3)( 6 ) + (1)( 3) + ( 2 )( 6 )
+ ( −3)( −3) + ( 5 )( −12 ) + ( 2 )( 3) = 0
„
Note that the current source
supplies power (-60w), and
the voltage source absorbs
power (+60w)
17
Question 4: Choosing an appropriate
Force Sensitive Resistor
„
A FSR has been wired up using a potential divider
‰
„
„
„
Force sensitive resistor (FSR) is a resistor with its resistance
changed according to the force applied to it
Maximum power rating of the FSR is 1 mW
Vcc = 12V
Suggest Rref that will make the circuit works
across all values of input force
‰
‰
vout < 0.5V when no force is applied
vout > 6V when force is applied
(force is ≥ 0.5N).
18
6
Solution 4
„
Power dissipated through FSR:
P = iv =
„
R fsr
R fsr
Vcc
⋅ Vcc
= Vcc2
R fsr + Rref
R fsr + Rref
( R fsr + Rref
)
2
By considering the derivative of P with respect to
Rfsr, maximum power dissipated at FSR when
Rfsr = Rref. Therefore, maximum power dissipated at
the FSR will be
Pmax = Vcc2
(R
ref
Rref
+ Rref
)
2
=
Vcc2 1
⋅
4 Rref
19
Solution 4
Vcc2 1
⋅
≤ 10−3 ⇒ Rref ≥ 36k Ω
4 Rref
„
Since Pmax ≤ 10−3 ⇒
„
Using Rref = 36k Ω , when force = 0N, since R fsr = 1M Ω
vout = Vcc
„
Rref
Rref + 1×106
= 0.42V < 0.5V
Furthermore, when Force ≤ 0.5 N and R fsr ≤ 10k Ω
vout ≥ Vcc
Rref
Rref + 10 × 103
= 9.39V
20
Solution 4
„
Meanwhile, the shut off voltage is < 0.5
vout = Vcc
„
Rref
Rref + 1× 106
< 0.5 ⇒ 2 (Vcc − 1) Rref < 106 ⇒ Rref < 43478Ω
Therefore, the range of acceptable Rref is
36000Ω ≤ Rref < 43478Ω
21
7