Duality; First-Order Circuits
ES – I Lecture – 10 (Ch.3)
Date 30/08/08
• A circuit has dual if, and only if, it is a planar network.
Given Circuit Dual Circuit
1. Series connection 1. Parallel connection
2. Parallel connection 2. Series connection
3. Voltage of x volts 3. Current of x amperes
4. Current of x amperes 4. Voltage of x volts
5. Resistance of x Ω 5. Conductance of x ʊ
6. Conductance of x ʊ 6. Resistance of x Ω
7. Capacitance of x 7. Inductance of x
farads
8. Inductance of x henries
8. Capacitance of x
henries farads
Rules for finding dual circuit :-
Rule 1:- Inside of each mesh, including the infinite region
surrounding the circuit, place a node.
Rule 2:- Suppose two of these nodes, say nodes a and b,
are in adjacent meshes. Then there is at least one element in
the boundary common to these two meshes. Place the dual
of each common element between nodes a and b.
Example 1:- Find the dual of the circuit shown below?
1H 2Ω
3Ω 4F 5Ω
6cos(3t) A ↑
The dual circuit can be redrawn as shown below :
c 4H
b 3ʊ d
+ 1F 2ʊ 5ʊ
6cos(3t) V
−
a
Example 2:- Find the dual of the circuit shown below?
3F
1H
2Ω 5Ω
+ 4F 6H
7sin(4t) V
−
The dual circuit can be redrawn as shown below :
2ʊ
b 1F
c
4H
3H
d
5ʊ
7sin(4t) V ↑ 6F
a
Example 3:- Find the dual of the circuit shown below?
1F 3H
4Ω
2Ω 6Ω
+ 8Ω
9cost V 5H 7F
−
The dual circuit can be redrawn as shown below :
5F
a e b
4ʊ 6ʊ
2ʊ
7H
d 3F
c
9cost A ↑ 1H 8ʊ
f
First-Order Circuits :-
There are two types of responses.
1. Natural Response (or Transient Response) :- It depends
on nature of elements and their interconnection.
2. Forced Response :- It depends up on excitation of the
circuit and is expressed as complementary function
Rg S t0 s ?
To find answer to this question ⎯
By K.C.L.,
iC + iR = 0
In terms of voltage, we write
C.(dv/dt) + (v/R) = 0
In this equation, variable is v(t) and therefore dv(t)/dt
i.e., first derivative.
• For this reason this equation is first-order differential
equation.
• Further, the describing equation is homogeneous, linear
differential equation, because every non-zero term is of
first degree in the dependent variable v and its derivative.
• Note that coefficients of v and its derivative are
constants.
To find v(t), that satisfies the describing equation and the
initial condition, let us rearrange describing equation as
(dv/dt) = −(1/RC).v
Separating variables i.e., dividing both sides with by “v”,
(1/v).(dv/dt) = −(1/RC)
Integrating both sides with respect to time,
∫ (1/v).(dv/dt).dt = ∫ −(1/RC).dt
By Chain rule of calculus, we change integrating variable
on Left Hand Side as
∫ (1/v).dv.dt/dv = ∫ (1/v).dv = ∫ −(1/RC).dt
∴ ln v(t) = −(1/RC).t + K
where K is a constant of integration.
Taking exponent on both sides,
eln v(t) = e (−t/RC + K)
∴ v(t) = v(0).e(−t/RC).eK
Use initial condition to determine K.
At t=0 s, v(0) = e0.eK = eK
Substituting eK = v(0) in expression for v(t), we get
v(t) = v(0).e(−t/RC) for t≥0 s
As shown earlier,
v(0) = [R/(R+Rg)] . V volts
Sketch of v(t) for t≥0 s is shown below.
v(t) = v(0).e-t/RC
v(0)
v(0).e-1=0.368.v(0)
v(0).e-2=0.135.v(0)
v(0).e-3=0.050.v(0)
t
0 1RC 2RC 3RC 4RC
We see that initially (at t=0 s) the capacitor is charged to
v(0) volts, and for t>0 s, the capacitor discharges through
the resistor exponentially.
• Resistor current for t≥0 s is
iR(t) = v(t)/R = v(0).e(−t/RC) / R
• Current through capacitor for t≥0 s is
iC(t) = −iR(t) = −(v(0)/R).e(−t/RC)
Or
iC(t) = C.(dv(t)/dt) = C.d(v(0).e(−t/RC))/dt
⇒ iC(t) = C.(−1/RC).v(0).e (−t/RC)
⇒ iC(t) = −(v(0)/R).e(− t/RC)
• Energy stored in capacitor is at t=0 s is
wC(t) = C.v2(0)/2 J
Since v(t) → ∞ as t → ∞
wC(t) → 0 as t → ∞
• Energy initially stored in the capacitor is eventually
dissipated as heat by resistor.
The power absorbed by the resistor is
pR(t) = R.iR2(t) = R.(v(0).e(−t/RC)/R)2
⇒ pR = (v2(0).e(−2t/RC)/R)
∴Total energy absorbed by R is
∞ ∞
wR = ∫ pR(t).dt = ∫ (v2(0).e(−2t/RC)/R).dt = C.v2(0)/2
0 0
⇒ wR = wC(0)
Now RC = Time constant of circuit = τ (seconds)
∴ v(t) = v(0).e(−t/ τ)
Natural Response :-
• For t≥0 s, the battery has no effect on the circuit.
• Therefore the circuit behaves on its own, i.e., behaves
naturally.
• The expressions for v(t) and i(t) for t≥0 s are said to
give/describe NATURAL RESPONSE of the circuit.
Example 4:- For the circuit shown the switch opens at time
t=0 s. Find v1(t), v2(t), i1(t), i2(t) and v(t) for all time?
3Ω
t=0 s
+ v1 −
→
+ (1/2) F i1 i2↓
+
+ (1/4) F v2 2Ω
10 V v
−
−
−
For t<0 s, the circuit is shown below :
3Ω
+ v1 −
→
+ (1/2) F i1 i2↓
+
+ (1/4) F v2 2Ω
10 V v
−
−
−
For dc, the capacitors are open circuited.
Thus i1(t) = i2(t) = 0A; Also v(t) = 10 V
By voltage division
v1(t) = 3.10/(3+2) = 6 V
v2(t) = 2.10/(3+2) = 4 V
∴v1(0) = 6 V, v2(0) = 4 V and
∴v(0) = v1(0) + v2(0) = 6+4 = 10 V
For t≥0 s, the circuit is shown below :
3Ω
0A 0A
+ v1 −
→ → →
+ (1/2) F i1 i2↓
+
+ (1/4) F v2 2Ω
10 V v
−
− 0A
← −
By K.C.L.,
(1/12).(dv1/dt) + (v1/3) = 0 & (1/4).(dv2/dt) + (v2/2) = 0
(dv1/dt) + 4.v1 = 0 & (dv2/dt) + 2.v2 = 0
∴ v1(t) = v1(0).e(−4t) = 6.e(−4t) V
v2(t) = v2(0).e(−2t) = 4.e(−2t) V
Thus,
i1(t) = (1/12).[d(6.e(−4t))/dt] = −2.e(−4t) A
and i2(t) = (1/4).[d(4.e(−2t))/dt] = −2.e(−2t) A
By K.V.L., v(t) = v1(t) + v2(t) = 6.e(−4t) + 4.e(−2t) V
Hence,
v1(t) = 6 V for t<0 s
v1(t) = 6.e(−4t) V for t≥0 s
v2(t) = 4 V for t<0 s
v2(t) = 4.e(−2t) V for t≥0 s
i1(t) = 0 A for t<0 s
i1(t) = −2.e(−4t) A for t≥0 s
i2(t) = 0 A for t<0 s
i2(t) = −2.e(−2t) A for t≥0 s
v(t) = 10 V for t<0 s
v(t) = 6.e(−4t) + 4.e(−2t) V for t≥0 s
Summary
• First Order Circuit R-C ckt, exhibits natural
response on removing excitation
• The exact response is determined by time
constant of the circuit
• Natural response refers to exponential decay of
energy stored in the capaitor through resistor
• The response of L-R ckt is determined in a
similar fashion.