EE101: RC and RL Circuits (with DC sources)
M. B. Patil
mbpatil@ee.iitb.ac.in
www.ee.iitb.ac.in/~sequel
Department of Electrical Engineering
Indian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Capacitors
i
conductor
insulator
conductor
Q
1111111111111
0000000000000
0000000000000
1111111111111
t
0000000000000
1111111111111
0000000000000
1111111111111
0000000000000
1111111111111
i
v
C=
ǫA
t
Q
Unit: Farad (F)
M. B. Patil, IIT Bombay
Capacitors
i
conductor
insulator
conductor
Q
1111111111111
0000000000000
0000000000000
1111111111111
t
0000000000000
1111111111111
0000000000000
1111111111111
0000000000000
1111111111111
i
v
C=
ǫA
t
Q
Unit: Farad (F)
* In practice, capacitors are available in a wide range of shapes and values, and
they differ significantly in the way they are fabricated.
(http://en.wikipedia.org/wiki/Capacitor)
M. B. Patil, IIT Bombay
Capacitors
i
conductor
insulator
conductor
Q
1111111111111
0000000000000
0000000000000
1111111111111
t
0000000000000
1111111111111
0000000000000
1111111111111
0000000000000
1111111111111
i
v
C=
ǫA
t
Q
Unit: Farad (F)
* In practice, capacitors are available in a wide range of shapes and values, and
they differ significantly in the way they are fabricated.
(http://en.wikipedia.org/wiki/Capacitor)
* To make C larger, we need (a) high , (b) large area, (c) small thickness.
M. B. Patil, IIT Bombay
Capacitors
i
conductor
insulator
conductor
Q
1111111111111
0000000000000
0000000000000
1111111111111
t
0000000000000
1111111111111
0000000000000
1111111111111
0000000000000
1111111111111
i
v
C=
ǫA
t
Q
Unit: Farad (F)
* In practice, capacitors are available in a wide range of shapes and values, and
they differ significantly in the way they are fabricated.
(http://en.wikipedia.org/wiki/Capacitor)
* To make C larger, we need (a) high , (b) large area, (c) small thickness.
* For a constant capacitance,
dQ
dv
dv
Q(t) = C v (t) ,
=C
, i.e, i(t) = C
.
dt
dt
dt
M. B. Patil, IIT Bombay
Capacitors
i
conductor
insulator
conductor
Q
1111111111111
0000000000000
0000000000000
1111111111111
t
0000000000000
1111111111111
0000000000000
1111111111111
0000000000000
1111111111111
i
v
C=
ǫA
t
Q
Unit: Farad (F)
* In practice, capacitors are available in a wide range of shapes and values, and
they differ significantly in the way they are fabricated.
(http://en.wikipedia.org/wiki/Capacitor)
* To make C larger, we need (a) high , (b) large area, (c) small thickness.
* For a constant capacitance,
dQ
dv
dv
Q(t) = C v (t) ,
=C
, i.e, i(t) = C
.
dt
dt
dt
* If v = constant, i = 0, i.e., a capacitor behaves like an open circuit in DC
conditions as one would expect from two conducting plates separated by an
insulator.
M. B. Patil, IIT Bombay
Example
i
20
i (mA)
Plot v, p, and W versus time
for the given source current.
Assume v(0) = 0 V, C = 5 mF.
0
−20
v
Example
i
20
i (mA)
Plot v, p, and W versus time
for the given source current.
Assume v(0) = 0 V, C = 5 mF.
0
−20
v
i(t) = C
v(t) =
dv
dt
1
C
Z
i(t) dt
Example
i
20
i (mA)
Plot v, p, and W versus time
for the given source current.
Assume v(0) = 0 V, C = 5 mF.
0
−20
v
i(t) = C
v (V)
8
dv
dt
1
v(t) =
C
Z
i(t) dt
4
0
−4
Example
i
20
i (mA)
Plot v, p, and W versus time
for the given source current.
Assume v(0) = 0 V, C = 5 mF.
0
−20
v
i(t) = C
v (V)
8
dv
dt
1
v(t) =
C
Z
i(t) dt
p(t) = v(t) × i(t)
4
0
−4
Example
i
20
i (mA)
Plot v, p, and W versus time
for the given source current.
Assume v(0) = 0 V, C = 5 mF.
0
−20
v
dv
dt
1
v(t) =
C
Z
i(t) dt
p(t) = v(t) × i(t)
4
0
−4
0.2
power (Watts)
i(t) = C
v (V)
8
0.1
0
−0.1
−0.2
Example
i
20
i (mA)
Plot v, p, and W versus time
for the given source current.
Assume v(0) = 0 V, C = 5 mF.
0
−20
v
dv
dt
1
v(t) =
C
Z
i(t) dt
p(t) = v(t) × i(t)
Z
W(t) = p(t) dt
4
0
−4
0.2
power (Watts)
i(t) = C
v (V)
8
0.1
0
−0.1
−0.2
Example
i
20
i (mA)
Plot v, p, and W versus time
for the given source current.
Assume v(0) = 0 V, C = 5 mF.
0
−20
v
dv
dt
Z
i(t) dt
p(t) = v(t) × i(t)
Z
W(t) = p(t) dt
4
0
−4
0.2
power (Watts)
1
v(t) =
C
0.1
0
−0.1
−0.2
0.2
energy (J)
i(t) = C
v (V)
8
0.1
0
0
1
2
3
time (sec)
4
5
6
Example
i
20
i (mA)
Plot v, p, and W versus time
for the given source current.
Assume v(0) = 0 V, C = 5 mF.
0
−20
v
dv
dt
Z
i(t) dt
p(t) = v(t) × i(t)
Z
W(t) = p(t) dt
Z
W(t) = p(t) dt
Z
dv
= C v dt
dt
Z
= C v dv
1
= C v2
2
4
0
−4
0.2
power (Watts)
1
v(t) =
C
0.1
0
−0.1
−0.2
0.2
energy (J)
i(t) = C
v (V)
8
0.1
0
0
1
2
3
time (sec)
4
5
6
M. B. Patil, IIT Bombay
Home work
i (mA)
i
20
v
0
1
2
time (sec)
M. B. Patil, IIT Bombay
Home work
i (mA)
i
20
v
0
1
2
time (sec)
* For the given source current, plot v (t), p(t), and W (t), assuming v (0) = 0 V ,
C = 5 mF .
M. B. Patil, IIT Bombay
Home work
i (mA)
i
20
v
0
1
2
time (sec)
* For the given source current, plot v (t), p(t), and W (t), assuming v (0) = 0 V ,
C = 5 mF .
* Verify your results with circuit simulation.
M. B. Patil, IIT Bombay
Inductors
core
Magnetic field lines
Symbol
v
i
L
Units: Henry (H)
M. B. Patil, IIT Bombay
Inductors
core
Magnetic field lines
Symbol
v
i
L
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
M. B. Patil, IIT Bombay
Inductors
core
Magnetic field lines
Symbol
v
i
L
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
»„
« –
dφ
d
d
µN i
* V =N
=N
(B · A) = N
A .
dt
dt
dt
l
di
Compare with v = L
.
dt
A
A
⇒ L = µ N2
= µr µ0 N 2 .
l
l
M. B. Patil, IIT Bombay
Inductors
core
Magnetic field lines
Symbol
v
i
L
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
»„
« –
dφ
d
d
µN i
* V =N
=N
(B · A) = N
A .
dt
dt
dt
l
di
Compare with v = L
.
dt
A
A
⇒ L = µ N2
= µr µ0 N 2 .
l
l
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
M. B. Patil, IIT Bombay
Inductors
core
Magnetic field lines
Symbol
v
i
L
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
»„
« –
dφ
d
d
µN i
* V =N
=N
(B · A) = N
A .
dt
dt
dt
l
di
Compare with v = L
.
dt
A
A
⇒ L = µ N2
= µr µ0 N 2 .
l
l
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
* For 99.8 % pure iron, µr ' 5, 000 .
For “supermalloy” (Ni: 79 %, Mo: 5 %, Fe): µr ' 106 .
M. B. Patil, IIT Bombay
Inductors
core
Magnetic field lines
Symbol
v
i
L
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
»„
« –
dφ
d
d
µN i
* V =N
=N
(B · A) = N
A .
dt
dt
dt
l
di
Compare with v = L
.
dt
A
A
⇒ L = µ N2
= µr µ0 N 2 .
l
l
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
* For 99.8 % pure iron, µr ' 5, 000 .
For “supermalloy” (Ni: 79 %, Mo: 5 %, Fe): µr ' 106 .
* If i = constant, v = 0, i.e., an inductor behaves like a short circuit in DC
conditions as one would expect from a highly conducting coil.
M. B. Patil, IIT Bombay
Inductors
core
Magnetic field lines
Symbol
v
i
L
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
»„
« –
dφ
d
d
µN i
* V =N
=N
(B · A) = N
A .
dt
dt
dt
l
di
Compare with v = L
.
dt
A
A
⇒ L = µ N2
= µr µ0 N 2 .
l
l
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
* For 99.8 % pure iron, µr ' 5, 000 .
For “supermalloy” (Ni: 79 %, Mo: 5 %, Fe): µr ' 106 .
* If i = constant, v = 0, i.e., an inductor behaves like a short circuit in DC
conditions as one would expect from a highly conducting coil.
* Note: B = µ H is an approximation. In practice, B may be a nonlinear function
of H, depending on the core material.
M. B. Patil, IIT Bombay
RC circuits with DC sources
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
B
RC circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
B
≡
i
VTh
v
C
B
M. B. Patil, IIT Bombay
RC circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
≡
i
VTh
B
v
C
B
* If all sources are DC (constant), VTh = constant .
M. B. Patil, IIT Bombay
RC circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
≡
i
VTh
B
v
C
B
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RTh C
dv
+v.
dt
M. B. Patil, IIT Bombay
RC circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
≡
i
VTh
B
v
C
B
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RTh C
dv
+v.
dt
* Homogeneous solution:
dv
1
+ v = 0 , where τ = RTh C is the “time constant.”
dt
τ
→ v (h) = K exp(−t/τ ) .
M. B. Patil, IIT Bombay
RC circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
≡
i
VTh
B
v
C
B
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RTh C
dv
+v.
dt
* Homogeneous solution:
dv
1
+ v = 0 , where τ = RTh C is the “time constant.”
dt
τ
→ v (h) = K exp(−t/τ ) .
* Particular solution is a specific function that satisfies the differntial equation. We
know that all time derivatives will vanish as t → ∞ , making i = 0, and we get
v (p) = VTh as a particular solution (which happens to be simply a constant).
M. B. Patil, IIT Bombay
RC circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
≡
i
VTh
B
v
C
B
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RTh C
dv
+v.
dt
* Homogeneous solution:
dv
1
+ v = 0 , where τ = RTh C is the “time constant.”
dt
τ
→ v (h) = K exp(−t/τ ) .
* Particular solution is a specific function that satisfies the differntial equation. We
know that all time derivatives will vanish as t → ∞ , making i = 0, and we get
v (p) = VTh as a particular solution (which happens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ ) + VTh .
M. B. Patil, IIT Bombay
RC circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
≡
i
VTh
B
v
C
B
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RTh C
dv
+v.
dt
* Homogeneous solution:
dv
1
+ v = 0 , where τ = RTh C is the “time constant.”
dt
τ
→ v (h) = K exp(−t/τ ) .
* Particular solution is a specific function that satisfies the differntial equation. We
know that all time derivatives will vanish as t → ∞ , making i = 0, and we get
v (p) = VTh as a particular solution (which happens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ ) + VTh .
* In general, v (t) = A exp(−t/τ ) + B , where A and B can be obtained from
known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
≡
B
i
VTh
v
C
B
* If all sources are DC (constant), we have
v (t) = A exp(−t/τ ) + B , τ = RC .
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
B
≡
i
VTh
v
C
B
* If all sources are DC (constant), we have
v (t) = A exp(−t/τ ) + B , τ = RC .
„
«
dv
1
* i(t) = C
= C × A exp(−t/τ ) −
≡ A0 exp(−t/τ ) .
dt
τ
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
B
≡
i
VTh
v
C
B
* If all sources are DC (constant), we have
v (t) = A exp(−t/τ ) + B , τ = RC .
„
«
dv
1
* i(t) = C
= C × A exp(−t/τ ) −
≡ A0 exp(−t/τ ) .
dt
τ
* As t → ∞, i → 0, i.e., the capacitor behaves like an open circuit since all
derivatives vanish.
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
C
B
≡
i
VTh
v
C
B
* If all sources are DC (constant), we have
v (t) = A exp(−t/τ ) + B , τ = RC .
„
«
dv
1
* i(t) = C
= C × A exp(−t/τ ) −
≡ A0 exp(−t/τ ) .
dt
τ
* As t → ∞, i → 0, i.e., the capacitor behaves like an open circuit since all
derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in
the circuit can be expressed as
x(t) = K1 exp(−t/τ ) + K2 ,
where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
B
RL circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
B
≡
i
VTh
v
B
M. B. Patil, IIT Bombay
RL circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
≡
i
VTh
B
v
B
* If all sources are DC (constant), VTh = constant .
M. B. Patil, IIT Bombay
RL circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
≡
i
VTh
B
v
B
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + L
di
.
dt
M. B. Patil, IIT Bombay
RL circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
≡
i
VTh
B
v
B
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + L
di
.
dt
* Homogeneous solution:
di
1
+ i = 0 , where τ = L/RTh
dt
τ
→ i (h) = K exp(−t/τ ) .
M. B. Patil, IIT Bombay
RL circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
≡
i
VTh
B
v
B
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + L
di
.
dt
* Homogeneous solution:
di
1
+ i = 0 , where τ = L/RTh
dt
τ
→ i (h) = K exp(−t/τ ) .
* Particular solution is a specific function that satisfies the differntial equation. We
know that all time derivatives will vanish as t → ∞ , making v = 0, and we get
i (p) = VTh /RTh as a particular solution (which happens to be simply a constant).
M. B. Patil, IIT Bombay
RL circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
≡
i
VTh
B
v
B
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + L
di
.
dt
* Homogeneous solution:
di
1
+ i = 0 , where τ = L/RTh
dt
τ
→ i (h) = K exp(−t/τ ) .
* Particular solution is a specific function that satisfies the differntial equation. We
know that all time derivatives will vanish as t → ∞ , making v = 0, and we get
i (p) = VTh /RTh as a particular solution (which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ ) + VTh /RTh .
M. B. Patil, IIT Bombay
RL circuits with DC sources
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
≡
i
VTh
B
v
B
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + L
di
.
dt
* Homogeneous solution:
di
1
+ i = 0 , where τ = L/RTh
dt
τ
→ i (h) = K exp(−t/τ ) .
* Particular solution is a specific function that satisfies the differntial equation. We
know that all time derivatives will vanish as t → ∞ , making v = 0, and we get
i (p) = VTh /RTh as a particular solution (which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ ) + VTh /RTh .
* In general, i(t) = A exp(−t/τ ) + B , where A and B can be obtained from
known conditions on i.
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
≡
B
i
VTh
v
B
* If all sources are DC (constant), we have
i(t) = A exp(−t/τ ) + B , τ = L/R .
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
B
≡
i
VTh
v
B
* If all sources are DC (constant), we have
i(t) = A exp(−t/τ ) + B , τ = L/R .
„
«
di
1
* v (t) = L
= L × A exp(−t/τ ) −
≡ A0 exp(−t/τ ) .
dt
τ
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
B
≡
i
VTh
v
B
* If all sources are DC (constant), we have
i(t) = A exp(−t/τ ) + B , τ = L/R .
„
«
di
1
* v (t) = L
= L × A exp(−t/τ ) −
≡ A0 exp(−t/τ ) .
dt
τ
* As t → ∞, v → 0, i.e., the inductor behaves like a short circuit since all
derivatives vanish.
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
RTh
A
A
Circuit
(resistors,
voltage sources,
current sources,
CCVS, CCCS,
VCVS, VCCS)
i
v
B
≡
i
VTh
v
B
* If all sources are DC (constant), we have
i(t) = A exp(−t/τ ) + B , τ = L/R .
„
«
di
1
* v (t) = L
= L × A exp(−t/τ ) −
≡ A0 exp(−t/τ ) .
dt
τ
* As t → ∞, v → 0, i.e., the inductor behaves like a short circuit since all
derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in
the circuit can be expressed as
x(t) = K1 exp(−t/τ ) + K2 ,
where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
R=1k
i
Vs
Vc
C = 1 µF
Vc (0) = 0 V
Vs
5V
0V
t
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
R=1k
i
Vs
Vc
C = 1 µF
Vc (0) = 0 V
Vs
5V
0V
t
* Vs changes from 0 V (at t = 0− ), to 5 V (at t = 0+ ). As a result of this
change, Vc will rise. How fast can Vc change?
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
R=1k
i
Vs
Vc
C = 1 µF
Vc (0) = 0 V
Vs
5V
0V
t
* Vs changes from 0 V (at t = 0− ), to 5 V (at t = 0+ ). As a result of this
change, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constant
rate of 1 V /1 µs = 106 V /s?
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
R=1k
i
Vs
Vc
C = 1 µF
Vc (0) = 0 V
Vs
5V
0V
t
* Vs changes from 0 V (at t = 0− ), to 5 V (at t = 0+ ). As a result of this
change, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constant
rate of 1 V /1 µs = 106 V /s?
* i =C
dVc
V
= 1 µF × 106
= 1 A.
dt
s
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
R=1k
i
Vs
Vc
C = 1 µF
Vc (0) = 0 V
Vs
5V
0V
t
* Vs changes from 0 V (at t = 0− ), to 5 V (at t = 0+ ). As a result of this
change, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constant
rate of 1 V /1 µs = 106 V /s?
dVc
V
= 1 µF × 106
= 1 A.
dt
s
* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.
* i =C
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
R=1k
i
Vs
Vc
C = 1 µF
Vc (0) = 0 V
Vs
5V
0V
t
* Vs changes from 0 V (at t = 0− ), to 5 V (at t = 0+ ). As a result of this
change, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constant
rate of 1 V /1 µs = 106 V /s?
dVc
V
= 1 µF × 106
= 1 A.
dt
s
* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.
* i =C
* We conclude that Vc (0+ ) = Vc (0− ) ⇒ A capacitor does not allow abrupt
changes in Vc if there is a finite resistance in the circuit.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
R=1k
i
Vs
Vc
C = 1 µF
Vc (0) = 0 V
Vs
5V
0V
t
* Vs changes from 0 V (at t = 0− ), to 5 V (at t = 0+ ). As a result of this
change, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constant
rate of 1 V /1 µs = 106 V /s?
dVc
V
= 1 µF × 106
= 1 A.
dt
s
* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.
* i =C
* We conclude that Vc (0+ ) = Vc (0− ) ⇒ A capacitor does not allow abrupt
changes in Vc if there is a finite resistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL .
M. B. Patil, IIT Bombay
Analysis of RC /RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.
For example,
Vs
(1) t < t1
(2) t1 < t < t2
0
t1
t2
(3) t > t2
M. B. Patil, IIT Bombay
Analysis of RC /RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.
For example,
Vs
(1) t < t1
(2) t1 < t < t2
0
t1
t2
(3) t > t2
* For any current or voltage x(t),
x(t) = A1 exp(−t/τ ) + B1 ,
x(t) = A2 exp(−t/τ ) + B2 ,
x(t) = A3 exp(−t/τ ) + B3 ,
write general expressions such as,
t < t1 ,
t1 < t < t2 ,
t > t2 .
M. B. Patil, IIT Bombay
Analysis of RC /RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.
For example,
Vs
(1) t < t1
(2) t1 < t < t2
0
t1
t2
(3) t > t2
* For any current or voltage x(t),
x(t) = A1 exp(−t/τ ) + B1 ,
x(t) = A2 exp(−t/τ ) + B2 ,
x(t) = A3 exp(−t/τ ) + B3 ,
write general expressions such as,
t < t1 ,
t1 < t < t2 ,
t > t2 .
* Work out suitable conditions on x(t) at specific time points using
M. B. Patil, IIT Bombay
Analysis of RC /RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.
For example,
Vs
(1) t < t1
(2) t1 < t < t2
0
t1
t2
(3) t > t2
* For any current or voltage x(t),
x(t) = A1 exp(−t/τ ) + B1 ,
x(t) = A2 exp(−t/τ ) + B2 ,
x(t) = A3 exp(−t/τ ) + B3 ,
write general expressions such as,
t < t1 ,
t1 < t < t2 ,
t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time
(long compared to τ ), all derivatives are zero.
dVc
diL
⇒ iC = C
= 0 , and VL = L
= 0.
dt
dt
M. B. Patil, IIT Bombay
Analysis of RC /RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.
For example,
Vs
(1) t < t1
(2) t1 < t < t2
0
t1
t2
(3) t > t2
* For any current or voltage x(t),
x(t) = A1 exp(−t/τ ) + B1 ,
x(t) = A2 exp(−t/τ ) + B2 ,
x(t) = A3 exp(−t/τ ) + B3 ,
write general expressions such as,
t < t1 ,
t1 < t < t2 ,
t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time
(long compared to τ ), all derivatives are zero.
dVc
diL
⇒ iC = C
= 0 , and VL = L
= 0.
dt
dt
(b) When a source voltage (or current) changes, say, at t = t0 ,
Vc (t) or iL (t) cannot change abruptly, i.e.,
Vc (t0+ ) = Vc (t0− ) , and iL (t0+ ) = iL (t0− ) .
M. B. Patil, IIT Bombay
Analysis of RC /RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.
For example,
Vs
(1) t < t1
(2) t1 < t < t2
0
t1
t2
(3) t > t2
* For any current or voltage x(t),
x(t) = A1 exp(−t/τ ) + B1 ,
x(t) = A2 exp(−t/τ ) + B2 ,
x(t) = A3 exp(−t/τ ) + B3 ,
write general expressions such as,
t < t1 ,
t1 < t < t2 ,
t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time
(long compared to τ ), all derivatives are zero.
dVc
diL
⇒ iC = C
= 0 , and VL = L
= 0.
dt
dt
(b) When a source voltage (or current) changes, say, at t = t0 ,
Vc (t) or iL (t) cannot change abruptly, i.e.,
Vc (t0+ ) = Vc (t0− ) , and iL (t0+ ) = iL (t0− ) .
* Compute A1 , B1 , · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
R
Vs
i
Vs
v
V0
C
0V
t
RC circuits: charging and discharging transients
R
Vs
i
Vs
v
V0
C
t
0V
Let v(t) = A exp(−t/τ ) + B, t > 0
(A)
RC circuits: charging and discharging transients
R
Vs
i
Vs
v
V0
C
t
0V
Let v(t) = A exp(−t/τ ) + B, t > 0
(A)
Conditions on v(t):
(1) v(0− ) = Vs (0− ) = 0 V
v(0+ ) ≃ v(0− ) = 0 V
Note that we need the condition at 0+ (and not at 0− )
because Eq. (A) applies only for t > 0.
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = V0
RC circuits: charging and discharging transients
R
Vs
i
Vs
v
V0
C
t
0V
Let v(t) = A exp(−t/τ ) + B, t > 0
(A)
Conditions on v(t):
(1) v(0− ) = Vs (0− ) = 0 V
v(0+ ) ≃ v(0− ) = 0 V
Note that we need the condition at 0+ (and not at 0− )
because Eq. (A) applies only for t > 0.
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = V0
Imposing (1) and (2) on Eq. (A), we get
t = 0+ : 0 = A + B ,
t → ∞: V0 = B .
i.e., A = V0 , B = −V0
RC circuits: charging and discharging transients
R
Vs
i
Vs
v
V0
C
t
0V
Let v(t) = A exp(−t/τ ) + B, t > 0
(A)
Conditions on v(t):
(1) v(0− ) = Vs (0− ) = 0 V
v(0+ ) ≃ v(0− ) = 0 V
Note that we need the condition at 0+ (and not at 0− )
because Eq. (A) applies only for t > 0.
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = V0
Imposing (1) and (2) on Eq. (A), we get
t = 0+ : 0 = A + B ,
t → ∞: V0 = B .
i.e., A = V0 , B = −V0
v(t) = V0 [1 − exp(−t/τ )]
RC circuits: charging and discharging transients
R
i
Vs
v
R
Vs
Vs
C
t
0V
Let v(t) = A exp(−t/τ ) + B, t > 0
(A)
Conditions on v(t):
(1) v(0− ) = Vs (0− ) = 0 V
v(0+ ) ≃ v(0− ) = 0 V
Note that we need the condition at 0+ (and not at 0− )
because Eq. (A) applies only for t > 0.
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = V0
Imposing (1) and (2) on Eq. (A), we get
t = 0+ : 0 = A + B ,
t → ∞: V0 = B .
i.e., A = V0 , B = −V0
v(t) = V0 [1 − exp(−t/τ )]
Vs
i
V0
v
V0
C
0V
t
RC circuits: charging and discharging transients
R
i
Vs
v
R
Vs
Vs
C
t
0V
Let v(t) = A exp(−t/τ ) + B, t > 0
(A)
Conditions on v(t):
(1) v(0− ) = Vs (0− ) = 0 V
v(0+ ) ≃ v(0− ) = 0 V
Note that we need the condition at 0+ (and not at 0− )
because Eq. (A) applies only for t > 0.
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = V0
Imposing (1) and (2) on Eq. (A), we get
t = 0+ : 0 = A + B ,
t → ∞: V0 = B .
i.e., A = V0 , B = −V0
v(t) = V0 [1 − exp(−t/τ )]
Vs
i
V0
v
V0
C
t
0V
Let v(t) = A exp(−t/τ ) + B, t > 0
(A)
RC circuits: charging and discharging transients
R
i
Vs
v
R
Vs
Vs
C
C
Let v(t) = A exp(−t/τ ) + B, t > 0
(A)
t
(A)
Conditions on v(t):
−
(1) v(0− ) = Vs (0− ) = V0
(1) v(0 ) = Vs (0 ) = 0 V
+
V0
0V
Conditions on v(t):
−
v
t
0V
Let v(t) = A exp(−t/τ ) + B, t > 0
Vs
i
V0
−
v(0+ ) ≃ v(0− ) = V0
v(0 ) ≃ v(0 ) = 0 V
+
Note that we need the condition at 0 (and not at 0 )
Note that we need the condition at 0+ (and not at 0− )
because Eq. (A) applies only for t > 0.
because Eq. (A) applies only for t > 0.
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = V0
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+ : 0 = A + B ,
t → ∞: V0 = B .
i.e., A = V0 , B = −V0
v(t) = V0 [1 − exp(−t/τ )]
−
RC circuits: charging and discharging transients
R
i
Vs
v
R
Vs
Vs
C
C
Let v(t) = A exp(−t/τ ) + B, t > 0
(A)
t
(A)
Conditions on v(t):
−
(1) v(0− ) = Vs (0− ) = V0
(1) v(0 ) = Vs (0 ) = 0 V
+
V0
0V
Conditions on v(t):
−
v
t
0V
Let v(t) = A exp(−t/τ ) + B, t > 0
Vs
i
V0
−
v(0+ ) ≃ v(0− ) = V0
v(0 ) ≃ v(0 ) = 0 V
+
Note that we need the condition at 0 (and not at 0 )
Note that we need the condition at 0+ (and not at 0− )
because Eq. (A) applies only for t > 0.
because Eq. (A) applies only for t > 0.
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = V0
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
Imposing (1) and (2) on Eq. (A), we get
+
t = 0 : 0 = A+B,
−
t = 0+ : V0 = A + B ,
t → ∞: V0 = B .
t → ∞: 0 = B .
i.e., A = V0 , B = −V0
i.e., A = V0 , B = 0
v(t) = V0 [1 − exp(−t/τ )]
RC circuits: charging and discharging transients
R
i
Vs
v
R
Vs
Vs
C
C
t
Let v(t) = A exp(−t/τ ) + B, t > 0
(A)
(A)
Conditions on v(t):
−
(1) v(0− ) = Vs (0− ) = V0
(1) v(0 ) = Vs (0 ) = 0 V
+
V0
0V
Conditions on v(t):
−
v
t
0V
Let v(t) = A exp(−t/τ ) + B, t > 0
Vs
i
V0
−
v(0+ ) ≃ v(0− ) = V0
v(0 ) ≃ v(0 ) = 0 V
+
Note that we need the condition at 0 (and not at 0 )
Note that we need the condition at 0+ (and not at 0− )
because Eq. (A) applies only for t > 0.
because Eq. (A) applies only for t > 0.
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = V0
(2) As t → ∞ , i → 0 → v(∞) = Vs (∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
Imposing (1) and (2) on Eq. (A), we get
+
t = 0 : 0 = A+B,
−
t = 0+ : V0 = A + B ,
t → ∞: V0 = B .
t → ∞: 0 = B .
i.e., A = V0 , B = −V0
i.e., A = V0 , B = 0
v(t) = V0 [1 − exp(−t/τ )]
v(t) = V0 exp(−t/τ )
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
R
Vs
i
Vs
v
V0
C
0V
Compute i(t), t > 0 .
t
RC circuits: charging and discharging transients
R
Vs
i
Vs
v
V0
C
0V
Compute i(t), t > 0 .
(A) i(t) = C
=
d
V0 [1 − exp(−t/τ )]
dt
CV0
V0
exp(−t/τ ) =
exp(−t/τ )
τ
R
t
RC circuits: charging and discharging transients
R
Vs
i
Vs
v
V0
C
0V
Compute i(t), t > 0 .
(A) i(t) = C
=
d
V0 [1 − exp(−t/τ )]
dt
CV0
V0
exp(−t/τ ) =
exp(−t/τ )
τ
R
(B) Let i(t) = A′ exp(−t/τ ) + B′ , t > 0 .
t = 0+ : v = 0 , Vs = V0 ⇒ i(0+ ) = V0 /R .
t → ∞: i(t) = 0 .
Using these conditions, we obtain
A′ =
V0
V0
, B′ = 0 ⇒ i(t) =
exp(−t/τ )
R
R
t
RC circuits: charging and discharging transients
R
i
Vs
R
Vs
v
Compute i(t), t > 0 .
d
(A) i(t) = C V0 [1 − exp(−t/τ )]
dt
=
CV0
V0
exp(−t/τ ) =
exp(−t/τ )
τ
R
(B) Let i(t) = A′ exp(−t/τ ) + B′ , t > 0 .
t = 0+ : v = 0 , Vs = V0 ⇒ i(0+ ) = V0 /R .
t → ∞: i(t) = 0 .
Using these conditions, we obtain
A′ =
Vs
C
0V
V0
V0
, B′ = 0 ⇒ i(t) =
exp(−t/τ )
R
R
Vs
i
V0
v
t
V0
C
0V
Compute i(t), t > 0 .
t
RC circuits: charging and discharging transients
R
i
Vs
R
Vs
v
Vs
C
0V
Vs
i
V0
v
t
Compute i(t), t > 0 .
d
(A) i(t) = C V0 [1 − exp(−t/τ )]
dt
(A) i(t) = C
CV0
V0
exp(−t/τ ) =
exp(−t/τ )
τ
R
(B) Let i(t) = A′ exp(−t/τ ) + B′ , t > 0 .
t = 0+ : v = 0 , Vs = V0 ⇒ i(0+ ) = V0 /R .
t → ∞: i(t) = 0 .
Using these conditions, we obtain
A′ =
V0
V0
, B′ = 0 ⇒ i(t) =
exp(−t/τ )
R
R
C
0V
Compute i(t), t > 0 .
=
V0
=−
d
V0 [exp(−t/τ )]
dt
CV0
V0
exp(−t/τ ) = − exp(−t/τ )
τ
R
t
RC circuits: charging and discharging transients
R
i
Vs
R
Vs
v
Vs
C
0V
Vs
i
V0
v
t
Compute i(t), t > 0 .
d
(A) i(t) = C V0 [1 − exp(−t/τ )]
dt
(A) i(t) = C
CV0
V0
exp(−t/τ ) =
exp(−t/τ )
τ
R
(B) Let i(t) = A′ exp(−t/τ ) + B′ , t > 0 .
C
t
0V
Compute i(t), t > 0 .
=
V0
=−
d
V0 [exp(−t/τ )]
dt
CV0
V0
exp(−t/τ ) = − exp(−t/τ )
τ
R
(B) Let i(t) = A′ exp(−t/τ ) + B′ , t > 0 .
t = 0+ : v = 0 , Vs = V0 ⇒ i(0+ ) = V0 /R .
t = 0+ : v = V0 , Vs = 0 ⇒ i(0+ ) = −V0 /R .
t → ∞: i(t) = 0 .
t → ∞: i(t) = 0 .
Using these conditions, we obtain
Using these conditions, we obtain
A′ =
V0
V0
, B′ = 0 ⇒ i(t) =
exp(−t/τ )
R
R
A′ = −
V0
V0
, B′ = 0 ⇒ i(t) = − exp(−t/τ )
R
R
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
R=1k
Vs
i
Vs
v
5V
C = 1 µF
0V
v(t) = V0 [1 − exp(−t/τ )]
i(t) =
V0
exp(−t/τ )
R
t
RC circuits: charging and discharging transients
R=1k
Vs
i
Vs
v
5V
C = 1 µF
0V
v(t) = V0 [1 − exp(−t/τ )]
i(t) =
5
V0
exp(−t/τ )
R
Vs
v (Volts)
v
0
t
RC circuits: charging and discharging transients
R=1k
Vs
i
Vs
v
5V
C = 1 µF
t
0V
v(t) = V0 [1 − exp(−t/τ )]
i(t) =
5
V0
exp(−t/τ )
R
Vs
v (Volts)
v
0
i (mA)
5
0
−2
0
2
4
time (msec)
6
8
RC circuits: charging and discharging transients
R=1k
i
Vs
v
R=1k
Vs
v (Volts)
0
i (mA)
5
0
2
4
time (msec)
0V
i(t) = −
Vs
0
5V
C = 1 µF
v(t) = V0 exp(−t/τ )
V0
exp(−t/τ )
R
v
−2
v
t
v(t) = V0 [1 − exp(−t/τ )]
5
Vs
C = 1 µF
0V
i(t) =
Vs
i
5V
6
8
V0
exp(−t/τ )
R
t
RC circuits: charging and discharging transients
R=1k
i
Vs
v
R=1k
Vs
Vs
C = 1 µF
v(t) = V0 exp(−t/τ )
V0
exp(−t/τ )
R
i(t) = −
V0
exp(−t/τ )
R
5
Vs
Vs
v (Volts)
v (Volts)
v
0
v
0
i (mA)
5
0
−2
0
2
4
time (msec)
5V
C = 1 µF
0V
v(t) = V0 [1 − exp(−t/τ )]
5
v
t
0V
i(t) =
Vs
i
5V
6
8
t
RC circuits: charging and discharging transients
R=1k
i
Vs
v
R=1k
Vs
Vs
C = 1 µF
5V
C = 1 µF
t
0V
v(t) = V0 [1 − exp(−t/τ )]
5
v
t
0V
i(t) =
Vs
i
5V
v(t) = V0 exp(−t/τ )
V0
exp(−t/τ )
R
i(t) = −
V0
exp(−t/τ )
R
5
Vs
Vs
v (Volts)
v (Volts)
v
v
5
0
i (mA)
0
i (mA)
0
0
−2
−5
0
2
4
time (msec)
6
8
−2
0
2
4
time (msec)
6
8
M. B. Patil, IIT Bombay
Significance of the time constant (τ )
x
e −x
1 − e −x
0.0
1.0
0.0
1.0
0.3679
0.6321
2.0
0.1353
0.8647
3.0
4.9787×10−2
0.9502
4.0
1.8315×10−2
0.9817
5.0
6.7379×10−3
0.9933
M. B. Patil, IIT Bombay
Significance of the time constant (τ )
x
e −x
1 − e −x
0.0
1.0
0.0
1.0
0.3679
0.6321
2.0
0.1353
0.8647
3.0
4.9787×10−2
0.9502
4.0
1.8315×10−2
0.9817
5.0
6.7379×10−3
0.9933
* For x = 5, e −x ' 0, 1 − e −x ' 1.
M. B. Patil, IIT Bombay
Significance of the time constant (τ )
x
e −x
1 − e −x
0.0
1.0
0.0
1.0
0.3679
0.6321
2.0
0.1353
0.8647
3.0
4.9787×10−2
0.9502
4.0
1.8315×10−2
0.9817
5.0
6.7379×10−3
0.9933
* For x = 5, e −x ' 0, 1 − e −x ' 1.
* In RC circuits, x = t/τ ⇒ When t = 5 τ , the charging (or discharging) process
is almost complete.
M. B. Patil, IIT Bombay
Significance of the time constant (τ )
x
e −x
1 − e −x
0.0
1.0
0.0
1.0
0.3679
0.6321
2.0
0.1353
0.8647
3.0
4.9787×10−2
0.9502
4.0
1.8315×10−2
0.9817
5.0
6.7379×10−3
0.9933
1
1 − exp(−x)
exp(−x)
0
0
1
2
3
x
4
5
6
* For x = 5, e −x ' 0, 1 − e −x ' 1.
* In RC circuits, x = t/τ ⇒ When t = 5 τ , the charging (or discharging) process
is almost complete.
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
R
i
Vs
v
R
Vs
C = 1 µF
0V
Vs
R = 100 Ω
v (Volts)
v (Volts)
R = 1 kΩ
0
1
2
3
4
time (msec)
t
5
0
0
5V
C = 1 µF
0V
v(t) = V0 exp(−t/τ )
R = 1 kΩ
−1
v
t
v(t) = V0 [1 − exp(−t/τ )]
5
Vs
i
5V
5
6
−1
R = 100 Ω
0
1
2
3
4
time (msec)
5
6
M. B. Patil, IIT Bombay
RL circuit: example
R1
Vs
R1 = 10 Ω
Vs
i
v
i(0) = 0 A, Find i(t).
R2 = 40 Ω
10 V
R2
L = 0.8 H
t0 = 0
t0
t1
t
t1 = 0.1 s
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
i(0) = 0 A, Find i(t).
R2
(2) t0 < t < t1
(3) t > t1
L = 0.8 H
t0 = 0
t0
t1
There are three intervals of constant Vs :
(1) t < t0
R2 = 40 Ω
10 V
t
t1 = 0.1 s
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
i(0) = 0 A, Find i(t).
R2
t1
There are three intervals of constant Vs :
(2) t0 < t < t1
(3) t > t1
RTh seen by L is the same in all intervals:
RTh = R1 k R2 = 8 Ω
Vs
R2
L = 0.8 H
t0 = 0
t0
(1) t < t0
R1
R2 = 40 Ω
10 V
τ = L/RTh
= 0.8 H/8 Ω
= 0.1 s
t
t1 = 0.1 s
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
i(0) = 0 A, Find i(t).
R2
t1
(1) t < t0
(2) t0 < t < t1
(3) t > t1
RTh seen by L is the same in all intervals:
RTh = R1 k R2 = 8 Ω
Vs
R2
L = 0.8 H
t0 = 0
t0
There are three intervals of constant Vs :
R1
R2 = 40 Ω
10 V
τ = L/RTh
= 0.8 H/8 Ω
= 0.1 s
t
t1 = 0.1 s
At t = t−
0 , v = 0 V, Vs = 0 V .
+
⇒ i(t−
0 ) = 0 A ⇒ i(t0 ) = 0 A .
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
i(0) = 0 A, Find i(t).
R2 = 40 Ω
10 V
R2
L = 0.8 H
t0 = 0
t0
t1
There are three intervals of constant Vs :
(1) t < t0
t
t1 = 0.1 s
At t = t−
0 , v = 0 V, Vs = 0 V .
+
⇒ i(t−
0 ) = 0 A ⇒ i(t0 ) = 0 A .
If Vs did not change at t = t1 ,
(2) t0 < t < t1
we would have
(3) t > t1
RTh seen by L is the same in all intervals:
Vs
10 V
RTh = R1 k R2 = 8 Ω
R1
Vs
R2
τ = L/RTh
= 0.8 H/8 Ω
= 0.1 s
t0
t1
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
i(0) = 0 A, Find i(t).
R2 = 40 Ω
10 V
R2
L = 0.8 H
t0 = 0
t0
t1
There are three intervals of constant Vs :
(1) t < t0
t
t1 = 0.1 s
At t = t−
0 , v = 0 V, Vs = 0 V .
+
⇒ i(t−
0 ) = 0 A ⇒ i(t0 ) = 0 A .
If Vs did not change at t = t1 ,
(2) t0 < t < t1
we would have
(3) t > t1
RTh seen by L is the same in all intervals:
Vs
10 V
RTh = R1 k R2 = 8 Ω
R1
Vs
R2
τ = L/RTh
= 0.8 H/8 Ω
= 0.1 s
t0
t1
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
Using i(t+
0 ) and i(∞), we can obtain
i(t), t > 0 (See next slide).
M. B. Patil, IIT Bombay
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
R2
t0
i (Amp)
1
0
0
0.2
0.4
0.6
time (sec)
R2 = 40 Ω
10 V
0.8
t1
L = 0.8 H
t
t0 = 0
t1 = 0.1 s
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
R2
t0
i (Amp)
1
0
0
0.2
0.4
0.6
time (sec)
In reality, Vs changes at t = t1 ,
and we need to work out the
solution for t > t1 separately.
R2 = 40 Ω
10 V
0.8
t1
L = 0.8 H
t
t0 = 0
t1 = 0.1 s
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
R2 = 40 Ω
10 V
R2
t0
t1
L = 0.8 H
t
t0 = 0
t1 = 0.1 s
For t0 < t < t1 , i(t) = 1 − exp(−t/τ ) Amp.
1
i (Amp)
Consider t > t1 .
0
0
0.2
0.4
0.6
time (sec)
In reality, Vs changes at t = t1 ,
and we need to work out the
solution for t > t1 separately.
0.8
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
R2 = 40 Ω
10 V
R2
t0
t1
L = 0.8 H
t
t0 = 0
t1 = 0.1 s
For t0 < t < t1 , i(t) = 1 − exp(−t/τ ) Amp.
1
Consider t > t1 .
i (Amp)
−
−1
i(t+
= 0.632 A (Note: t1 /τ = 1).
1 ) = i(t1 ) = 1 − e
i(∞) = 0 A.
0
0
0.2
0.4
0.6
time (sec)
In reality, Vs changes at t = t1 ,
and we need to work out the
solution for t > t1 separately.
0.8
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
R2 = 40 Ω
10 V
R2
t0
t1
L = 0.8 H
t
t0 = 0
t1 = 0.1 s
For t0 < t < t1 , i(t) = 1 − exp(−t/τ ) Amp.
1
Consider t > t1 .
i (Amp)
−
−1
i(t+
= 0.632 A (Note: t1 /τ = 1).
1 ) = i(t1 ) = 1 − e
i(∞) = 0 A.
Let i(t) = A exp(−t/τ ) + B.
0
0
0.2
0.4
0.6
time (sec)
In reality, Vs changes at t = t1 ,
and we need to work out the
solution for t > t1 separately.
0.8
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
R2 = 40 Ω
10 V
R2
t0
t1
L = 0.8 H
t
t0 = 0
t1 = 0.1 s
For t0 < t < t1 , i(t) = 1 − exp(−t/τ ) Amp.
1
Consider t > t1 .
i (Amp)
−
−1
i(t+
= 0.632 A (Note: t1 /τ = 1).
1 ) = i(t1 ) = 1 − e
i(∞) = 0 A.
Let i(t) = A exp(−t/τ ) + B.
0
It is convenient to rewrite i(t) as
0
0.2
0.4
0.6
time (sec)
In reality, Vs changes at t = t1 ,
and we need to work out the
solution for t > t1 separately.
0.8
i(t) = A′ exp[−(t − t1 )/τ ] + B.
RL circuit: example
R1
R1 = 10 Ω
Vs
i
Vs
v
R2 = 40 Ω
10 V
R2
t0
t1
L = 0.8 H
t
t0 = 0
t1 = 0.1 s
For t0 < t < t1 , i(t) = 1 − exp(−t/τ ) Amp.
1
Consider t > t1 .
i (Amp)
−
−1
i(t+
= 0.632 A (Note: t1 /τ = 1).
1 ) = i(t1 ) = 1 − e
i(∞) = 0 A.
Let i(t) = A exp(−t/τ ) + B.
0
It is convenient to rewrite i(t) as
0
0.2
0.4
0.6
time (sec)
In reality, Vs changes at t = t1 ,
0.8
i(t) = A′ exp[−(t − t1 )/τ ] + B.
Using i(t+
1 ) and i(∞), we get
i(t) = 0.693 exp[−(t − t1 )/τ ] A.
and we need to work out the
solution for t > t1 separately.
M. B. Patil, IIT Bombay
RL circuit: example
R1
Vs
v
R2
i(t) = 0.693 exp[−(t − t1 )/τ ] A.
i (Amp)
1
0
0.2
0.4
0.6
time (sec)
R2 = 40 Ω
10 V
t0
0
R1 = 10 Ω
Vs
i
0.8
t1
L = 0.8 H
t
t0 = 0
t1 = 0.1 s
RL circuit: example
R1
Vs
R1 = 10 Ω
Vs
i
v
R2 = 40 Ω
10 V
R2
t0
i(t) = 0.693 exp[−(t − t1 )/τ ] A.
L = 0.8 H
t
t1
t0 = 0
t1 = 0.1 s
Combining the solutions for t0 < t < t1 and t > t1 ,
we get
1
i (Amp)
i (Amp)
1
0
0
0.2
0.4
0.6
time (sec)
0.8
0
0
0.2
0.4
0.6
time (sec)
0.8
(SEQUEL file: ee101_rl1.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
t=0
i
5k
R1
R3 = 5 k
ic
1k
R2
6V
5 µF
vc
RC circuit: example
t=0
i
5k
R1
R3 = 5 k
5k
ic
1k
R2
6V
5 µF
vc
5k
ic
i
5k
1k
5 µF
ic
i
vc
AND
5k
5 µF
6V
t<0
t>0
vc
RC circuit: example
t=0
i
5k
R1
R3 = 5 k
5k
ic
1k
R2
6V
5 µF
vc
5k
ic
i
5k
1k
5 µF
ic
i
vc
AND
5k
5 µF
6V
t<0
t = 0− : capacitor is an open circuit, ⇒ i(0− ) = 6 V/(5 k + 1 k) = 1 mA.
t>0
vc
RC circuit: example
t=0
i
5k
R1
R3 = 5 k
5k
ic
1k
R2
5 µF
vc
5k
ic
i
5k
6V
1k
5 µF
vc
AND
5k
5 µF
6V
t<0
t = 0− : capacitor is an open circuit, ⇒ i(0− ) = 6 V/(5 k + 1 k) = 1 mA.
vc (0− ) = 6 V − 1 mA × R2 = 5 V ⇒ vc (0+ ) = 5 V.
ic
i
t>0
vc
RC circuit: example
t=0
i
5k
R1
R3 = 5 k
5k
ic
1k
R2
5 µF
vc
5k
ic
i
5k
6V
1k
5 µF
vc
t<0
⇒ i(0+ ) = 5 V/(5 k + 5 k) = 0.5 mA.
AND
5k
5 µF
6V
t = 0− : capacitor is an open circuit, ⇒ i(0− ) = 6 V/(5 k + 1 k) = 1 mA.
vc (0− ) = 6 V − 1 mA × R2 = 5 V ⇒ vc (0+ ) = 5 V.
ic
i
t>0
vc
RC circuit: example
t=0
i
5k
R1
R3 = 5 k
5k
ic
1k
R2
5 µF
vc
5k
ic
i
5k
6V
1k
5 µF
ic
i
vc
AND
5k
5 µF
6V
t<0
t = 0− : capacitor is an open circuit, ⇒ i(0− ) = 6 V/(5 k + 1 k) = 1 mA.
vc (0− ) = 6 V − 1 mA × R2 = 5 V ⇒ vc (0+ ) = 5 V.
⇒ i(0+ ) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = A exp(-t/τ ) + B for t > 0, with τ = 10 k × 5 µF = 50 ms.
t>0
vc
RC circuit: example
t=0
i
5k
R1
R3 = 5 k
5k
ic
1k
R2
5 µF
vc
5k
ic
i
5k
6V
1k
5 µF
vc
t<0
vc (0− ) = 6 V − 1 mA × R2 = 5 V ⇒ vc (0+ ) = 5 V.
⇒ i(0+ ) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = A exp(-t/τ ) + B for t > 0, with τ = 10 k × 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ ) mA.
AND
5k
5 µF
6V
t = 0− : capacitor is an open circuit, ⇒ i(0− ) = 6 V/(5 k + 1 k) = 1 mA.
Using i(0+ ) and i(∞) = 0 A, we get
ic
i
t>0
vc
RC circuit: example
t=0
i
R3 = 5 k
1k
5k
R2
R1
5k
ic
5 µF
vc
5k
ic
i
5k
1k
5 µF
vc
t<0
t = 0− : capacitor is an open circuit, ⇒ i(0− ) = 6 V/(5 k + 1 k) = 1 mA.
vc (0− ) = 6 V − 1 mA × R2 = 5 V ⇒ vc (0+ ) = 5 V.
⇒ i(0+ ) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = A exp(-t/τ ) + B for t > 0, with τ = 10 k × 5 µF = 50 ms.
Using i(0+ ) and i(∞) = 0 A, we get
i(t) = 0.5 exp(-t/τ ) mA.
1
i (mA)
0
time (sec)
AND
5k
5 µF
6V
6V
0
ic
i
0.5
t>0
vc
RC circuit: example
t=0
i
R3 = 5 k
1k
5k
R2
R1
5k
ic
5 µF
5k
ic
i
vc
1k
5k
vc
5 µF
ic
i
5k
AND
5 µF
vc
6V
6V
t<0
t>0
t = 0− : capacitor is an open circuit, ⇒ i(0− ) = 6 V/(5 k + 1 k) = 1 mA.
vc (0− ) = 6 V − 1 mA × R2 = 5 V ⇒ vc (0+ ) = 5 V.
⇒ i(0+ ) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = A exp(-t/τ ) + B for t > 0, with τ = 10 k × 5 µF = 50 ms.
Using i(0+ ) and i(∞) = 0 A, we get
i(t) = 0.5 exp(-t/τ ) mA.
1
0
5
ic (mA)
i (mA)
vc (V)
0
0
0
time (sec)
0.5
−0.5
0
time (sec)
0.5
0
time (sec)
0.5
RC circuit: example
t=0
i
R3 = 5 k
1k
5k
R2
R1
5k
ic
5 µF
5k
ic
i
vc
1k
5k
vc
5 µF
ic
i
5k
AND
5 µF
vc
6V
6V
t<0
t>0
t = 0− : capacitor is an open circuit, ⇒ i(0− ) = 6 V/(5 k + 1 k) = 1 mA.
vc (0− ) = 6 V − 1 mA × R2 = 5 V ⇒ vc (0+ ) = 5 V.
⇒ i(0+ ) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = A exp(-t/τ ) + B for t > 0, with τ = 10 k × 5 µF = 50 ms.
Using i(0+ ) and i(∞) = 0 A, we get
i(t) = 0.5 exp(-t/τ ) mA.
(SEQUEL file: ee101_rc2.sqproj)
1
0
5
ic (mA)
i (mA)
vc (V)
0
0
0
time (sec)
0.5
−0.5
0
time (sec)
0.5
0
time (sec) M. B. Patil,0.5
IIT Bombay
RC circuits: home work
10 Ω
i2
10 V
i1
10 Ω
ic
200 µF
vc
M. B. Patil, IIT Bombay
RC circuits: home work
10 Ω
i2
10 V
i1
10 Ω
ic
200 µF
vc
* Given vc (0) = 0 V , find vc (t) for t > 0. Using this vc (t), find i1 , i2 , ic for t > 0.
Plot vc , i1 , i2 , ic versus t.
M. B. Patil, IIT Bombay
RC circuits: home work
10 Ω
i2
10 V
i1
10 Ω
ic
200 µF
vc
* Given vc (0) = 0 V , find vc (t) for t > 0. Using this vc (t), find i1 , i2 , ic for t > 0.
Plot vc , i1 , i2 , ic versus t.
* Find i1 , i2 , ic directly (i.e., without getting vc ) by finding the initial and final
conditions for each of them (i1 (0+ ) and i1 (∞), etc.) and then using them to
compute the coefficients in the general expression,
x(t) = A exp(−t/τ ) + B.
M. B. Patil, IIT Bombay
RC circuits: home work
10 Ω
i2
10 V
i1
10 Ω
ic
200 µF
vc
* Given vc (0) = 0 V , find vc (t) for t > 0. Using this vc (t), find i1 , i2 , ic for t > 0.
Plot vc , i1 , i2 , ic versus t.
* Find i1 , i2 , ic directly (i.e., without getting vc ) by finding the initial and final
conditions for each of them (i1 (0+ ) and i1 (∞), etc.) and then using them to
compute the coefficients in the general expression,
x(t) = A exp(−t/τ ) + B.
* Verify your results with SEQUEL (file: ee101 rc3.sqproj).
M. B. Patil, IIT Bombay
RC circuits: home work
2Ω
t=0
24 V
3Ω
i1
ic
5Ω
1 mF
vx
vc
0.1 vx
M. B. Patil, IIT Bombay
RC circuits: home work
2Ω
t=0
24 V
3Ω
i1
ic
5Ω
1 mF
vx
vc
0.1 vx
* Find vc (0− ), vc (∞).
M. B. Patil, IIT Bombay
RC circuits: home work
2Ω
t=0
24 V
3Ω
i1
ic
5Ω
1 mF
vx
vc
0.1 vx
* Find vc (0− ), vc (∞).
* Find RTh as seen by the capacitor for t > 0.
M. B. Patil, IIT Bombay
RC circuits: home work
2Ω
3Ω
i1
ic
5Ω
1 mF
t=0
24 V
vx
vc
0.1 vx
* Find vc (0− ), vc (∞).
* Find RTh as seen by the capacitor for t > 0.
* Solve for vc (t) and i1 (t), t > 0.
M. B. Patil, IIT Bombay
RC circuits: home work
2Ω
3Ω
i1
ic
5Ω
1 mF
t=0
24 V
vx
vc
0.1 vx
* Find vc (0− ), vc (∞).
* Find RTh as seen by the capacitor for t > 0.
* Solve for vc (t) and i1 (t), t > 0.
* Verify your results with SEQUEL (file: ee101 rc4.sqproj).
M. B. Patil, IIT Bombay
RL circuits: home work
20 Ω
t=0
5V
20 Ω
i
L=0.1 H
10 V
M. B. Patil, IIT Bombay
RL circuits: home work
20 Ω
t=0
5V
20 Ω
i
L=0.1 H
10 V
* Find i(0− ), i(∞).
M. B. Patil, IIT Bombay
RL circuits: home work
20 Ω
t=0
5V
20 Ω
i
L=0.1 H
10 V
* Find i(0− ), i(∞).
* Find RTh as seen by the inductor for t > 0.
M. B. Patil, IIT Bombay
RL circuits: home work
20 Ω
t=0
5V
20 Ω
i
L=0.1 H
10 V
* Find i(0− ), i(∞).
* Find RTh as seen by the inductor for t > 0.
* Solve for i(t), t > 0.
M. B. Patil, IIT Bombay
RL circuits: home work
20 Ω
t=0
5V
20 Ω
i
L=0.1 H
10 V
* Find i(0− ), i(∞).
* Find RTh as seen by the inductor for t > 0.
* Solve for i(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rl2.sqproj).
M. B. Patil, IIT Bombay