EE101: RC and RL Circuits (with DC sources)

EE101: RC and RL Circuits (with DC sources)

M. B. [email protected]

www.ee.iitb.ac.in/~sequel

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT Bombay

Capacitors

v

i

Q

Q

insulator

i

conductor

conductor

Unit: Farad (F)

C =ǫ A

tt

* In practice, capacitors are available in a wide range of shapes and values, andthey differ significantly in the way they are fabricated.(http://en.wikipedia.org/wiki/Capacitor)

* To make C larger, we need (a) high ε, (b) large area, (c) small thickness.

* For a constant capacitance,

Q(t) = C v(t) ,dQ

dt= C

dv

dt, i.e, i(t) = C

dv

dt.

* If v = constant, i = 0, i.e., a capacitor behaves like an open circuit in DCconditions as one would expect from two conducting plates separated by aninsulator.


Capacitors

v

i

Q

Q

insulator

i

conductor

conductor

Unit: Farad (F)

C =ǫ A

tt




Q(t) = C v(t) ,dQ

dt= C

dv

dt, i.e, i(t) = C

dv

dt.



Capacitors

v

i

Q

Q

insulator

i

conductor

conductor

Unit: Farad (F)

C =ǫ A

tt




Q(t) = C v(t) ,dQ

dt= C

dv

dt, i.e, i(t) = C

dv

dt.



Capacitors

v

i

Q

Q

insulator

i

conductor

conductor

Unit: Farad (F)

C =ǫ A

tt




Q(t) = C v(t) ,dQ

dt= C

dv

dt, i.e, i(t) = C

dv

dt.



Capacitors

v

i

Q

Q

insulator

i

conductor

conductor

Unit: Farad (F)

C =ǫ A

tt




Q(t) = C v(t) ,dQ

dt= C

dv

dt, i.e, i(t) = C

dv

dt.



Example

i

v

0

20

−20

i (m

A)

for the given source current.

Plot v, p, and W versus time

Assume v(0) = 0 V, C= 5 mF.

i(t) = Cdv

dt

v(t) =1

C

∫i(t) dt

8

4

0

−4

v (V

)

p(t) = v(t)× i(t)

0.2

0.1

0

−0.1

−0.2

pow

er (

Wat

ts)

W(t) =∫

p(t) dt

0.2

0.1

0ener

gy (

J)

time (sec)

0 1 2 3 4 5 6

W(t) =∫

p(t) dt

= C∫

vdv

dtdt

= C∫

v dv

=1

2C v2


Example

i

v

0

20

−20

i (m

A)




i(t) = Cdv

dt

v(t) =1

C

∫i(t) dt

8

4

0

−4

v (V

)

p(t) = v(t)× i(t)

0.2

0.1

0

−0.1

−0.2

pow

er (

Wat

ts)

W(t) =∫

p(t) dt

0.2

0.1

0ener

gy (

J)

time (sec)

0 1 2 3 4 5 6

W(t) =∫

p(t) dt

= C∫

vdv

dtdt

= C∫

v dv

=1

2C v2


Example

i

v

0

20

−20

i (m

A)




i(t) = Cdv

dt

v(t) =1

C

∫i(t) dt

8

4

0

−4

v (V

)

p(t) = v(t)× i(t)

0.2

0.1

0

−0.1

−0.2

pow

er (

Wat

ts)

W(t) =∫

p(t) dt

0.2

0.1

0ener

gy (

J)

time (sec)

0 1 2 3 4 5 6

W(t) =∫

p(t) dt

= C∫

vdv

dtdt

= C∫

v dv

=1

2C v2


Example

i

v

0

20

−20

i (m

A)




i(t) = Cdv

dt

v(t) =1

C

∫i(t) dt

8

4

0

−4

v (V

)

p(t) = v(t)× i(t)

0.2

0.1

0

−0.1

−0.2

pow

er (

Wat

ts)

W(t) =∫

p(t) dt

0.2

0.1

0ener

gy (

J)

time (sec)

0 1 2 3 4 5 6

W(t) =∫

p(t) dt

= C∫

vdv

dtdt

= C∫

v dv

=1

2C v2


Example

i

v

0

20

−20

i (m

A)




i(t) = Cdv

dt

v(t) =1

C

∫i(t) dt

8

4

0

−4

v (V

)

p(t) = v(t)× i(t)

0.2

0.1

0

−0.1

−0.2

pow

er (

Wat

ts)

W(t) =∫

p(t) dt

0.2

0.1

0ener

gy (

J)

time (sec)

0 1 2 3 4 5 6

W(t) =∫

p(t) dt

= C∫

vdv

dtdt

= C∫

v dv

=1

2C v2


Example

i

v

0

20

−20

i (m

A)




i(t) = Cdv

dt

v(t) =1

C

∫i(t) dt

8

4

0

−4

v (V

)

p(t) = v(t)× i(t)

0.2

0.1

0

−0.1

−0.2

pow

er (

Wat

ts)

W(t) =∫

p(t) dt

0.2

0.1

0ener

gy (

J)

time (sec)

0 1 2 3 4 5 6

W(t) =∫

p(t) dt

= C∫

vdv

dtdt

= C∫

v dv

=1

2C v2


Example

i

v

0

20

−20

i (m

A)




i(t) = Cdv

dt

v(t) =1

C

∫i(t) dt

8

4

0

−4

v (V

)

p(t) = v(t)× i(t)

0.2

0.1

0

−0.1

−0.2

pow

er (

Wat

ts)

W(t) =∫

p(t) dt

0.2

0.1

0ener

gy (

J)

time (sec)

0 1 2 3 4 5 6

W(t) =∫

p(t) dt

= C∫

vdv

dtdt

= C∫

v dv

=1

2C v2


Example

i

v

0

20

−20

i (m

A)




i(t) = Cdv

dt

v(t) =1

C

∫i(t) dt

8

4

0

−4

v (V

)

p(t) = v(t)× i(t)

0.2

0.1

0

−0.1

−0.2

pow

er (

Wat

ts)

W(t) =∫

p(t) dt

0.2

0.1

0ener

gy (

J)

time (sec)

0 1 2 3 4 5 6

W(t) =∫

p(t) dt

= C∫

vdv

dtdt

= C∫

v dv

=1

2C v2


Home work

v

i i (mA)

0 1 2

20

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.


Home work

v

i i (mA)

0 1 2

20

time (sec)




Home work

v

i i (mA)

0 1 2

20

time (sec)




Inductors

core Magnetic field lines

v

i L

Symbol

Units: Henry (H)

* An inductor is basically a conducting coil wound around a “core.”

* V = Ndφ

dt= N

d

dt(B · A) = N

d

dt

»„µN i

l

«A

–.

Compare with v = Ldi

dt.

⇒ L = µN2 A

l= µrµ0 N2 A

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 DCconditions as one would expect from a highly conducting coil.

* Note: B = µH is an approximation. In practice, B may be a nonlinear functionof H, depending on the core material.


Inductors


v

i L

Symbol

Units: Henry (H)


* V = Ndφ

dt= N

d

dt(B · A) = N

d

dt

»„µN i

l

«A

–.


dt.

⇒ L = µN2 A

l= µrµ0 N2 A

l.






Inductors


v

i L

Symbol

Units: Henry (H)


* V = Ndφ

dt= N

d

dt(B · A) = N

d

dt

»„µN i

l

«A

–.


dt.

⇒ L = µN2 A

l= µrµ0 N2 A

l.






Inductors


v

i L

Symbol

Units: Henry (H)


* V = Ndφ

dt= N

d

dt(B · A) = N

d

dt

»„µN i

l

«A

–.


dt.

⇒ L = µN2 A

l= µrµ0 N2 A

l.






Inductors


v

i L

Symbol

Units: Henry (H)


* V = Ndφ

dt= N

d

dt(B · A) = N

d

dt

»„µN i

l

«A

–.


dt.

⇒ L = µN2 A

l= µrµ0 N2 A

l.






Inductors


v

i L

Symbol

Units: Henry (H)


* V = Ndφ

dt= N

d

dt(B · A) = N

d

dt

»„µN i

l

«A

–.


dt.

⇒ L = µN2 A

l= µrµ0 N2 A

l.






Inductors


v

i L

Symbol

Units: Henry (H)


* V = Ndφ

dt= N

d

dt(B · A) = N

d

dt

»„µN i

l

«A

–.


dt.

⇒ L = µN2 A

l= µrµ0 N2 A

l.






RC circuits with DC sources

i

Cv

A

B

Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

i

Cv

A

B

≡ VTh

RTh

* If all sources are DC (constant), VTh = constant .

* KVL: VTh = RTh i + v → VTh = RThCdv

dt+ v .

* Homogeneous solution:

dv

dt+

1

τv = 0 , where τ = RTh C is the “time constant.”

→ v (h) = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making i = 0, and we getv (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 fromknown conditions on v .



i

Cv

A

B


i

Cv

A

B

≡ VTh

RTh



dt+ v .


dv

dt+

1


→ v (h) = K exp(−t/τ) .


* v = v (h) + v (p) = K exp(−t/τ) + VTh .




i

Cv

A

B


i

Cv

A

B

≡ VTh

RTh



dt+ v .


dv

dt+

1


→ v (h) = K exp(−t/τ) .


* v = v (h) + v (p) = K exp(−t/τ) + VTh .




i

Cv

A

B


i

Cv

A

B

≡ VTh

RTh



dt+ v .


dv

dt+

1


→ v (h) = K exp(−t/τ) .


* v = v (h) + v (p) = K exp(−t/τ) + VTh .




i

Cv

A

B


i

Cv

A

B

≡ VTh

RTh



dt+ v .


dv

dt+

1


→ v (h) = K exp(−t/τ) .


* v = v (h) + v (p) = K exp(−t/τ) + VTh .




i

Cv

A

B


i

Cv

A

B

≡ VTh

RTh



dt+ v .


dv

dt+

1


→ v (h) = K exp(−t/τ) .


* v = v (h) + v (p) = K exp(−t/τ) + VTh .




i

Cv

A

B


i

Cv

A

B

≡ VTh

RTh



dt+ v .


dv

dt+

1


→ v (h) = K exp(−t/τ) .


* v = v (h) + v (p) = K exp(−t/τ) + VTh .




i

Cv

A

B


i

Cv

A

B

≡ VTh

RTh



dt+ v .


dv

dt+

1


→ v (h) = K exp(−t/τ) .


* v = v (h) + v (p) = K exp(−t/τ) + VTh .



RC circuits with DC sources (continued)

i i

Cv

Cv

A

B

A

B


≡ VTh

RTh

* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RC .

* i(t) = Cdv

dt= C × A exp(−t/τ)

„− 1

τ

«≡ A′ exp(−t/τ) .

* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since allderivatives vanish.

* Since the circuit in the black box is linear, any variable (current or voltage) inthe circuit can be expressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).



i i

Cv

Cv

A

B

A

B


≡ VTh

RTh


* i(t) = Cdv


„− 1

τ

«≡ A′ exp(−t/τ) .





i i

Cv

Cv

A

B

A

B


≡ VTh

RTh


* i(t) = Cdv


„− 1

τ

«≡ A′ exp(−t/τ) .





i i

Cv

Cv

A

B

A

B


≡ VTh

RTh


* i(t) = Cdv


„− 1

τ

«≡ A′ exp(−t/τ) .




RL circuits with DC sources

v

iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

A

B

v

i

A

B

≡ VTh

RTh


* KVL: VTh = RTh i + Ldi

dt.


di

dt+

1

τi = 0 , where τ = L/RTh

→ i (h) = K exp(−t/τ) .

* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making v = 0, and we geti (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 fromknown conditions on i .



v


A

B

v

i

A

B

≡ VTh

RTh



dt.


di

dt+

1


→ i (h) = K exp(−t/τ) .






v


A

B

v

i

A

B

≡ VTh

RTh



dt.


di

dt+

1


→ i (h) = K exp(−t/τ) .






v


A

B

v

i

A

B

≡ VTh

RTh



dt.


di

dt+

1


→ i (h) = K exp(−t/τ) .






v


A

B

v

i

A

B

≡ VTh

RTh



dt.


di

dt+

1


→ i (h) = K exp(−t/τ) .






v


A

B

v

i

A

B

≡ VTh

RTh



dt.


di

dt+

1


→ i (h) = K exp(−t/τ) .






v


A

B

v

i

A

B

≡ VTh

RTh



dt.


di

dt+

1


→ i (h) = K exp(−t/τ) .






v


A

B

v

i

A

B

≡ VTh

RTh



dt.


di

dt+

1


→ i (h) = K exp(−t/τ) .





RL circuits with DC sources (continued)

vv

i iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)

A

B

A

B

≡ VTh

RTh

* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/R .

* v(t) = Ldi

dt= L× A exp(−t/τ)

„− 1

τ

«≡ A′ exp(−t/τ) .

* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since allderivatives vanish.




vv


A

B

A

B

≡ VTh

RTh


* v(t) = Ldi


„− 1

τ

«≡ A′ exp(−t/τ) .





vv


A

B

A

B

≡ VTh

RTh


* v(t) = Ldi


„− 1

τ

«≡ A′ exp(−t/τ) .





vv


A

B

A

B

≡ VTh

RTh


* v(t) = Ldi


„− 1

τ

«≡ A′ exp(−t/τ) .




RC circuits: Can Vc change “suddenly?”

i

t0 V

5 V

C = 1 µFVc

Vs

Vs

R = 1 k

Vc(0)=0 V

* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of thischange, Vc will rise. How fast can Vc change?

* For example, what would happen if Vc changes by 1 V in 1 µs at a constantrate of 1 V /1 µs = 106 V /s?

* i = CdVc

dt= 1 µF × 106 V

s= 1 A .

* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.

* We conclude that Vc (0+) = Vc (0−)⇒ A capacitor does not allow abruptchanges in Vc if there is a finite resistance in the circuit.

* Similarly, an inductor does not allow abrupt changes in iL.



i

t0 V

5 V

C = 1 µFVc

Vs

Vs

R = 1 k

Vc(0)=0 V



* i = CdVc

dt= 1 µF × 106 V

s= 1 A .






i

t0 V

5 V

C = 1 µFVc

Vs

Vs

R = 1 k

Vc(0)=0 V



* i = CdVc

dt= 1 µF × 106 V

s= 1 A .






i

t0 V

5 V

C = 1 µFVc

Vs

Vs

R = 1 k

Vc(0)=0 V



* i = CdVc

dt= 1 µF × 106 V

s= 1 A .






i

t0 V

5 V

C = 1 µFVc

Vs

Vs

R = 1 k

Vc(0)=0 V



* i = CdVc

dt= 1 µF × 106 V

s= 1 A .






i

t0 V

5 V

C = 1 µFVc

Vs

Vs

R = 1 k

Vc(0)=0 V



* i = CdVc

dt= 1 µF × 106 V

s= 1 A .






i

t0 V

5 V

C = 1 µFVc

Vs

Vs

R = 1 k

Vc(0)=0 V



* i = CdVc

dt= 1 µF × 106 V

s= 1 A .





Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant.For example,

(1) t < t1

(3) t > t2

(2) t1 < t < t2

0 t2t1

Vs

* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , 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.

⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .

(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,

Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .

* Compute A1, B1, · · · using the conditions on x(t).




(1) t < t1

(3) t > t2

(2) t1 < t < t2

0 t2t1

Vs




⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .


Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .





(1) t < t1

(3) t > t2

(2) t1 < t < t2

0 t2t1

Vs




⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .


Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .





(1) t < t1

(3) t > t2

(2) t1 < t < t2

0 t2t1

Vs




⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .


Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .





(1) t < t1

(3) t > t2

(2) t1 < t < t2

0 t2t1

Vs




⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .


Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .





(1) t < t1

(3) t > t2

(2) t1 < t < t2

0 t2t1

Vs




⇒ iC = CdVc

dt= 0 , and VL = L

diL

dt= 0 .


Vc (t+0 ) = Vc (t−0 ) , and iL(t+

0 ) = iL(t−0 ) .



RC circuits: charging and discharging transients

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

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.

As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., A = V0 , B = −V0

t = 0+: 0 = A + B ,

t→∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):



v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V


t = 0+: V0 = A + B ,

i.e., A = V0 , B = 0

t→∞: 0 = B .

v(t) = V0 exp(−t/τ)



i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V



As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0


i.e., A = V0 , B = −V0

t = 0+: 0 = A + B ,

t→∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):



v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V


t = 0+: V0 = A + B ,

i.e., A = V0 , B = 0

t→∞: 0 = B .




i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V



As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0


i.e., A = V0 , B = −V0

t = 0+: 0 = A + B ,

t→∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):



v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V


t = 0+: V0 = A + B ,

i.e., A = V0 , B = 0

t→∞: 0 = B .




i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V



As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0


i.e., A = V0 , B = −V0

t = 0+: 0 = A + B ,

t→∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):



v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V


t = 0+: V0 = A + B ,

i.e., A = V0 , B = 0

t→∞: 0 = B .




i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V



As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0


i.e., A = V0 , B = −V0

t = 0+: 0 = A + B ,

t→∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):



v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V


t = 0+: V0 = A + B ,

i.e., A = V0 , B = 0

t→∞: 0 = B .




i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V



As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0


i.e., A = V0 , B = −V0

t = 0+: 0 = A + B ,

t→∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):



v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V


t = 0+: V0 = A + B ,

i.e., A = V0 , B = 0

t→∞: 0 = B .




i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V



As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0


i.e., A = V0 , B = −V0

t = 0+: 0 = A + B ,

t→∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):



v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V


t = 0+: V0 = A + B ,

i.e., A = V0 , B = 0

t→∞: 0 = B .




i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V



As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0


i.e., A = V0 , B = −V0

t = 0+: 0 = A + B ,

t→∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):



v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V


t = 0+: V0 = A + B ,

i.e., A = V0 , B = 0

t→∞: 0 = B .




i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V



As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0


i.e., A = V0 , B = −V0

t = 0+: 0 = A + B ,

t→∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):



v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V


t = 0+: V0 = A + B ,

i.e., A = V0 , B = 0

t→∞: 0 = B .




i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):

v(0−) = Vs(0−) = 0 V

v(0+) ≃ v(0−) = 0 V



As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0


i.e., A = V0 , B = −V0

t = 0+: 0 = A + B ,

t→∞: V0 = B .

v(t) = V0 [1− exp(−t/τ)]

i

t0 V

Cv

R

Vs

Vs

V0

(A)Let v(t) = A exp(−t/τ) + B, t > 0

(1)

(2)

Conditions on v(t):



v(0−) = Vs(0−) = V0

v(0+) ≃ v(0−) = V0

As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V


t = 0+: V0 = A + B ,

i.e., A = V0 , B = 0

t→∞: 0 = B .




i

t0 V

Cv

R

Vs

Vs

V0

Compute i(t), t > 0 .

(A) i(t) = Cd

dtV0 [1− exp(−t/τ)]

=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)

Using these conditions, we obtain

(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 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0

τexp(−t/τ) = −V0

Rexp(−t/τ)


(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t→∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)



i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd


=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)


(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 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0


Rexp(−t/τ)


(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t→∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)



i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd


=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)


(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 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0


Rexp(−t/τ)


(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t→∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)



i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd


=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)


(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 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0


Rexp(−t/τ)


(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t→∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)



i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd


=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)


(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 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0


Rexp(−t/τ)


(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t→∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)



i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd


=CV0

τexp(−t/τ) =

V0

Rexp(−t/τ)


(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 .

A′ =V0

R, B′ = 0 ⇒ i(t) =

V0

Rexp(−t/τ)

i

t0 V

Cv

R

Vs

Vs

V0


(A) i(t) = Cd

dtV0 [exp(−t/τ)]

= −CV0


Rexp(−t/τ)


(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .

t→∞: i(t) = 0 .

t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .

A′ = −V0

R, B′ = 0 ⇒ i(t) = −V0

Rexp(−t/τ)



i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (V

olts

) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V


i(t) = −V0

Rexp(−t/τ)

v (V

olts

)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8



i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (V

olts

) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V


i(t) = −V0

Rexp(−t/τ)

v (V

olts

)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8



i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (V

olts

) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V


i(t) = −V0

Rexp(−t/τ)

v (V

olts

)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8



i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (V

olts

) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V


i(t) = −V0

Rexp(−t/τ)

v (V

olts

)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8



i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (V

olts

) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V


i(t) = −V0

Rexp(−t/τ)

v (V

olts

)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8



i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V

v(t) = V0 [1− exp(−t/τ)]

i(t) =V0

Rexp(−t/τ)

0

5

v (V

olts

) v

Vs

5

0

i (m

A)

time (msec)

−2 0 2 4 6 8

i

t0 V

C = 1 µFvVs

R = 1 kVs

5 V


i(t) = −V0

Rexp(−t/τ)

v (V

olts

)

5

0

v

Vs

i (m

A)

0

−5

time (msec)

−2 0 2 4 6 8


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) processis almost complete.

0

1

x

0 1 2 3 4 5 6

exp(−x)

1− exp(−x)



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.


0

1

x

0 1 2 3 4 5 6

exp(−x)

1− exp(−x)



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.


0

1

x

0 1 2 3 4 5 6

exp(−x)

1− exp(−x)



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.


0

1

x

0 1 2 3 4 5 6

exp(−x)

1− exp(−x)



i i

time (msec) time (msec)

t0 V t0 V

0

5

v (V

olts

)

v (V

olts

)

5

0

−1 0 1 2 3 4 5 6 −1 0 1 2 3 4 5 6

C = 1 µF C = 1 µFv vVs Vs

R RVs Vs

5 V 5 V

R = 1 kΩ

R = 100 Ω

R = 1 kΩ

R = 100 Ω

v(t) = V0 exp(−t/τ)v(t) = V0 [1− exp(−t/τ)]


RL circuit: example

i

v

t

10 V

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s

i(0) = 0 A, Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1

There are three intervals of constant Vs:

R2

R1

Vs

RTh seen by L is the same in all intervals:

τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8 Ω

⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .

If Vs did not change at t = t1,

we would have

t1t0

Vs

i(t), t > 0 (See next slide).

Using i(t+0 ) and i(∞), we can obtain


RL circuit: example

i

v

t

10 V

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s

i(0) = 0 A, Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1


R2

R1

Vs


τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8 Ω

⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .


we would have

t1t0

Vs




RL circuit: example

i

v

t

10 V

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s

i(0) = 0 A, Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1


R2

R1

Vs


τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8 Ω

⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .


we would have

t1t0

Vs




RL circuit: example

i

v

t

10 V

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s

i(0) = 0 A, Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1


R2

R1

Vs


τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8 Ω

⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .


we would have

t1t0

Vs




RL circuit: example

i

v

t

10 V

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s

i(0) = 0 A, Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1


R2

R1

Vs


τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8 Ω

⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .


we would have

t1t0

Vs




RL circuit: example

i

v

t

10 V

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s

i(0) = 0 A, Find i(t).

(1) t < t0

(2) t0 < t < t1

(3) t > t1


R2

R1

Vs


τ = L/RTh

= 0.1 s

= 0.8 H/8Ω

RTh = R1 ‖ R2 = 8 Ω

⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .

At t = t−0 , v = 0 V, Vs = 0 V .

10 V

t

v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .


we would have

t1t0

Vs




RL circuit: example

i

v

t

10 V

i (A

mp)

1

0

0time (sec)

0.2 0.4 0.6 0.8

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s

and we need to work out the

solution for t > t1 separately.

In reality, Vs changes at t = t1,

Consider t > t1.

For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.

i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.

Let i(t) = A exp(−t/τ) + B.

It is convenient to rewrite i(t) as

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.


RL circuit: example

i

v

t

10 V

i (A

mp)

1

0

0time (sec)

0.2 0.4 0.6 0.8

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s




Consider t > t1.


i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.



i(t) = A′ exp[−(t− t1)/τ ] + B.


i(t) = 0.693 exp[−(t− t1)/τ ] A.


RL circuit: example

i

v

t

10 V

i (A

mp)

1

0

0time (sec)

0.2 0.4 0.6 0.8

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s




Consider t > t1.


i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.



i(t) = A′ exp[−(t− t1)/τ ] + B.


i(t) = 0.693 exp[−(t− t1)/τ ] A.


RL circuit: example

i

v

t

10 V

i (A

mp)

1

0

0time (sec)

0.2 0.4 0.6 0.8

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s




Consider t > t1.


i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.



i(t) = A′ exp[−(t− t1)/τ ] + B.


i(t) = 0.693 exp[−(t− t1)/τ ] A.


RL circuit: example

i

v

t

10 V

i (A

mp)

1

0

0time (sec)

0.2 0.4 0.6 0.8

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s




Consider t > t1.


i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.



i(t) = A′ exp[−(t− t1)/τ ] + B.


i(t) = 0.693 exp[−(t− t1)/τ ] A.


RL circuit: example

i

v

t

10 V

i (A

mp)

1

0

0time (sec)

0.2 0.4 0.6 0.8

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s




Consider t > t1.


i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.



i(t) = A′ exp[−(t− t1)/τ ] + B.


i(t) = 0.693 exp[−(t− t1)/τ ] A.


RL circuit: example

i

v

t

10 V

i (A

mp)

1

0

0time (sec)

0.2 0.4 0.6 0.8

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s




Consider t > t1.


i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.



i(t) = A′ exp[−(t− t1)/τ ] + B.


i(t) = 0.693 exp[−(t− t1)/τ ] A.


RL circuit: example

i

v

t

10 V

i (A

mp)

1

0

0 0.2 0.4 0.6 0.8time (sec)

i(t) = 0.693 exp[−(t− t1)/τ ] A.

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s

(SEQUEL file: ee101_rl1.sqproj)

0

1

0 0.2 0.4 0.6 0.8

i (A

mp)

time (sec)

Combining the solutions for t0 < t < t1 and t > t1,

we get


RL circuit: example

i

v

t

10 V

i (A

mp)

1

0

0 0.2 0.4 0.6 0.8time (sec)

i(t) = 0.693 exp[−(t− t1)/τ ] A.

t1t0

R2

R1

Vs

VsR1 = 10 Ω

R2 = 40 Ω

L = 0.8 H

t0 = 0

t1 = 0.1 s

(SEQUEL file: ee101_rl1.sqproj)

0

1

0 0.2 0.4 0.6 0.8

i (A

mp)

time (sec)

Combining the solutions for t0 < t < t1 and t > t1,

we get


RC circuit: example

1 k

t=0i

6 V

5 kR1 5 µF

ic

vc

R3 = 5 k

R2

AND

i i

5 k 1 k

6 V

5 k 5 k

5 k

5 µF

ic

t < 0

5 µF

ic

vc

t > 0

vc

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) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.

i(t) = 0.5 exp(-t/τ) mA.

Using i(0+) and i(∞) = 0 A, we get

0

1

i (mA)

time (sec) 0 0.5

0

−0.50

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)

(SEQUEL file: ee101_rc2.sqproj)


RC circuit: example

1 k

t=0i

6 V

5 kR1 5 µF

ic

vc

R3 = 5 k

R2 AND

i i

5 k 1 k

6 V

5 k 5 k

5 k

5 µF

ic

t < 0

5 µF

ic

vc

t > 0

vc


vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0

+) = 5 V.

⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.


i(t) = 0.5 exp(-t/τ) mA.


0

1

i (mA)

time (sec) 0 0.5

0

−0.50

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)



RC circuit: example

1 k

t=0i

6 V

5 kR1 5 µF

ic

vc

R3 = 5 k

R2 AND

i i

5 k 1 k

6 V

5 k 5 k

5 k

5 µF

ic

t < 0

5 µF

ic

vc

t > 0

vc


vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0

+) = 5 V.

⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.


i(t) = 0.5 exp(-t/τ) mA.


0

1

i (mA)

time (sec) 0 0.5

0

−0.50

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)



RC circuit: example

1 k

t=0i

6 V

5 kR1 5 µF

ic

vc

R3 = 5 k

R2 AND

i i

5 k 1 k

6 V

5 k 5 k

5 k

5 µF

ic

t < 0

5 µF

ic

vc

t > 0

vc


vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0

+) = 5 V.

⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.


i(t) = 0.5 exp(-t/τ) mA.


0

1

i (mA)

time (sec) 0 0.5

0

−0.50

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)



RC circuit: example

1 k

t=0i

6 V

5 kR1 5 µF

ic

vc

R3 = 5 k

R2 AND

i i

5 k 1 k

6 V

5 k 5 k

5 k

5 µF

ic

t < 0

5 µF

ic

vc

t > 0

vc


vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0

+) = 5 V.

⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.


i(t) = 0.5 exp(-t/τ) mA.


0

1

i (mA)

time (sec) 0 0.5

0

−0.50

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)



RC circuit: example

1 k

t=0i

6 V

5 kR1 5 µF

ic

vc

R3 = 5 k

R2 AND

i i

5 k 1 k

6 V

5 k 5 k

5 k

5 µF

ic

t < 0

5 µF

ic

vc

t > 0

vc


vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0

+) = 5 V.

⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.


i(t) = 0.5 exp(-t/τ) mA.


0

1

i (mA)

time (sec) 0 0.5

0

−0.50

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)



RC circuit: example

1 k

t=0i

6 V

5 kR1 5 µF

ic

vc

R3 = 5 k

R2 AND

i i

5 k 1 k

6 V

5 k 5 k

5 k

5 µF

ic

t < 0

5 µF

ic

vc

t > 0

vc


vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0

+) = 5 V.

⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.


i(t) = 0.5 exp(-t/τ) mA.


0

1

i (mA)

time (sec) 0 0.5

0

−0.50

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)



RC circuit: example

1 k

t=0i

6 V

5 kR1 5 µF

ic

vc

R3 = 5 k

R2 AND

i i

5 k 1 k

6 V

5 k 5 k

5 k

5 µF

ic

t < 0

5 µF

ic

vc

t > 0

vc


vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0

+) = 5 V.

⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.


i(t) = 0.5 exp(-t/τ) mA.


0

1

i (mA)

time (sec) 0 0.5

0

−0.50

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)



RC circuit: example

1 k

t=0i

6 V

5 kR1 5 µF

ic

vc

R3 = 5 k

R2 AND

i i

5 k 1 k

6 V

5 k 5 k

5 k

5 µF

ic

t < 0

5 µF

ic

vc

t > 0

vc


vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0

+) = 5 V.

⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.


i(t) = 0.5 exp(-t/τ) mA.


0

1

i (mA)

time (sec) 0 0.5

0

−0.50

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)



RC circuit: example

1 k

t=0i

6 V

5 kR1 5 µF

ic

vc

R3 = 5 k

R2 AND

i i

5 k 1 k

6 V

5 k 5 k

5 k

5 µF

ic

t < 0

5 µF

ic

vc

t > 0

vc


vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0

+) = 5 V.

⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.


i(t) = 0.5 exp(-t/τ) mA.


0

1

i (mA)

time (sec) 0 0.5

0

−0.50

5

time (sec)

time (sec)

0 0.5 0 0.5

ic (mA)vc (V)



RC circuits: home work

i1i2

vc

ic

10 Ω

10 Ω

200 µF10V

* 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 finalconditions for each of them (i1(0+) and i1(∞), etc.) and then using them tocompute the coefficients in the general expression,x(t) = A exp(−t/τ) + B.

* Verify your results with SEQUEL (file: ee101 rc3.sqproj).



i1i2

vc

ic

10 Ω

10 Ω

200 µF10V






i1i2

vc

ic

10 Ω

10 Ω

200 µF10V






i1i2

vc

ic

10 Ω

10 Ω

200 µF10V






t=0 i1

vx 1 mF5Ω vc

ic

2 Ω 3 Ω

0.1 vx24V

* Find vc (0−), vc (∞).

* Find RTh as seen by the capacitor for t > 0.

* Solve for vc (t) and i1(t), t > 0.




t=0 i1

vx 1 mF5Ω vc

ic

2 Ω 3 Ω

0.1 vx24V

* Find vc (0−), vc (∞).






t=0 i1

vx 1 mF5Ω vc

ic

2 Ω 3 Ω

0.1 vx24V

* Find vc (0−), vc (∞).






t=0 i1

vx 1 mF5Ω vc

ic

2 Ω 3 Ω

0.1 vx24V

* Find vc (0−), vc (∞).






t=0 i1

vx 1 mF5Ω vc

ic

2 Ω 3 Ω

0.1 vx24V

* Find vc (0−), vc (∞).





RL circuits: home work

t=0 i

10 VL=0.1 H5 V

20Ω20Ω

* 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).



t=0 i

10 VL=0.1 H5 V

20Ω20Ω

* Find i(0−), i(∞).






t=0 i

10 VL=0.1 H5 V

20Ω20Ω

* Find i(0−), i(∞).






t=0 i

10 VL=0.1 H5 V

20Ω20Ω

* Find i(0−), i(∞).






t=0 i

10 VL=0.1 H5 V

20Ω20Ω

* Find i(0−), i(∞).





Documents

EE101: RC and RL Circuits (with DC sources)