INC 112 Basic Circuit Analysis

Week 8RC Circuits

RC Circuits

AC

+

-

u(t)

i(t)

RC

The response of RC circuits can be categorized into two parts:

• Transient Response• Forced Response

Transient response comes from the dynamic of R,C.Forced response comes from the voltage source.

Source-Free RC Circuitsi(t)

R C

+

v

-

+

v

-

Capacitor has some energy stored so thatThe initial voltage at t=0 is V0

0)0( Vv Initial condition

Find i(t) from R, C

0)()(

)()(

RCtv

dttdv

Rtv

dttdvC

0)()( ti

LR

dttdi

tLR

eIti

0)(

Compare with the solution of RL circuits.The solution of RC circuits can be obtained with the same method.

Source-free RL Source-free RC

0)(1)( tv

RCdttdv

tRCeVtv1

0)(

tRCeVtv1

0)(

t

v(t)

V0

tRCe

RV

Rtvti

10)()(

or

tRCe

RV

dttdvCti

10)()(

t

i(t)

V0/R

Time Constant

The product RC is time constant for RC circuits

RC Unit: second

tt

RC eVeVtv

0

1

0)(

Forced RC Circuitst=0

R

V C+

vc(t)-

i(t)+ vR(t) -

C has an initial voltage of 0

fromdttdvCti c

c)()( dtti

Ctv cc )(1)(

Use KVL, we got

dttiC

RtiV

tvtvV

c

cR

)(1)(

)()(

dttiC

RtiV c )(1)(From

Differentiate both sides

0)(1)(

)(1)(0

tiRCdt

tdi

tiCdt

tdiR

tRCeIti1

0)(

Solve first-order differential equation

Where I0 is the initial current of the circuit

C has an initial voltage = 0,

But from KVL,

therefore, and

So,

0)0( cv

)()( tvtvV cR

VvR )0(RVi )0(

tRCe

RVti

1

)(

tRC

R VeRtitv1

)()(

)1()()(11 tRC

tRC

RC eVVeVtvVtv

Force Response

Natural Response

t

i(t)

V/R

t

vR(t)

V

tRCe

RVti

1

)(

tRC

R Vetv1

)(

t

vC(t)

V

)1()(1 tRC

C eVtv

Note: Capacitor’s voltagecannot abruptly change

How to Solve Problems? (RC)• Start by finding the voltage of the capacitor first

• Assume the response that we want to find is in form of

t

ekk

21

• Find the time constant τ (may use Thevenin’s)

• Solve for k1, k2 using initial conditions and status at the stable point

• From the voltage, find other values that the problem ask using KCL, KVL

Examplet=0

1M

5V 1uF+

vc(t)-

i(t)

1M

1V

Switch open for a long time before t=0, find and sketch i(t)

First, we start by finding vc(t)

The initial condition of C is vc(0) = 1V

The stable condition of C is vc(∞) = 3V

Assume vc(t) in form of t

c ekktv

21)(

Find the time constant after t=0 by Thevenin’s, viewing C as a load

KMMReq 5001||1

Therefore, the time constant is

sec5.01500 FKRC

Find k1, k2 using vc(0) = 1, vc(∞) = 3

At t=0, vc(0) = 1 V 211 kk

At t = ∞, vc(∞) = 3 V 03 1 k

t

C ekktv

21)(

Therefore, k1=3, k2 = -2

We can find i(t) by using Ohm’s law on the resistor

tC etv 223)(

AeAMe

Me

Mtvti

tt

tc

22

2

22122

1)23(5

1)(5)(

t

i(t)

4A

2A

Aeti t222)(

Example

t=01K

5V

3uF+

vc(t)-

i(t)

1K

1K

The switch was opened for a long time before t=0, Find i(t)

Start with vc(t)

The initial condition of C is vc(0) = 5V

The final stable condition of C comes from voltage divider,which is vc(∞) = 5*(1/1+0.5) = 3.33V

Assume vc(t) in form of t

c ekktv

21)(

Find the time constant after t=0 by Thevenin’s, viewing C as a load

33.3331||1||1 KKKReq

Therefore, the time constant is

sec1333.333 mFRC

Find k1, k2 using vc(0) = 5, vc(∞) = 3.33

At t=0, vc(0) = 5 V 215 kk

At t = ∞, vc(∞) = 3.33 V 033.3 1 k

t

C ekktv

21)(

Therefore, k1=3.33, k2 = 1.66

We can find i(t) by using Ohm’s law on the resistor

tC etv 100066.133.3)(

mAeAKe

Ke

Ktvti

tt

tc

10001000

1000

66.166.1166.166.1

1)66.133.3(5

1)(5)(

t

i(t)

1.66mA

mAeti t100066.166.1)(

t

vc(t)

tC etv 100066.133.3)(

5V

3.33V