Energy stored in capacitor

V = VB

V = VB

+ Q

- Q

Stored charge

V = VB

V = VB

+ Q

- Q

10m

A R

Stored charge produces the current

V = VB

V = VB

+ Q

- Q

V = VB

V = VB

+ Q

- Q

V = VB

V = VB

+ Q

- Q

Charging capacitor

Energy stored in capacitor

Electric Energy = Charge x Voltage: W = Q × V

This formula would be true for capacitors, had the

charge between plates been transferred at a constant voltage.

For capacitors, Q = C × V

As the charge Q on the plate increases,

the voltage V increases too (and vice versa).

Suppose the capacitor was charged to the voltage V0:

V

Q

V0

Q0

WC

The stored energy is the area under

the Q – V line.0 0

2C

V QW =

Note that, Q0 = C V0: 20

2C

CVW =

V0 is the voltage on the charged capacitor

Example problem 1

0

of

5

The capacitor of 1 mF has been charged to 100 V.

What energy is stored in the capacitor in Joules?

Timed response

120120

20

2C

CVW =

Commercial capacitors

0d AC

d

ε ε⋅=

20 mF capacitor

Capacitor bank (series-parallel)

V = VBV = VB

C1 C2

Parallel connection of capacitors

Capacitors C1 and C2 are connected in parallel: both terminal of each

capacitor are connected to the same wires.

The voltage on each of the capacitors is the same, VB

The charge on the capacitor C1, Q1 = C1×VB

The charge on the capacitor C2, Q2 = C2×VB

Total charge stored in both capacitors:

QTot = Q1 + Q2 = (C1 + C2) × VB

The equivalent capacitance, Ceq = QTot/ VB = C1 + C2

Cpar= C1 + C2

Q1 = C1 V1

Q2 = C2 V2

V = VBV = VB

C1 C2

Parallel connection of capacitors

We can also find the equivalent capacitance from the KCL.

For the 1st and 2nd capacitors, ;; 2211

t

VCI

t

VCI

∂

∂=

∂

∂=

IT I1 I2

According to the KCL, the total current IT = I1 + I2.

Substituting the values for I1 and I2:

( ) ;t

VC

t

VCC

t

VC

t

VCI parT

∂

∂=

∂

∂+=

∂

∂+

∂

∂= 2121

Cpar= C1 + C2

V = VBV = VB

C1 C2

V1 V2

Series connection of capacitors

Capacitors C1 and C2 are connected in series:

the charge Q on C1 and C2 is the same; For each of the capacitors,

From the KVL, VTot= V1 + V2.

1 1 2 2; ;Q C V Q C V= =

The total charge, on the C1 – C2 combination is still Q.

The equivalent capacitance is defined as: Q = CEqS VTot

1/CSer= 1/C1 + 1/C2

+=

21

21

CC

CCCS

12 1

2

CV V

C=

11 2 1 1 1 1

2

( ) ; ;EqS EqS

CQ C V V C V V Q C V

C

= + = + =

11

2

1 ;EqS

CC C

C

+ =

Example problem 2

0

of

5

Three capacitors 2 mF each are connected in parallel.What is the total capacitance in mF?

Timed response

120120

Example problem 3

0

of

5

Three capacitors 2 mF each are connected in series.What is the total capacitance in mF?

Timed response

120120

Example problem 4

0

of

5

Parallel-plate capacitor has a capacitance of 9 nF.A thin metal plate has been inserted in the middle between the top and bottom plates.What is the capacitance (in nF) of the capacitor now?

Timed response

120120

0d AC

d

ε ε=

td

VdCI C

C ×=

R

VI R

R =

C

R

Series R-C circuit

VB

Transients in R-C circuit

The first moment after closing the switch, the voltage across the capacitor = 0;

The capacitor behaves as a short-circuit;

The current at t=0, I0 = VB/R;

After all the transients are over (t Yh) , I = 0

0|

CC

d V V VI C C

dt dt

+ −

→

−= × = ×

Commutation rule for capacitors

C

RVB

Consider a capacitor right before and right

after commutation in an arbitrary circuit.

The capacitor voltage (charge) does not

have to be zero before the commutation.

VC

timeCommutation event

V_V+

If VC changes instantaneously after the commutation, the current in the

connected circuit would be infinitely high:

If V+ is different from V- when dt 0,

then IC ∞

The capacitor voltage does not change after commutation: VC- = VC+

Graphs showing the current and

voltage for a capacitor charging

( )t

B RCV

I t eR

−

=

( ) 1t

RCC BV t V e

− = × −

When t = 3 ×τRC,

VC = 0.95VB ;

τRC = R×C

Cap

aci

tor

volt

age

Graphs showing the current and

voltage for a capacitor discharging

Capacitor starting voltage is VB

( )t

B RCV

I t eR

−

=

( )t

RCC BV t V e

−

=

When t = 3 τRC,

VC = 0.05VB=5% VB

τRC = R×C

General formula for step response of an arbitrary R-C circuit

C

R

R-C circuit

VS

( ) /

0

t

C CF C CFv v v v eτ−= + − τ = RC

R is the total resistance connected to the capacitor after commutation

(al the sources are zeroed to find the equivalent total resistance)

VC0 is the capacitor voltage right after (or right before) the commutation;

VCF is the capacitor voltage long time after all the transient processes are over.

C=0.5 mF

R=10 k

VB=4.5 V

S1

S2

Example 1

Delayed alarm circuit

Sensor switch: S1.

Electronic switch S2 triggers the alarm system when the voltage

across it exceeds the preset threshold value VT.

R = 10k; C = 0.5 mF; VB = 4.5 V.

Assume the S2 resistance infinitely high.

The required time delay between the switch S1 turn-on and triggering

switch S2 must be tt = 3s.

What threshold voltage VT must the switch S2 be tuned to?

C=0.5 mF

R=10 k

VB=4.5 V

S1

S2

Example 1

Delayed alarm circuit

( ) /

0

t

C CF C CFv v v v eτ−= + −

VC0 = 0; VCF = VB; τ = R×C;

/t

C B Bv v v eτ−= − The required alarm triggering time: tt=3s

The required switch threshold voltage vSW = vc(tr):

3 /(10 3 0.5 3) 0.64.5 4.5 4.5 4.5 2e e

swv e e V− × − −= − = − × =

Example 2

The switch in the circuit shown in Fig. 7.25 has been in position a for a long time. At t = 0 the switch is moved to position b.What is the vC time dependence at t>0?

( ) /

0

t

C CF C CFv v v v eτ−= + −

vC0 = V60Ω = −40V *60/(60+20) = -30 V

/ 0.290 120 t

Cv e V−= −

vF = 90 V; τ = RC = 400 kΩ ∗ 0.5µF = 0.2 s.