is this the write forumula

[jenko]

100v supply, value 20µf

charge formula = C=Q/V = Capacitence = charge/voltage

Capacitence = 20µf/100 = 0.2?

is this correct?



is energy the same as stored energy ?

energy stored: 1/2CV² = 1/2 Q²/C = 1/2 VQ

is this the correct formula is there an easier lay out i dont under stand how to work this out? :S

any help would be app!

Added after 8 minutes:

is this correct :

w=1/2 (vxc)
100*0.020=2
2x1/2=1watt ===== 1 watt of energy? is this correct?

 

[MSRA]

100v supply, value 20µf

charge formula = C=Q/V = Capacitence = charge/voltage

Capacitence = 20µf/100 = 0.2?

is this correct?

20uf is capacitance ...not the charge....u put it under charge.....if it is charge it would be 20uC......

 

[Kral]

jenko,
MSRA's answer to the first part of your question is correct. Energy is the same as stored energy.
The most straighforward way to calculate energy is to use E = 1/2(CV^2).
.
The equation E = 1/2(Q^2/C) is correct.
The equation E = 1/2VQ is correct.
.
E = 1/2(CV^2) = 1/2(2E-5) X 100^2 =
1/2(2E-5 X 1E4) = (2E-1)/2 = .1 Joules.
.
Note that the unit of energy is Joules, not Watts. Power (Watts) is the rate of flow of energy wrt time. It is the same as Joules/second.
Regards,
Kral

 

[laktronics]

Hi,
There are a few set of equations used to solve problems regarding capacitors, there is no shortcut, you have to know them to solve the problems. These are :

1. Q [ charge in Coulomb] = V [Voltage in Volts] x C [Capacitance in Farads]

2. Energy Stored in a Capacitor

E [in Joules] = 1/2 x C [Farads] x V^2 [ Volts squared]

3. When Capacitors are paralleled,
- Total capacitance Ctp = Sum of the individual capacitor values.

Ctp [ in Farads] = C1[Farads] + C2 [Farads] + C3 [Farads]

- The Voltage across each capacitor will be same and equal to applied
Voltage Vp.

Vp = V1 = V2 = V3

- The charge on each capacitor will be different, higher capacitor has more
charge than lower capacitor. The total charge Qtp = Q1 + Q2 + Q3 and is
equal to the charge supplied by the source.

4. When Capacitrs are connected in series,

- the total capacotance Cts will be smaller than the smallest in the circuit and
Cts is calculated from

1/Cts = 1/C1 +1/C2 + 1/C3

- The voltage across each capacitor will be different, the smaller capacitor will
have more voltage than the larger ones. The applied voltage Vts is
distributed across all series capacitors such that

Vts = V1 + V2 + V3 ; all in Volts

- The Charge [Coulomb] on each Capacitor will be same and equal to the
total Charge Qts drawn from the supply.

Qts = Q1 = Q2 =Q3.

Eventhough, the basic units of charge is Coulomb and that of Capacitor is Farad,in practical circuits fractional units of MicroFarad, NanoFarad and PicoFarad are commonly used for Capacitors and MicroCoulomb and smaller fractional units are used for Charge.

You should always specify the correct units for various variables used in Physics. It is also a good practice to note down commonly used equations in a reference book as you go on studying, which will be handy while solving your home work.

Regards,
Laktronics

 

[subharpe]

Hi I think you should be a little careful about the diensions and the units.

 

[adeel.sid]

Formula is rite but u are using it in wrong way!!
C-satnds for "capacitance".Its unit is "Farad(f)"
Q-stands for charge.its unit is "Coulomb(C)
V-Potential Difference acroee Capacitor.Its unit is Volt(V)

 

[avi_e-]

Hi Friend,

I suggest to follow Latronics.

Regards,
Avinash.S.

 

[smith123]

Energy stored: 1/2CV²

 

[gabin]

hi
the formula's for energy are correct.
You should be little careful about the units

 

[Asif2008]

yes dear the formula is C=Q/V