# Resistors in front of capacitors

1. May 5, 2010

### vjk2

Most arrangements I've seen have a power source hooked up to a resistor hooked up to a capacitor. Like this

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html

Why? I mean, why can't we directly hook the capacitor to the emf?

sounds stupid, I know.

Also, resistors follow Ohms, V=IR in their consumption of current and voltage. How does a capacitor consume current and voltage?

2. May 5, 2010

### sophiecentaur

You can.
But every 'emf' source will have a finite source resistance. The C will charge up v quickly but not instntly. RC is still the time constant.
The current flowing into a C depends upon the value of the series R. The Charge on the capacitor is related to the Voltage across it
Q = CV
where C is the capacity
The rate of charging (the current) depends upon the value of and voltage across the series resistor - hence the exponential shape of a charge or discharge curve.

3. May 17, 2010

### karthicknar

A capacitor will not consume power (product of voltage and current) instead it stores energy in electrostatic form. If a capacitor is directly connected to a voltage source, it is charged up instantaneously (assuming ideal voltage source and capacitor). The current flow will be a spike at t=0, and voltage across c will be v at t=0.

4. May 17, 2010

### sophiecentaur

That particular model will radiate an equal amount of energy to that which is stored 'in' the capacitor. I say that because there will be half C Vsquared stored in the energy but the emf will be V all the time. That would mean that, for conservation of Energy, the rest of the physical circuit must act as an antenna, appearing in series with the C to account for the other half of the energy supplied by the emf. There will be a burst of 'ringing' at some RF frequency, during which the energy will be radiated / dissipated into space.
It's another of those 'Irresistible Force and Immovable Object' scenarios for trapping the unwary student.

5. May 17, 2010

### n.karthick

Ohh I didn't get it. To my small knowledge, once the capacitor charges to v, the current flow stops from the source and hence the power. So law of conservation of energy wont come in to picture here. There is no compulsion for capacitor to radiate energy or the remaining circuit to act as antenna. The formula half c v squared just indicates energy stored in capacitor and there is no another half to be radiated in to space. Clarify me if i am wrong.

6. May 17, 2010

### Integral

Staff Emeritus
You have shown a series RC circuit, the cap IS connected to the battery. It does not make any difference which order the components are in.

As for resistance, for caps and inductors we speak of the Impedance. The impedance is dependent upon the frequency.

7. May 17, 2010

### uart

Yes you're wrong. The net charge transferred to the capacitor during charging is Q=CV, which means that the net energy drawn from the supply is W = QV = CV^2. But this is exactly twice the energy stored on the capacitor therefore an equal amount (to that stored) is also lost. It's lost either in the series resistance (no matter how small) and additionally in radiation from the wires as the current oscillates in series resonance with the self inductance of the wires (in the case where the series resistance is sufficiently small.

8. May 17, 2010

### uart

BTW. Anyone who wishes to conceptualize circuits as having instantaneous changes in voltage and infinitely sharp spikes in current needs to be aware of the properties of the "Dirac delta function". In particular they should understand that it's ratio of mean squared value to mean value is infinite. They should understand that this means it is impossible to exist in nature and that approximations to it may therefore dissipate arbitrarily large energy into arbitrarily small resistances, even when the actual impulse value (area) is small.

9. May 17, 2010

### sophiecentaur

I was referring to the model in the preceding post in which a capacitor was connected directly to an (ideal) emf source (zero series resistance, which is not possible). Energy has to go somewhere (the emf must be V, whatever current is flowing).
In the simple RC case, half the energy supplied by the source is dissipated in the series resistor. Except that there will always be a finite amount of energy radiated into space (very low for 'normal' component values, of course.

@uart
In this case we are more likely dealing with the results of the Heaviside (step) function, I think.

10. May 17, 2010

### uart

No. I was talking about the current, the derivative of the Heaviside step function. An impulse has unbounded energy per unit area (under curve), eg per unit charge in the case of current flow.

Last edited: May 17, 2010