# Capacitor voltages and Inductor currents

1. Sep 19, 2005

### Corneo

I have been taught that the voltage in a capacitor and the current in an inductor cannot change instanteously. This is important especially when solving differential equations for a circuit network. Can someone explain to me why these events cannot happen? To the extent of my knowledge for these events to happen, there is a need of infinite energy.

2. Sep 19, 2005

### SGT

The current in a capacitor is $$i = C\frac{dv}{dt}$$ and the voltage in an inductor is $$v = L\frac{di}{dt}$$.
So, for an instantaneous change of voltage (dt = 0) in a capacitor, you need an infinite current. In the same way, for an instantaneous change of current in an inductor, you need an infinite voltage.
Notice that there is an infinite mean power, but a finite energy, since dt = 0.

3. Sep 19, 2005

### EvLer

he-he...
we were just told last week in lecture that capacitor charge (or voltage) CAN change instantaneously if input signal is delta function.

4. Sep 20, 2005

### SGT

Yes, the delta 'function' or impulse, is a signal with infinite amplitude and zero duration. Because of this it has finite energy. So a current impulse can instantaneously change the voltage of a capacitor and a voltage impulse can instantaneously change the current in an inductor.

5. Sep 20, 2005

### chroot

Staff Emeritus
Of course, delta functions don't exist in the real world. There's no such thing as an impulse with infinite amplitude and zero duration in the real world.

- Warren

6. Sep 21, 2005

### SGT

You are right. Delta function is an artificial construct that allows us to solve for the current fed to an ideal capacitor, initially discharged, connected to an ideal voltage source of V volts. According to Kirchoff's voltage law, the voltage in the capacitor rises instantaneously from 0 to V volts. This is only possible if the current charging the capacitor is $$i_C(t) = CV\cdot\delta(t)$$.
Since no ideal elements exists, there is no impulse. Assuming a real voltage source with an output resistance and a real capacitor with an associated resistance, the real current will be $$i_C(t) = \frac{V}{R}\cdot e^{-\frac{t}{RC}}$$, where R is the combined resistance of source and capacitor.

7. Sep 29, 2005

### leright

physically, the voltage is not changing instantly...however, since this change occurs very abruptly (but no instantly) it can be ROUGHLY modeled using a dirac delta function. Nothing behaves EXACTLY like a delta function representation, but calculations can be greatly simplified by using them.