# Capacitors and Inductors at t=0

I am a little confused on how capacitors and inductors act when t=0. For example lets say we have the following circuit:

Is(S) = 8/(s+2)

Whats the current I(L) at t=0?

What if the inductor was a capacitor, Would it change the current at I(C)?

EDIT: After some reading, I think your supposed to treat the circuit as a DC circuit at t=0.

Then I(S) = 0 A

an inductor would act as a short

a capacitor would act as open.

If the circuit above has an inductor then I(L) = 0 A

If the circuit above has a capacitor instead of inductor then I(C) = 0 A

• Welcome to EE.SE! What was the current in the inductor just before t=0 ? I think that info is also required to answer this question. – AJN Dec 4 '20 at 17:57
• The current was 0, I have added a graph of the current over time – user270240 Dec 4 '20 at 18:00
• Are $I_s$ and $i_s$ supposed to mean the same thing? (Maybe I'm being overly pedantic, but my EE classes were usually very particular about case, often having $I_X$, $i_X$, and $i_x$ with different meanings in the same problem). – Justin Dec 4 '20 at 19:10
• In this case I think they mean the same thing – user270240 Dec 4 '20 at 19:18
• Where’s the switch that opens or closes at t=0? – Chu Dec 5 '20 at 0:39

Intuitively , what you are seeing is the integration of an exponential waveform.

• Solving for $$\I_L(t),V_L(t)\$$ , you know $$\I_L(0)=0A, V_L(0)=32V\$$, what is $$\I_L(t)\$$ max ? ,
• at t=0 V=32V all the step current is into the resistor then the inductor starts to integrate current as the resistor decays in current and voltage.
• eventually the current slowly reaches a maximum dI/dt=o which means V=0 yet the current must be dissipated thru R so the voltage keep decaying to some negative lower voltage
• as I decays to 0 the negative voltage also decays to 0

## other thoughts.

In the time domain, for L, V(t)= L dI(t)/dt or you can integrate both sides for a current source as you have.

in the s domain for an ideal L, Z(s)= s L
at steady state s.

What happens when s=0 (dc !), You know a constant voltage starts with zero current when applied to an open circuit inductor and in theory ramps up to infinity, yet in practice the DCR series resistance limits current not shown.

So the steady impedance at DC is equivalent to 0 Ohms yet the transient impedance is infinite (meaning transient change in impedance starting up , or a change in voltage) because initially the current does not change.

This is simply the property of any integrator and the inductor integrates current with the applied voltage just like the capacitor integrates the opposite (charge voltage is integrated from current changes.) So the capacitor has the inverse impedance characteristics in the s domain to the inductor.

## Now tackle your problem.

Given a Time domain current that steps from 0 to 8A then decays (-2t) exponentially to 0 with t/τ = 2t or τ=0.5 = "tau" the greek symbol used for exponential Time Constant

## So what the L is the value ?

• (sorry for puny pun)for T = 4 second exponential time constant.

Ask your Prof why show I(s) when the impedance is not constant in the s domain?

See what he /she says. (hehe) This is a time domain problem.

The s term in the L(s) may be confusing you because s domain is implies steady state spectrum (s) for which this is clearly not , it's a transient DC or "baseband signal" and the inductor impedance is constantly changing.

• At t=0 the output voltage steps to 8A x 4Ω= 32V then exponential decays
• As the inductor integrates up in current to a peak at which point dV/dt=0 and the thus the Voltage = 0 but the resistor has been draining the current, so the inductor current decays slower as required (4s) and the that negative dI/dt means the voltage swings below 0 (negative) but much less than the step voltage. - then - then voltage decays from -ve towards 0 as the inductor current decays exponentially to 0 with tau=L/R

When voltage is first applied to an inductor (which previously had no current flowing through it), the initial current is 0.

When current is first applied to a capacitor (which previously had no voltage accross it), the initial voltage across it is 0.

What's the current I(L) at t=0?

It could be anything. It could be +100 A, or -1.3 mA, or whatever. The way you have your post right now, you have not provided enough data for us to answer that question. Inductor currents are (independent) state variables, in circuit analysis. In other words: none of the other pieces of data you have provided determine how large iL(t=0) has to be.

What if the inductor was a capacitor, Would it change the current at I(C)?

Similarly, it could also be anything. In this case, the independent state variable would be the initial capacitor voltage (vC(t=0)) which, together with the circuit it would be attached to, would determine how large iC(t=0) would be. Since you have also not provided how large vC(t=0) would be, I cannot tell you how large iC(t=0) would be.