W9_parallel_resonance.eps

284 Week 8: Faraday’s Law and Induction

Power

Let’s track the flow of energy in this circuit. Remember, the power delivered to/used by any given
circuit element isP=V IwhereVis the voltage gain/drop across the element andIis the current
through it (which we now know).

The power provided by the battery (positive):

PV=V 0 I(t) =V

02

R

(

1 −e−(

RL)t)

(615)

Wow, that was easy!

The power burned in the resistor (negative – remember, this is energy that is all turned into
(joule) heat(ing):

PR = VRI(t) = (−I(t)R)I(t) =−I(t)^2 R

= −

V 02

R

(

1 −e−(

RL)t)^2

= −

V 02

R

(

1 − 2 e−(

RL)t

+e−(^2

RL)t)

(616)

which is a bit more complicated, but still not terrible. Note that I stuck a minus sign in front because
this is power beingremovedfrom the system by the voltagedropacross the resistor. With this sign
choice, we are guaranteed to have energy conserved, as we will see below.

The power delivered to the inductor (negative, but where does thisenergy go? See the next
topic...):

PL = VLI(t) = (−L

dI

dt

)I(t)

= −

{

L

V 0

R

(

R

L

)

e−(

RL)t

}

V 0

R

(

1 −e−(

RL)t)

= −V

02

R

(

e−(

RL)t

−e−^2 (

RL)t)

(617)

Note that we used the fact that

VL(t) =−L

dI

dt

=−V 0 e−(

RL)t

(618)

is the voltage drop across the inductor just as:

VR(t) =−IR=−V 0

(

1 −e−(

RL)t)

(619)

is the voltage drop across the resistor.

You can easily verify that these three add up to zero, so energy is conserved, but of course how
could itnotbe conserved? Take Kirchoff’s rule for this circuit above and multiply itbyI(t):

V 0 −IR−LdI

dt

= 0

(V 0 −IR−L

dI

dt

)I(t) = 0

V 0 I(t)−I(t)^2 R−LdI

dt

I(t) = 0

PV+PR+PL = 0 (620)

(where the signs all hopefully make sense to you). Thewhole pointof Kirchoff’s Loop Rule is that
it guarantees energy conservation around circuit loops, so we shouldn’t really be surprised when it