4—Differential Equations 81Now integrate this to get
L
∫
dI
V 0 −IR
=t+C, or −
L
R
ln(V 0 −IR) =t+C
Solve for the currentIto get
RI(t) =V 0 −e−(L/R)(t+C) (4.38)
Now does this make sense? Look at the dimensions and you see that itdoesn’t, at least not yet. The
problem is the logarithm on the preceding line where you see that its units don’t make sense either.
How can this be? The differential equation that you started with is correct, so how did the units get
messed up? It goes back to the standard equation for integration,
∫dx/x= lnx+C
Ifxis a length for example, then the left side is dimensionless, but this right side is the logarithm of
a length. It’s a peculiarity of the logarithm that leads to this anomaly. You can write the constant of
integration asC=−lnC′whereC′is another arbitrary constant, then
∫
dx/x= lnx+C= lnx−lnC′= ln
x
C′
IfC′is a length this is perfectly sensible dimensionally. To see that the dimensions in Eq. (4.38) will
work themselves out (this time), put on some initial conditions. SetI(0) = 0so that the circuit starts
with zero current.
R.0 =V 0 −e−(L/R)(0+C) implies e−(L/R)(C)=V 0
RI(t) =V 0 −V 0 e−Lt/R or I(t) = (1−e−Lt/R)V 0 /R
and somehow the units have worked themselves out. Logarithms do this, but you still better check.
The current in the circuit starts at zero and climbs gradually to its final valueI=V 0 /R.
4.8 Circuits
The methods of section4.1apply to simple linear circuits, and the use of complex algebra as in that
section leads to powerful and simple ways to manipulate such circuit equations. You probably remember
the result of putting two resistors in series or in parallel, but what about combinations of capacitors or
inductors under the same circumstances? And what if you have some of each? With the right tools,
all of these questions become the same question, so it’s not several different techniques, but one.
If you have an oscillating voltage source (a wall plug), and you apply it to a resistor or to a
capacitor or to an inductor, what happens? In the first case,V =IRof course, but what about the
others? The voltage equation for a capacitor isV =q/Cand for an inductor it isV =LdI/dt. A
voltage that oscillates at frequencyωisV =V 0 cosωt, but using this trigonometric function forgoes
all the advantages that complex exponentials provide. Instead, assume that your voltage source is
V =V 0 eiωtwith the real part understood. Carry this exponential through the calculation, and take
the real part only at the end — often you won’t even need to do that.