Capacitors and inductors

ENGR 40M lecture notes — July 21, 2017

Chuan-Zheng Lee, Stanford University

Unlike the components we’ve studied so far, in capacitors and inductors, the relationship between current
and voltage doesn’t depend only on the present. Capacitors and inductors store electrical energy—capacitors
in an electric field, inductors in a magnetic field. This enables a wealth of new applications, which we’ll see
in coming weeks.

Quick reference
Capacitor Inductor

Symbol

Stores energy in electric field magnetic field

Value of component capacitance, C inductance, L
(unit) (farad, F) (henry, H)
dv di
I–V relationship i=C v=L
dt dt
At steady state, looks like open circuit short circuit

General behavior
In order to describe the voltage–current relationship in capacitors and inductors, we need to think of voltage
and current as functions of time, which we might denote v(t) and i(t). It is common to omit (t) part, so v
and i are implicitly understood to be functions of time.
The voltage v across and current i through a capacitor with capacitance C are related by the equation

C
dv i
i=C ,
dt + −
v

where dv 1
dt is the rate of change of voltage with respect to time. From this, we can see that an sudden change
in the voltage across a capacitor—however minute—would require infinite current. This isn’t physically
possible, so a capacitor’s voltage can’t change instantaneously. More generally, capacitors oppose changes in
voltage—they tend to “want” their voltage to change “slowly”.
Similarly, in an inductor with inductance L,

L
di i
v=L .
dt + v −

An inductor’s current can’t change instantaneously, and inductors oppose changes in current.
Note that we’re following the passive sign convention, just like for resistors.
1 That is, the derivative of voltage with respect to time. If you haven’t studied calculus, think of this as the slope of the
curve at a given time t, if you draw a graph of voltage against time.
Combinations in series and parallel
Inductors combine similarly to resistors:

L1
L1 L2
L2

 −1
1 1 L1 L2
Ls = L1 + L2 Lp = + =
L1 L2 L1 + L2

Capacitors, however, are the other way round:

C1
C1 C2
C2

 −1
1 1 C1 C2
Cs = + =
C1 C2 C1 + C2 Cp = C1 + C2

Steady state analysis

Steady state refers to the condition where voltage and current are no longer changing. Most circuits, left
undisturbed for sufficiently long, eventually settle into a steady state. In a circuit that is in steady state,
dv di
dt = 0 and dt = 0 for all voltages and currents in the circuit—including those of capacitors and inductors.

Thus, at steady state, in a capacitor, i = C dv di

dt = 0, and in an inductor, v = L dt = 0. That is, in steady
state, capacitors look like open circuits, and inductors look like short circuits, regardless of their capacitance
or inductance.

in steady state, looks like

in steady state, looks like

(This might seem trivial now, but we’ll use this fact repeatedly in more complex situations later.)

Practical matters
Manufacturers typically specify a voltage rating for capacitors, which is the maximum voltage
that is safe to put across the capacitor. Exceeding this can break down the dielectric in the +
capacitor.
Capacitors are not, by nature, polarized : it doesn’t normally matter which way round you −
connect them. However, some capacitors are polarized—in particular, electrolytic capacitors,
where the insulator is a very thin oxide layer that would be depleted if a negative voltage is
polarized
applied. In schematics, if a capacitor is polarized, we use a special symbol for it, as shown
capacitor
right.

2
Examples
Example 1. In the (contrived) circuit below, at t = 0, the voltage across the capacitor is 0 V. The switch
begins on the lower throw, moves to the upper throw at t = 0 ms, then moves back to the lower throw at
t = 1 ms. Draw a graph of the voltage across the capacitor as a function of time.

1 mA 250 nF

Example 2. In the (contrived) circuit below, at t = 0 ms, va = 10 V. The switch begins on the lower throw,
and moves to the upper throw at t = 0 ms. When does the capacitor’s voltage to reach 0 V?

va

500 µA 2 µF

Example 3. You wish to replace the following network of capacitors with a single capacitor. What should
its value be?

300 nF 600 nF

350 nF

150 nF

Example 4. In the (contrived) circuit below, at t = 0, the current through the inductor is 100 mA upwards.
The switch begins on the lower throw, moves to the upper throw at t = 0 ms, then moves back to the lower
throw at t = 1 ms. Draw a graph of the current through the inductor iL , with reference direction going
downwards, as a function of time.

2.7 V + 10 mH
−
iL

3
Example 5. Find the total inductance of this inductor network.

15 µH 15 µH

50 µH

45 µH

Example 6. The circuit below has reached steady state. What is the voltage vx ?

4 kΩ

+ vx
− 3V

2.7 nF 8 kΩ

Example 7. The circuit below has reached steady state.

(a) What is the voltage vx ?
(b) What is the voltage across the capacitor? (Be sure to specify the direction.)

33 nF
vx

+ 680 Ω 120 kΩ
− 6V

Example 8. The circuit below has reached steady state. What is the voltage vx ?

4 nH

3.3 nF 10 kΩ
1 mH
+ vx
− 5V

2.7 nF 15 kΩ