Direct Current Circuits
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
1
Ammeters and Voltmeters
!  A device used to measure current is called an ammeter
!  A device used to measure potential difference is called a
voltmeter
!  To measure the current, the ammeter must be placed in the
circuit in series
Voltmeter in parallel
High resistance
!  To measure the potential
difference, the voltmeter
must be wired in parallel
with the component
across which the potential
difference is to be measured
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
Ammeter in series
Low resistance
2
Real-life Voltmeter in a Simple Circuit
!  Consider a simple circuit
consisting of a source of emf with
potential difference Vemf = 150. V
and a resistor with resistance
R = 100 kΩ.
!  A voltmeter with resistance RV = 10.0 MΩ is connected
across the resistor.
PROBLEM 1:
!  What is the current in the circuit before the voltmeter is
connected?
SOLUTION 1:
!  Ohm’s Law allows us to calculate the current before:
Vemf
150. V
−3
i=
=
=
1.50⋅10
A = 1.50 mA
3
R
100.⋅10 Ω
February 19, 2014
Chapter 26
6
Real-life Voltmeter in a Simple Circuit
PROBLEM 2:
!  What is the current in the circuit
after the voltmeter is connected?
SOLUTION 2:
!  The equivalent resistance of the
resistor and the voltmeter connected in parallel is given by:
1
1
1
= +
Req R RV
!  The equivalent resistance is:
3
6
100.⋅10
Ω
10.0⋅10
Ω)
(
)(
RRV
4
Req =
=
=
9.90⋅10
Ω = 99.0 kΩ
3
6
R + RV
100.⋅10 Ω +10.0⋅10 Ω
February 19, 2014
Chapter 26
7
Real-life Voltmeter in a Simple Circuit
!  The current after connecting the voltmeter is:
Vemf
150. V
−3
i=
=
=
1.52⋅10
A = 1.52 mA
4
R
99.0⋅10 Ω
!  So the current in the circuit increases by 0.02 mA, when the
voltmeter is connected because the parallel combination of
the resistor and the voltmeter has a lower resistance than
that of the resistor alone.
!  The effect is small.
!  The larger the resistance of the voltmeter, the better.
February 19, 2014
Chapter 26
8
RC Circuits
!  So far we have dealt with circuits containing sources of emf
and resistors
!  The currents in these circuits did not vary in time
!  Now we will study circuits that contain capacitors as well as
sources of emf and resistors
!  These RC circuits have currents that vary with time
!  Consider a circuit with
•  a source of emf, Vemf,
•  a resistor R,
•  a capacitor C
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
9
Charging a Capacitor
!  We start with the capacitor uncharged
!  We close the switch and current begins to
flow, building up opposite charges on the
plates of the capacitor and creating a
potential difference across the capacitor
!  When the capacitor is fully charged, no
more current flows
!  The potential difference across the plates
is equal to the potential difference
provided by the source of emf
!  The magnitude of the total charge on each
plate is
qtot = CVemf
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
10
Charging a Capacitor
!  While the capacitor is charging, we can
analyze the the current flowing by
applying Kirchhoff’s Loop Rule
q (t )
Vemf −VR −VC = Vemf − i (t ) R −
=0
C
!  The change of the charge on the capacitor
is the current
The term Vc is negative since
dq (t )
the top plate of the capacitor
i (t ) =
is connected to the positive dt
!  We can rewrite our equation as
dq (t ) q (t )
R
+
= Vemf or
higher potential - terminal of
the battery. Thus analyzing
counter-clockwise leads to a
drop in voltage across the
capacitor!
dt
C
dq (t ) q (t ) Vemf
dq
+
=
Differential equation relating q and
dt
RC
R
dt
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
11
Charging a Capacitor
!  The solution for this differential equation is
t
− ⎞
⎛
q (t ) = qmax ⎜ 1− e τ ⎟
⎝
⎠
!  The constant τ is the time constant given by
τ = RC
!  The constant qmax is given by
qmax = CVemf
!  So we can write the solution to our differential equation
t
−
⎛
⎞
RC
q (t ) = CVemf ⎜ 1− e ⎟
⎝
⎠
Math Reminder:
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
12
Charging a Capacitor
!  At t = 0 we have
0
−
⎛
⎞
RC
q ( 0 ) = CVemf ⎜ 1− e ⎟ = 0
⎝
⎠
!  At t = ∞ we have
∞
−
⎛
⎞
RC
q ( ∞ ) = CVemf ⎜ 1− e ⎟ = CVemf
⎝
⎠
!  The current flowing through the circuit is the time
derivative of the charge
dq(t ) ⎛ Vemf ⎞ − RCt
i (t ) =
=⎜
⎟⎠ e
⎝
dt
R
!  At t = 0 we have
i (0) =
Vemf
R
!  At t = ∞ we have
∞
⎛ Vemf ⎞ − RC
i (∞) = ⎜
e
=0
⎟
⎝
⎠
February 19, 2014
R
Physics for Scientists & Engineers 2, Chapter 26
13
Charging a Capacitor
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
14
Discharging a Capacitor
!  Now we consider a circuit containing
only one resistor and a fully charged
capacitor
!  The charge on the capacitor is qmax
!  Now we short the resistor across the
capacitor by moving the switch from
position 1 to position 2
!  Current flows until the capacitor is
completely discharged
!  While the capacitor is discharging,
Kirchhoff’s Loop Rule gives us
q (t )
−i (t ) R −VC = −i (t ) R −
=0
C
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
15
Discharging a Capacitor
!  Using the definition of current, we can
rewrite our equation as
dq(t ) q(t )
R
dt
+
=0
C
!  The solution to this differential equation is
q (t ) = qmax e
−
t
RC
!  At t = 0 we have
q ( 0 ) = qmax e
−
0
RC
= qmax
!  At t = ∞ we have
q ( ∞ ) = qmax e
February 19, 2014
−
∞
RC
=0
Physics for Scientists & Engineers 2, Chapter 26
16
Discharging a Capacitor
!  The current flowing through the circuit is the time
derivative of the charge
qmax ) − RCt
(
dq (t )
i (t ) =
=−
e
dt
RC
!  At t = 0 we have
qmax ) − RC0
qmax )
(
(
i (0) = −
e
=−
RC
RC
!  At t = ∞ we have
∞
qmax ) − RC
(
i (t ) = −
e
=0
RC
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
17
Discharging a Capacitor
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
18
Time Required to charge a Capacitor
!  Consider a circuit consisting of a 12.0 V battery, a 50.0 Ω
resistor, and a 100.0 μF capacitor wired in series
!  The capacitor is initially completely discharged
!  PROBLEM
!  How long after the circuit is closed will it take to charge the
capacitor to 90% of its maximum charge?
SOLUTION
!  The charge on the capacitor as a function of time is given by
t
−
⎛
⎞
RC
q (t ) = qmax ⎜ 1− e ⎟
⎝
⎠
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 26
20
Time Required to charge a Capacitor
!  We want to know the time until q(t)/qmax = 0.90
t
t
−
−
⎛
⎞
q
t
( ) = 0.90 ⇒ 0.10=e RC
RC
1−
e
=
⎜⎝
⎟⎠ q
max
!  Taking the natural log of both sides gives us
⎛ − RCt ⎞
ln ( 0.10 ) =ln ⎜ e ⎟
⎝
⎠
t
⇒ ln ( 0.10 ) = −
RC
!  So the time to discharge the capacitor to 90% of its
maximum charge is
t = −RC ln ( 0.10 ) = − (50.0 Ω )(100.0⋅10−6 F )( −2.30 )
t = 0.0115 s = 11.5 ms
February 19, 2014
Math Reminder: ln(ex)=x
Physics for Scientists & Engineers 2, Chapter 26
21
Summary: RC Circuit
!  Charging a capacitor:
t
− ⎞
⎛
q(t ) = q0 ⎜1 − e τ ⎟
⎝
⎠
dq ⎛ Vemf
i=
=⎜
dt ⎝ R
⎞
⎟e
⎠
⎛ t ⎞
⎜−
⎟
⎝ RC ⎠
!  Discharging a capacitor:
q = q0 e
⎛ t ⎞
⎜−
⎟
⎝ RC ⎠
dq ⎛ q0 ⎞
i=
=⎜
⎟e
dt ⎝ RC ⎠
!  Time constant:
⎛ t ⎞
⎜−
⎟
⎝ RC ⎠
τ = RC
Physics for Scientists & Engineers 2
22
Rate of Energy Storage in a
Capacitor
!  A resistor with R = 2.50 MΩ and a capacitor with
C = 1.25 μF are connected in series with a battery for which
Vemf = 12.0 V
!  At t = 2.50 s after the circuit is closed, what is the rate at
which energy is being stored in the capacitor?
SOLUTION
THINK
!  When the circuit is closed, the capacitor begins to charge
!  The rate at which energy is stored in the capacitor is given
by the time derivative of the amount of energy stored in the
capacitor, which is a function of the charge on the capacitor
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 25
23
Rate of Energy Storage in a
Capacitor
SKETCH
!  The circuit contains a battery, a resistor, and a capacitor in
series.
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 25
24
Rate of Energy Storage in a
Capacitor
RESEARCH
!  The charge on the capacitor as a function of time is
t
−
⎛
⎞
RC
q (t ) = CVemf ⎜ 1− e ⎟
⎝
⎠
!  The energy stored in a capacitor with charge q is
1 q2
U=
2C
!  The time derivative of the stored energy is
dU d ⎛ 1 q 2 ⎞ q (t ) dq (t )
= ⎜
=
⎟
dt dt ⎝ 2 C ⎠
C dt
!  The time derivative of the charge is the current
dq (t ) ⎛ Vemf ⎞ − RCt
i (t ) =
=⎜
e
⎟
⎝ R ⎠
dt
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 25
25
Rate of Energy Storage in a
Capacitor
SIMPLIFY
!  Combining our equations
t
−
⎛
⎞
RC
CVemf ⎜ 1− e ⎟
⎝
⎠ ⎛ Vemf ⎞ − RCt
dU q (t )
=
i (t ) =
⎜⎝
⎟⎠ e
dt
C
C
R
t
t
2
−
−
⎛
⎞
dU Vemf
RC
RC
=
e ⎜ 1− e ⎟
dt
R
⎝
⎠
CALCULATE
!  We first calculate the time constant
RC = ( 2.50⋅106 Ω )(1.25⋅10−6 F ) = 3.125 s
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 25
26
Rate of Energy Storage in a
Capacitor
!  Now we calculate the rate of change of energy stored in the
capacitor
2.50 s
2.50 s
−
⎛
⎞
12.0 V ) − 3.125
dU
(
−5
s
3.125 s
=
e
1
−
e
=
1.425
⋅10
W
6
⎜
⎟
dt 2.50 ⋅10 Ω
⎝
⎠
2
DOUBLE-CHECK (Energy Conservation)
!  The current at t = 2.50 is
2.50 s
−
12.0
V
⎛
⎞ 3.125 s
−6
i (t ) = ⎜
e
=
2.16⋅10
A
⎟
6
⎝ 2.50⋅10 Ω ⎠
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 25
27
Rate of Energy Storage in a
Capacitor
!  The rate of energy dissipation at t = 2.50 is
2
dU 2
−6
P=
= i R = ( 2.16 ⋅10 A ) ( 2.50 ⋅106 Ω ) = 1.16 ⋅10−5 W
dt
!  The rate at which the battery delivers energy to the circuit at
t = 2.50 s is
P=
dU
= iVemf = ( 2.16 ⋅10−6 A ) (12.0 V ) = 2.59 ⋅10−5 W
dt
!  The energy supplied by the battery is equal to the sum of the
energy dissipated in the resistor plus the energy stored in
the capacitor
February 19, 2014
Physics for Scientists & Engineers 2, Chapter 25
28