Chapter 28
Direct Current (DC) Circuits
In chapter 27 we learned how to determine the current through
a resistor using Ohm’s law
In chapter 28 we will learn how to analyze more complicated
circuits which contain several resistors and batteries using
Kirchhoff’s rules which are based on:
(a) energy conservation and
(b) charge conservation
In addition, we will study the charging and discharging of a
capacitor through a resistor as a function of time
(28-1)
Battery is a device that can maintain a constant potential
difference E between its terminals. E is known as the
“electromotive force (emf)” of the battery.
Units of emf: Volts Note: When we draw a circuit diagram
we assume that the connecting wires have negligible resistance
Circuit
Mechanical analog
A
A
B
pump
(28-2)
High
reservoir
Low
reservoir
B
Ideal versus real batteries
V
An ideal battery maintains a
constant potential difference V
across its terminals no matter
what the current I it provides is.
V=E
In real batteries the voltage V across the terminals decreases
with increasing I as follows:
V = E -Ir
where r is known as the internal
resistance of the battery. If we solve the equation above for I
E
I=
we get:
(28-3)
R+r
V
E
Slope = -r
V
O
I
V = E –Ir
This equation represents a straight line with
slope = -r and y-axis intercept = E (of the form: y = a +bx)
The power P provided by the battery is: P = E I = I(R+r)I
P = I2R + I2r.
The first term appears as heat on R. The
second term appears as heat on the battery.
(28-4)
We will analyze complicated circuits using Kirchhoff’s rules
(a) Kirchhoff’s loop rule (KLR)
(b) Kirchhoff’s junction rule (KJR)
KLR The potential change in traversing a complete loop in a
circuit is equal to zero
∆V = 0
∑
One can see that this is the case in the simple circuit below
From Ohm’s law: I = E/R
(assuming that r = 0) → E = IR
→ E – IR = 0
This is indeed KLR
(28-5)
Kirchhoff’s Junction Rule (KJR) : The algebraic sum of all
the currents at any junction is equal to zero
∑ Ii = 0
Sign convention:
i
If a current arrives at a junction
we count it as positive
I1
I4
P
If a current leaves from a
junction it is counted as negative
Example: Apply KJR at point P
I3
I2
I1 + I2 + I3 - I4 = 0
Note: KJR is a restatement of
charge conservation (28-6)
Details for the application of KLR
∑ ∆V = 0
(±E)
(±IR)
(±Q/C)
(28-7)
Gustav Robert Kirchhoff
1824-1887
I1
I3
I1
I3
KLR (abcda):
-E2 +I3R 3 –I2R 2 = 0
KLR (adefa):
I2R 2 + I1R 1 –E1 = 0
Example: Write
Kirchhoff’s
equations for the
circuit to the left so
that you can
determine al the
currents in the
circuit
KLR (fabcdef): -E1 – E2 +I3R3 + I1R 1 = 0
KJR (a): -I1 + I2 + I3 = 0
(28-8)
Resistors connected in series
R1, R 2, R 3 have a common current I
Ohm's law for R1 , R1 , R1
V1 = I1R1 , V2 = I 2 R2 , V3 = I 3 R3
E = V1 + V2 + V3 = I ( R1 + R2 + R3 )
V1
V3
E
→I =
R1 + R2 + R3
E
Ohm's law for Req : Req = →
I
Req = R1 + R2 + R3
In general for N resistors in series
Req = R1 + R 2 + …+RN
(28-9)
V2
Resistors connected in parallel. R 1, R 2, R3 have common
voltage E applied across them
Ohm's law for R1 , R1 , R1
E
E
E
I1 =
, I2 =
, I3 =
R1
R2
R3
Total current I = I1 + I 2 + I 3
 1
1
1 
I =E  +
+ 
 R1 R2 R3 
1
I
Ohm's law for Req :
=
R eq E
→
1
1
1
1
=
+
+
R eq R1 R2 R3
(28-10)
Example (28-4) Page 757
(28-11)
Find Req for the circuit given below
R1 = 100 Ω
R2 = 200 Ω
R3 = 300 Ω R 4 = 400 Ω
R5 = 500 Ω R 6 = 600 Ω
R2 and R3 are in series. We
replace them with R 7
R7 = (200 + 300) Ω = 500 Ω
R5 and R6 are also in series
We replace them with R 8
R8 = R 5 + R 6 = (500+600) Ω
R8 = 1100 Ω
Form fig. (a) we can see that R7, R4, and R8 are connected in parallel
we replace them with R9
1/R9 = 1/R7 + 1/R4 + 1/R8
1/R9 = 1/500 + 1/400 + 1/1100 Ω -1 = 5.4×10-3 Ω -1
thus R9 = 185 Ω
From fig.(b) we can see that R1 and R9 are connected in series. We
replace them with Req = R1 + R9 = 100 + 185 Ω = 285 Ω
(28-12)
Example 28-5 Find the currents in the circuit below
• Choose current directions
• Do not worry about mistakes in the current direction
because the method is self-correcting
• If a current comes out positive the chosen direction in the
figure is correct . If a current value comes out negative the
correct direction is opposite to that in the figure
I1
I1
I3
I3
(28-13)
KLR (abcda):
-E2 +I3R3 –I2R2 = 0
-12+80I3 –10I2 = 0
KLR (adefa):
(1)
I2R2 + I1R1 –E1 = 0
10I2 +100I1 –6 = 0
(2)
KJR (a): -I1 + I2 + I3 = 0
(3)
E1 = 6 V
E2 = 12 V
R1 = 100 Ω
I1
I1
I3
I3
R2 = 10 Ω
R3 = 80 Ω
(28-14)
-12 + 80I3 – 10I2 = 0
(1)
10I2 + 100I1 – 6 = 0
(2)
-I1 + I2 + I3 = 0
(3)
System of 3 equations and 3 unknowns
Form eqs.3 we have: I1 = I2 + I3
Substitute I1 in eqs(2)
Equation 2 becomes: 10I2 +100(I2 + I3) – 6 = 0
110I2 +100I3 = 6 (4) Equations 1 and 4 contain only I2 and I3
–10I2 +80I3 = 12 (1)
110I2 +100I3 = 6
(4)
×11
×1
eqs(1) ×11 + eqs(1) ×1 =
(-110+110)I2 +(880+100)I3 = 132+6 → 980I3 = 138 → I3 = 0.14A
From eqs.1 we have: I2 = (80I3-12)/10 = -0.073A
From eqs.3 we have: I1 = I2 + I3 = 0.068A
Note: The direction of I2 is opposite to that in the figure (28-15)
R
+
E
I
We wish to measure I in the
simple circuit to the left. We will
use an ammeter which is device
that measures current.
The ammeter must be connected in series so that the same
current I’ that passes through R also passes through the
ammeter. In general I’ ≠ I
I’
(28-16)
R
I’
+
RA
I’= E /(R + R A) and
We want
I’ ≈ I
Example: R = 100 Ω
E
I
RA = ammeter’s resistance
I = E/R
This is the case if : R A << R
RA = 0.1 Ω
Conclusion: A good ammeter should have a resistance
RA that is small compared with all other resistances in the
circuit
(28-17)
We wish to measure the voltage
V across the resistor R in the
circuit to the left. For this
purpose we will use a
voltmeter which is a device
that measures the potential
difference between two points
V=E
R
+
E
I
The voltmeter must be connected
I’
V’
in parallel to the resistor so that the
resistor and the voltmeter have the
same potential difference V’ across
them. In general V’ ≠ V
(28-18)
I’ = I1 + I2
I1 = E /R
If
I2 = E /RV
I2 << I1
V=E
R
then
+
I1 = E /R
E
I = E/R
and V’ = I1R = E
I2
I’
I1
V’
Under these conditions the voltmeter
reads the correct value V’ = E
I2 << I1
if
RV >> R
A good voltmeter should have a
large resistance
Example:
RV = 100000
R = 100 Ω,
Ω
(28-19)
A schematic diagram of a circuit is
given to the right. The real
diagrams are given below. These are
considerably messier
R2
R3
R1
E
+
(28-20)
RC circuits Charging and discharging of a capacitor C
through a resistor R
Assume that the
capacitor is not
charged. At t = 0
we throw the switch
from the middle
position to
position “a”. The battery will charge the capacitor C
through the resistor R. After C is fully charged we throw
the switch from position “a” to position “b”. The battery is
disconnected from the circuit and the capacitor discharges
through R. Charging and discharging of C is not
instantaneous but is described by the time constant
τ = RC
τ depends on both C and R
(28-21)
R
a
+
d
E
I
b
C
+Q
We wish to describe the
capacitor charge Q(t) as
function of time t
-Q
c
KLR along (abcda): − IR − VC + E = 0
Q
dQ
VC =
I=
→
C
dt
dQ Q
R
+ = E Divide RHS and LHS by R →
dt C
dQ Q E
+
=
dt RC R
This is a linear differential equation of first order. The
initial condition is: Q(t=0)
(28-22)
dQ Q E
+
=
dt RC R
The solution of this differential equation is:
(
Q (t ) = E C 1 − e − t /τ
)
where the time constant τ = RC The time constant tells us
what is the time scale of charging. Q(t = 1τ) = 0.63 (EC)
Q(t = 2τ) = 0.86(EC)
Q(t = 3τ) = 0.95(EC)
The capacitor is practically
fully charged within a few
time constants
(28-23)
(
Q (t ) = E C 1 − e −t /τ
)
dQ E − t / τ
Current I =
= e
where τ = RC
dt R
Note 1:
Note 2:
I(t = 1τ) = 0.37 (E/R)
I(t = ∞) = 0
I drops to zero within a few time constants
(28-24)
R
+
-
E
I
Energy conservation
C
+Q
-Q
(
Q(t ) = E C 1 − e − t /τ
)
Q(∞ )=E C
Wbat = work done by battery during charging = EQ = CE 2
Q 2 CE 2
U C = energy stored in the capacitor =
=
2C
2
U R = total energy dissipated on R as heat
UR = ?
Energy conservation: Wbat = UC + U R → U R = Wbat − UC →
2
2
C
E
C
E
U R = CE 2 −
=
2
2
(28-25)
Capacitor discharging: Initial condition: Q(t=0) = Qo
dQ
I=
,
dt
KLRfor (abcda): − IR −VC = 0
dQ Q
R
+ =0
dt C
a
→
dQ Q
→
+
= 0 Solution: Q ( t ) = Qo e −t / τ
dt RC
b
d
Q
VC =
C
c
where τ = RC
(28-26)
Initial condition:
Q(t=0) = Qo
Solution:
Q ( t ) = Qo e −t /τ
Q(t = ∞) = 0
Q(t = 1τ) = 0.37Qo
Q is practically zero within a few time
constants
dQ −Qo − t /τ
Current I =
=
e
dt
RC
Note: The negative sign of I
is telling us that we chose
the wrong current direction
(28-27)