Chapter 26: DC Circuits
Direct Current (“DC”): a current that is constant in magnitude and
“sense” of flow.
Alternating Current (“AC”): a current that alternates in sense of flow. An
alternating current is not constant in magnitude or sense.
In a broader interpretation of the terms AC and DC, we use “AC” to refer
to anything (current, voltage) that changes with time and “DC” to refer to
anything (current, voltage) that is constant. Hence, a “DC voltage” is a
voltage that doesn’t change – in magnitude or polarity – with time.
Incidentally, the polarity of a voltage refers to which end is positive and
which is negative.
In this chapter, we will deal only with DC voltages and currents. We’ll
return to the subject of AC in Chapter 31.
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Ch. 26: DC Circuits
Resistors in Series and Parallel
We saw in Chapter 24 how caps combine in series and parallel. Do
resistors combine according to the same rules?
Resistors in Series
Suppose two resistors, R1 and R2 , are connected in series across a
battery that supplies an emf ε . Let the node between the two resistors
be called “Node A.” Let I1 be the current through R1 and I 2 be the
current through R2 .
Now think about what would happen at Node A if I1 and I 2 were different.
Suppose, for example, that I1 = 5A and I 2 = 2A . Then we’d have more
charge per second coming into Node A than we’d have leaving Node A.
This means charge would have to be piling up at Node A or leaving the
circuit at Node A. Because neither of these things is happening, I1 can’t
be bigger than I 2 .
In a similar way, if I 2 were bigger than I1 — for example, I1 = 2A and
I 2 = 5A — then charge would have to be being supplied at Node A. But
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Ch. 26: DC Circuits
charge is not being supplied at Node A. Therefore, I 2 can’t be bigger
than I1 .
Resistors in series have the same current through them.
From now on, then, I’ll drop the subscripts “1” and “2” and just call the
current through both resistors “ I ”.
What single equivalent resistance, Req , could replace the series
combination of R1 and R2 and allow the same current I to flow? Well,
consider Ohm’s law applied to R1 and R2 . For R1 , I get:
V1 = IR1
and for R2 :
V2 = IR2
Adding these two equations, I get:
V1 + V2 = I ( R1 + R2 )
But V1 and V2 must add up to the emf supplied by the battery:
V1 + V2 = ε
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Ch. 26: DC Circuits
So:
ε = I (R + R )
1
This looks like Ohm’s law,
2
ε = IR
,
for a single resistor Req connected across the battery, with Req given by:
Req = R1 + R2
If we had N resistors R1 , R2 ,…, RN in series across the battery, applying
Ohm’s law to each would give:
V1 = IR1
eq
V2 = IR2
VN = IRN
These voltages would still have to add up to ε , as before, so:
ε = I ( R1 + R2 + + RN )
This looks like Ohm’s law for a single resistor Req connected across the
battery, with
Req = R1 + R2 + + RN
(1)
4
Ch. 26: DC Circuits
Resistors in Parallel
Consider two resistors, R1 and R2 , connected in parallel across a battery
supplying emf ε . What’s the equivalent resistance of this parallel
combination?
Let I be the total current supplied by the battery. There is a node where
the positive terminal of the battery is connected to the top ends of R1 and
R2 ; call this “Node A.” When the current I reaches this node, it must
split. Part of the current goes through R1 and part of it goes through R2 .
Let I1 be the current that goes through R1 and I 2 be the current that goes
through R2 . Applying Ohm’s law to R1 and R2 gives:
ε = I1R1
ε = I 2 R2
Notice:
Resistors in parallel have the same voltage across them.
Rearranging the two equations immediately above, we have:
I1 = ε R1
I 2 = ε R2
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Ch. 26: DC Circuits
Adding these two gives:
⎛ 1
1 ⎞
I1 + I 2 = ε ⎜ + ⎟
⎝ R1 R2 ⎠
If charge is not piling up at Node A or being supplied at Node A, then I1
and I 2 must add up to the total current supplied by the battery, I . So:
⎛ 1
1 ⎞
I =ε ⎜ + ⎟
⎝ R1 R2 ⎠
⎛ 1 ⎞
ε = I ⎜⎜ 1 + 1 ⎟⎟
⎝ R1 R2 ⎠
This looks like Ohm’s law for a single resistor Req connected across the
battery, with:
1
Req = 1 1
R1 + R2
For this special case of two resistors in parallel, we can rewrite Req as:
RR
Req = 1 2
(2)
R1 + R2
(product over sum)
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Ch. 26: DC Circuits
If we had N resistors R1 , R2 ,…, RN in parallel across the battery, applying
Ohm’s law to each would give:
I1 =
I2 =
IN
ε
R1
ε
R2
ε
=
RN
But these currents would still have to add up to the total current supplied
by the battery. Therefore, adding the equations immediately above
would give:
⎛ 1
1
1 ⎞
I =ε ⎜ +
+ +
⎟
R
R
R
2
N ⎠
⎝ 1
⎛
⎞
1
ε = I ⎜⎜ 1 + 1 + + 1 ⎟⎟
RN ⎠
⎝ R1 R2
This looks like Ohm’s law for a single equivalent resistor:
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Ch. 26: DC Circuits
Req =
1
R1
+
1
R2
1
+
+
1
RN
(3)
To simplify (3) slightly, we usually write the equation for Req in terms of
the reciprocal of Req :
1
1
1
1
= +
+ +
(4)
Req R1 R2
RN
Finally, note that if we had N equal resistors, R , in parallel, the
equivalent resistance would be, from (4):
1
1 1
1 N
= + + + =
Req R R
R R
N times
R
(5)
N
The equivalent resistance would be one- N th of any one of the resistors.
Req =
Comparing (1) through (5) with the corresponding results for caps
(Chapter 24), we see:
• Resistors in series combine like caps in parallel.
• Resistors in parallel combine like caps in series.
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Ch. 26: DC Circuits
Kirchhoff’s Rules
Named for the German physicist Gustav Kirchhoff (1824-1887),
Kirchhoff’s rules are rules for analyzing more complicated circuits. There
are two rules:
1. Kirchhoff’s current rule (also called “Kirchhoff’s junction rule”)
2. Kirchhoff’s voltage rule (also called “Kirchhoff’s loop rule”)
Kirchhoff’s Current Rule (KCR)
The sum of the currents into any node must equal the sum of the
currents out of that node:
∑ Iin = ∑ I out
(6)
This just follows from conservation of charge applied at each node.
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Ch. 26: DC Circuits
Kirchhoff’s Voltage Rule (KVR)
The algebraic sum of all the voltages as you go once around any closed
loop must equal zero.
(7)
∑ ΔV = 0
closed
loop
Kirchhoff’s voltage rule follows from conservation of energy. To see
this, consider two resistors, R1 and R2 , connected across a battery
suppying emf ε .
In an infinitesimal time dt , an infinitesimal charge dQ crosses the battery
from – to + (i.e., from Node A to Node B). This charge gains an
infinitesimal amount of electric potential energy given by:
dU elec = ε dQ
The total change in energy ( ΔU elec ) A→ B of a total charge Q crossing the
battery in any finite duration is, then:
( ΔU elec ) A→B = ∫ dU elec = ∫ ε dQ
Q
0
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Ch. 26: DC Circuits
But all the charge Q crosses the same emf ε , so ε is a constant, as far
as the integral is concerned:
( ΔU elec ) A→B = ε ∫ dQ = Qε
Q
0
Now consider a total charge Q going clockwise once around the circuit,
starting at the negative terminal of the battery. Conservation of energy
demands:
( ΔU elec ) A→B + ( ΔU elec ) B→C + ( ΔU elec )C →D + ( ΔU elec ) D→E + ( ΔU elec ) E → A = 0
Or:
( ΔU elec ) A→B + ( ΔU elec )C →D + ( ΔU elec ) D→E = 0
Qε + Q ( ΔV )C → D + Q ( ΔV ) D→ E = 0
ε + ( ΔV )
C →D
+ ( ΔV ) D→ E = 0
This is Kirchhoff’s voltage rule.
Applying Ohm’s law for R1 and R2 , we can rewrite the equation
immediately above:
ε − IR1 − IR2 = 0
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Ch. 26: DC Circuits
RC Circuits
When we speak of RC circuits, we mean circuits that contain resistors
and capacitors.
Charging A Capacitor
Consider a battery supplying emf ε connected by means of a switch S
to a resistor R and capacitor C in series. (See link to separate “RC
Circuits” handout on web page.) As the cap charges up through R , the
magnitude of the charge on the capacitor plates will grow with time. To
indicate that q is a function of time, we write it q ( t ) . The voltage across
the cap will also be a function of time, vC ( t ) , as will the voltage across
the resistor, vR ( t ) , and the current, i ( t ) .
Note that we use lowercase letters to indicate quantities that are timevarying.
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Ch. 26: DC Circuits
How exactly do the charge and the current depend on time? To answer
this, we need to find expressions for q ( t ) and i ( t ) . For the details of how
this is done, see the “RC Circuits” link on the web page. The results turn
out to be:
t
−
⎛
⎞
q ( t ) = Qmax ⎜1 − e RC ⎟ ,
⎝
⎠
in which:
Qmax = Cε ,
(8)
(9)
and:
i ( t ) = I 0e
in which:
I0 =
−
ε
R
t
RC
,
(10)
(11)
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Ch. 26: DC Circuits
The RC Time Constant
The factor RC that shows up in the exponential terms in (8) and (10) is
called the RC time constant. It is a time that characterizes how long it
takes for the capacitor to charge up. Using (8), in a time equal to one
RC , the charge rises from zero to:
RC
−
⎛
⎞
−1
RC
=
−
q ( RC ) = Qmax ⎜ 1 − e
Q
1
e
(
) ≈ ( 0.63) Qmax
⎟
max
⎝
⎠
So after one RC , the cap is about 63% fully charged.
Also after one RC , the current is, from (10):
−
RC
RC
i ( RC ) = I 0 e
= I 0 e −1 ≈ ( 0.37 ) I 0
So the current has fallen to about 37% of its initial value.
The product RC does have units of time. If R is in ohms and C is in
farads, then the unit for RC is:
⎛ V ⎞⎛ C ⎞ ⎛ V ⎞ ⎛ C ⎞
RC ⇒ ( Ω )( F ) = ⎜ ⎟⎜ ⎟ = ⎜
⎜ ⎟=s
⎟
⎝ A ⎠⎝ V ⎠ ⎝ C s ⎠ ⎝ V ⎠
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Ch. 26: DC Circuits
The “5-RC Rule”
Looking at (8), we see that it would take literally forever for q ( t ) to equal
Qmax . How do we know when we’ve waited long enough that we can say
that the cap is essentially fully charged? A good “rule of thumb” is to wait
for five RC time constants. At t = 5 RC , (8) gives:
5 RC
−
⎛
⎞
q ( 5 RC ) = Qmax ⎜ 1 − e RC ⎟ = Qmax (1 − e −5 ) ≈ ( 0.99 ) Qmax
⎝
⎠
Thus, after five RC time constants, the cap is about 99% fully charged.
The current at t = 5 RC is, from (10):
i ( 5 RC ) = I 0 e −5 ≈ ( 0.007 ) I 0
Thus, after five RC time constants, the current is down to about 0.7% of
its initial value.
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Ch. 26: DC Circuits
Discharging A Capacitor
Now suppose we have a cap already charged up and we want to
discharge it. To do this, we connect a resistor across the cap in order to
give some path for charge to get off of the positive plate and onto the
negative plate. (Refer to the “RC Circuits” link on the web page.)
Let the magnitude of the initial charge on the plates be called Q0 . Again,
the charge on the plates and the current will be functions of time. We
would like to know how, exactly, q ( t ) and i ( t ) depend on time.
The details are worked out in the separate “RC Circuits” handout. The
results (after a little calculus) turn out to be:
q ( t ) = Q0 e
−
t
RC
(12)
and:
i ( t ) = I 0e
−
t
RC
,
(13)
in which:
Q0
(14)
RC
Both the charge and the current drop asymptotically to zero as t → ∞ .
I0 =
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Ch. 26: DC Circuits
In one RC constant, we find, from (12) and (13):
q ( RC ) = Q0 e −1 ≈ ( 0.37 ) Q0
i ( RC ) = I 0 e −1 ≈ ( 0.37 ) I 0
so both the charge and the current have fallen to about 37% of their
initial values.
Looking at (12) and (13), we see that we would have to wait literally
forever for the cap to discharge fully. How do we decide when we’ve
waited long enough that we can say the cap is essentially fully
discharged?
The “5-RC Rule”
In a time equal to five RC time constants, (12) and (13) give:
q ( 5 RC ) = Q0 e −5 ≈ ( 0.007 ) Q0
i ( 5 RC ) = I 0 e −5 ≈ ( 0.007 ) I 0
so both the charge and the current are down to about 0.7% of their initial
values.
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