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DC circuits
Physics Department, New York
City College of Technology
Key words
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Electromotive force
(emf)
Terminal voltage
Resistors in parallel
and in series
Kirchhoff’s rules
Junction rule
Loop rule
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Capacitors in series
and in parallel
RC cuicuits
emf
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Electromotive force (emf) refers to the
potential difference between the
terminals of a source when no current
flows out. Its symbol is .
Terminal voltage
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Terminal voltage is the
potential difference
between the terminals
of a source when
current flows, and is
calculated as V    Ir
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 is the emf
r is the internal
resistance of the battery
Example #1
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A 65-Ω resistor is connected
to the terminals of a battery
whose emf is 12V and
whose internal resistance is
0.5Ω. Calculate (a) the
current in the circuit, (b) the
terminal voltage of the
battery, and (c) the power
dissipated in the resistor R
and in the battery's internal
resistance r.
Example #1—continued
(a)
Since Vab    Ir ,

12V
I

 0.183 A
R  r 65  0.5
(b)
Vab    Ir  12V  (0.183 A)(0.5)  11.9V
(c)
The power dissipated in R is
PR  I 2 R  (0.183 A) 2 (65)  2.18W ,
and in r is
Pr  I 2 R  (0.183 A) 2 (0.5)  0.02W .
Resistors in series
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The equivalent resistance for resistors
in series is
Req  R1  R2  R3  ...
Resistors in parallel
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The equivalent resistance for resistors
in parallel is 1  1  1  1  ...
Req
R1
R2
R3
Voltage drop along wire
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Disc 18, #1
Disc 18, #2
Disc 18, #6
Series/parallel resistors
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Disc 17, #23
Disc 17, #24
Example #2
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Two 100Ω resistors are
connected (a) in parallel,
and (b) in series, to a
24V battery. What is the
current through each
resistor and what is the
equivalent resistance of
each circuit?
Example #2—continued
(a)
The total current I from the battery splits to flow
through each resistor, so
I  I1  I 2
V
24V
V
24V
I1 

 0.24 A, I 2 

 0.24 A
R1 100
R2 100
I  I1  I 2  0.48 A
1
1
1
2
1




,
Req 100 100 100 50
so Req  50
Example #2—continued
(b)
I is the same in both resistors, and
V  V1  V2
V  IR1  IR2  I ( R1  R2 )
V
24V
I

 0.12 A
R1  R2 100  100
V 24.0V
Req  
 200,
I 0.12 A
or Req  R1  R2  200
Kirchhoff’s rules
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The junction rule: at any junction point, the
sum of all currents entering the junction
must equal the sum of all currents leaving
the junction. It is based on the conservation
of electric charge.
The loop rule: the sum of the changes in
potential around any closed path of a circuit
must be zero. It is based on the
conservation of energy.
Example #3
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Calculate the
currents I1, I2,
and I3.
Example #3—continued
Apply Kirchhoff' s junction rule at point a,
I 3  I1  I 2 .
(a)
Apply Kirchhoff' s loop rule to the upper loop,
 30 I1  45  41I 3  0.
(b)
Apply Kirchhoff' s loop rule to the outer loop,
 30 I1  (20  1) I 2  80  0.
(c)
Example #3—continued
From Eq. (c), we get
80  30I1
I2 
 3.8  1.4I1
21
From Eq. (b), we get
45  30I1
I3 
 1.1  0.73I1
41
(d)
(e)
Example #3—continued
Substitute (d) and (e) into Eq. (a),
I 1 I 3  I 2  1.1  0.73I1  3.8  1.4 I1.
I1  0.87 A
I 2  2.6 A
I 3  1.7 A
Capacitors in series and in
parallel
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The equivalent capacitance for
capacitors in series is
1
1
1
1



 ...
Ceq C1 C 2 C3
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The equivalent capacitance for
capacitors in parallel is
Ceq  C1  C 2  C3  ...
RC circuits
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In the charging
process,
Vc   (1  e t / RC )
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The time constant is
  RC
R
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In the discharging
process,
ε
C
Vc  V0 e  t / RC
Switch
RC charging curve
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Disc 18, #28