Exercise
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Suppose E = 12V, R=10 Ω
and r=2Ω
Potential across battery’s
terminals is
R
10Ω
Vb − Va = E
= (12V )
= 10V
R+r
10Ω + 2Ω
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V across terminals only equal
to E if no internal resistance
(r =0) or no current (i =0)
Review
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Kirchhoff’s loop rule – in traversing
a circuit loop the sum of the
changes in V is zero, ∆V =0
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Resistance rule – Move through
resistor in direction of current
V =-iR (+ to -; higher to lower), in
opposite direction V =+iR (- to +;
up the hill).
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Emf rule – Move through emf device
V =+ E going – to +, in opposite
direction V =- E .
Resistors in Series
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Resistors in series (one
path from b to a)
Have identical currents, i,
through them
Use Kirchhoff’s loop rule
E − iR1 − iR2 − iR3 = 0
E
i=
R1 + R2 + R3
Resistors in Series
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Resistors in series
The 3 resistors act the
same as an equivalent
resistor Req.
E
E
i=
=
R1 + R2 + R3 Req
Req = R1 + R2 + R3
Resistors in Series
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Resistors in series
Resistors have identical
currents, i
Sum of V s across
resistors = applied V
Req is sum of all resistors
n
Req = ∑ R j
j =1
Exercise
• If R1>R2>R3, rank greatest first
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A) current through resistors
i is same for all, tie
V = iR
Resistors in Parallel
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Resistors in parallel
Have same V across them
Arbitrarily choose direction
for currents in each branch
Write down current relation
for each resistor
V
i1 =
R1
V
i2 =
R2
V
i3 =
R3
Resistors in Parallel
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Resistors in parallel (more
than one path from a to b)
Apply Kirchhoff’s junction
rule at point a
Substitute current values
i = i1 + i2 + i3
⎛ 1
1
1 ⎞
⎟⎟
i = V ⎜⎜
+
+
R3 ⎠
⎝ R1 R 2
Resistors in Parallel
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Resistors in parallel
Replace 3 resistors with
equivalent resistor, Req
⎛ 1
1
1
1 ⎞
⎟⎟
= ⎜⎜
+
+
R eq ⎝ R1 R 2 R 3 ⎠
Exercise
Battery with potential V supplies
current i to 2 identical resistors
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What is V across and i
through either of the resistors if
they are connected in
z A) Series – What is constant?
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i is same, V1 = V /2
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B) Parallel – What is constant?
V is same, i1 = i /2
V = iR
Resistors and Capacitors
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Resistors
Series
R eq =
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n
1
1
=∑
Ceq j =1 C j
n
∑R
j =1
j
Parallel
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n
Capacitors
Series
1
1
=∑
Req j =1 R j
Parallel
C eq =
n
∑C
j =1
j
How to Analyze Complex Circuits
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Kirchhoff’s junction rule (or current law) –
z From
conservation of charge
z Sum of currents entering a junction is equal
to sum of currents leaving that junction
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Kirchhoff’s loop rule (or voltage law) –
z From
conservation of energy
z Sum of changes in potential going around a
complete circuit loop equals zero
Kirchhoff’s Rule #1
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Arbitrarily label currents,
using different subscript
for each branch
Using conservation of
charge at each junction
iin = iout
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At point d
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At point b
i1 + i3 = i2
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At point a
i1 = i1
i1 + i3 = i2
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At point c
i2 = i2
Exercise
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What is i of the circuit?
• Use Kirchhoff’s
loop rule
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Clockwise from point a gives
− E 2 − ir2 − iR − ir1 + E1 = 0
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Counterclockwise from point a gives
− E1 + ir1 + iR + ir2 + E 2 = 0
Exercise
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Solve for i
− E1 + ir1 + iR + ir2 + E 2 = 0
E1 − E 2
i=
R + r1 + r2
E1 = 4.4V E2 = 2.1V
r1 = 2.3Ω
r2 = 1.8Ω
R = 5.5Ω
i = 0.2396A ≈ 240mA
Exercise
A real battery has ( =12V and r = 2Ω.
Is the V across the terminals greater
than, less than or equal to 12V if the
current in the battery is
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A) from – to + terminal
LESS THAN
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Va + E − ir = Vb
B) from + to GREATER THAN
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C) i = 0
EQUAL TO 12V
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Va + E + ir = Vb
i