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
Circuits
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What is i through the
battery?
Label currents. New
label for every branch.
Pick any arbitrary
direction.
i through R1 or R4 is
same as for battery
Can use loop rule
E − i1 R1 − i2 R2 − i1 R4 = 0
+ -
- +
+ -
Circuits
E − i1 R1 − i2 R2 − i1 R4 = 0
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Equation has too many
unknowns so need to
apply loop rule again
Take the loop through
R2 and R3
− i3 R3 + i2 R2 = 0
i1 = i2 + i3
− (i1 − i2 ) R3 + i2 R2 = 0
i1 R3
i2 = −
( R3 + R2 )
Circuits
Now solve for i1:
E−i R −i R −i R = 0
1 1 2 2 1 4
iR R
E−i R − 1 3 2 −i R = 0
1 1 (R + R ) 1 4
2
3
E
i1 =
R2 R3
R1 +
+ R4
( R2 + R3 )
Circuits
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What is current Voltage
lost (V) in R2 ?
Recall that V=iR
V2 = i 2 R2
E − i1 R1 − i2 R2 − i1 R4 = 0
i1 R1 + i1 R4 − E
i2 =
R2
E
i1 =
R2 R3
R1 +
+ R4
( R2 + R3 )
RC - circuit
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Circuits where current
varies with time
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RC series circuit – a resistor
and capacitor are in series
with a battery and a switch
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At t =0 switch is open and
capacitor is uncharged so
q =0
RC - circuit
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z
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Close the switch at point a
Charge flows (current)
from battery to capacitor,
increasing q on plates and
V across plates
When VC equal Vbattery flow
of charge stops (current is
zero) and charge on
capacitor is
q = CV = CE
RC - circuit
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Want to know how q and
V of capacitor and i of
the circuit change with
time when charging the
capacitor
Apply loop rule, traversing
clockwise from battery
q
E − iR − = 0
C
i
RC - circuit
q
E − iR − = 0
C
z
z
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i
Contains 2 of the variables
we want i and q
Remember
dq
i =
dt
Substituting gives
dq
q
R
+
=E
dt
C
RC - circuit
dq
q
R
+
=E
dt
C
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Need a function which
satisfies initial condition
q = 0 at t = 0 and final
condition of q = C E at
t=∞
For charging a capacitor
(
q = CE 1 − e
−t RC
)
RC - circuit
(
q = CE 1 − e
z
z
−t RC
)
Want current as a
function of time
For charging a capacitor
dq
⎛ E ⎞ −t
= ⎜ ⎟e
i =
dt
⎝ R ⎠
RC
RC - circuit
(
q = CE 1 − e
z
z
−t RC
)
Want V across the
capacitor as function of
time
For charging a capacitor
VC
(
q
−t
=
= E 1− e
C
RC
)
RC - circuit
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Want to know how q of
capacitor and i of the
circuit change with time
when discharging the
capacitor
At new time t = 0, throw
switch to point b and
discharge capacitor
through resistor R
i
RC - circuit
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Apply the loop rule again
but this time no battery
q
− iR −
= 0
C
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Substituting for i again
gives differential equation
dq q
R
+ =0
dt C
i
RC - circuit
dq q
R
+ =0
dt C
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Solution must satisfy initial
condition that q0 = CV0
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For discharging a capacitor
q = q0e
−t RC
i
RC - circuit
q = q0e
z
−t RC
i
Find i for discharging
capacitor with initial
condition at i0 = V0/R =
q0/RC at t = 0
dq
⎛ q0 ⎞ −t RC
= −⎜
i=
⎟e
dt
⎝ RC ⎠
Negative sign
means charge
is decreasing
Circuits
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Charging capacitor
(
q = CE 1 − e
−t RC
⎛ E ⎞ −t
i = ⎜ ⎟e
⎝ R ⎠
z
RC
z
)
Discharging capacitor
q = q0 e
− t RC
⎛ q0 ⎞ −t RC
i = −⎜
⎟e
⎝ RC ⎠
Define capacitive time constant –
greater τ, greater (dis)charging time
τ = RC