Transcript PPT - LSU Physics & Astronomy

Physics 2113
Jonathan Dowling
Lecture 20: FRI 09 OCT
Circuits II
How to Solve Multi-Loop Circuits
Monster Maze
If all resistors have a
resistance of 4, and
all batteries are ideal
and have an emf of
4V, what is the
current through R?
i®
Step 1: Find loop that only
goes through R and through
batteries but no other resistors.
Step 2: Assume i is clockwise
on the loop and walk the loop
using the loop rules.
E + E + E - E - iR = 0
Step 3: Solve for i.
2E - iR = 0
i = 2E / R = 2A
Step 4: If i is negative,
reverse the direction of the
current in the diagram.
What If We Chose Wrong
Direction for i ???
If all resistors have a
resistance of 4, and
all batteries are ideal
and have an emf of
4V, what is the
current through R?
i ® ¬i
Step 1: Find loop that only
goes through R and through
batteries but no other resistors.
Step 2: Assume i is counterclockwise on the loop and
walk the loop using the loop
rules.
+E - E - E - E - iR = 0
Step 3: Solve for i.
-2E - iR = 0
i = -2E / R = -2A
Step 4: If i is negative,
reverse the direction of the
current in the diagram.
One Battery? Simplify!
Resistors
Key formula: V=iR
In series: same current dQ/dt
Req=∑Rj
In parallel: same voltage
1/Req= ∑1/Rj
P = iV = i2R = V2/R
Capacitors
Q=CV
same charge Q
1/Ceq= ∑1/Cj
same voltage
Ceq=∑Cj
U = QV/2 = Q2/2C = CV2
Too Many Batteries?
Loop & Junction!
One Battery: Simplify First
1
R
eq-par
123
1
1
1
= +
+
R1 R2 R3
Three Batteries: Straight to Loop
& Junction Rules:
Circuit CAN’T Be Simplified
Because R1 and R2 are neither in
Series Nor in Parallel Because
Batteries Between Them.
Apply Loop Rule
Around every loop moving in direction of current add +E
if you cross a battery from minus to plus, –E if plus to
minus, and –iR for each resistor. Then sum to Zero:
+E1 –E2 – iR1 – iR2 = 0. Assume current and loop
clockwise.
®i
R1
E1
+
–
-
-
+
–
R2
Conservation of ENERGY!
E2
Apply Junction Rule
At every junction sum the ingoing currents and outgoing
currents and set them equal.
i1 = i2 + i3
i1
i2
i3
Conservation of CHARGE!
Equations to Unknowns
Continue applying loop and junction until
you have as many equations as unknowns!
Given: E1 , E2 , i1 , R1 , R2
+E1 –E2 – i1R1 – i2 R2 = 0
and
i1 = i2 + i3
Solve for i2 , i3
Example
Find the equivalent resistance between points
(a) F and H and
(b) F and G.
(Hint: For each pair of points, imagine that a battery is
connected across the pair.)
Compile R’s in Series
Compile equivalent R’s in Parallel
Series
H
F
F
H
Parallel
Slide Rule
F
H
Example
Assume the batteries are ideal, and have emf
E1=8V, E2=5V, E3=4V, and R1=140
R2=75and R3=2.
What is the current in each branch?
What is the power delivered by each battery?
Which point is at a higher potential, a or b?
Apply loop rule three times
and junction rule twice.
Example
• What’s the current
through resistor R1?
• What’s the current
through resistor R2?
• What’s the current
through each battery?
Apply loop rule three times
and junction rule twice.
Too many batteries: loop and junction!
There are three loops and two junctions.
Assume all currents are clockwise.
Both junctions give same result:
i1 = i2 + i3
i1 ®
¬ i2
Walk three loops clockwise from a to a:
-E1 - i1R2 + E3 + E2 = 0
¬ i3 ® i3
-8V
i1 = E3 +ER22 -E1 = 4V +5V
= +0.13A
75W
-E1 - i1R2 + E3 - i3 R1 = 0
Finally reverse i3.
5V
= -0.04A
-E2 - i3 R1 = 0 i3 = -E2 / R1 = - 140W
Three equations and three unknowns i1, i2, i3. Tough in general but
trick here is to realize red and green loop equations depend on
only one of the i’s. Solve for i1 and i3 and use junction to get i2.
i2 = i1 - i3 = (+0.13A) - (-0.04A) = +0.17A
i1 ®
E1=8V, E2=5V, E3=4V, and
R1=140R2=75and R3=2.
¬ i2
What is the current in each branch?
i1 = +0.13A i2 = +0.17A i3 = +0.04A
What is the power delivered by
each battery? P = iV = iE
® i3
Which point is at a higher potential, Va = -E1 - i1 R2 + E3 + E2 = 0
a or b?
V = -E - i R < 0
b
Va > Vb
What’s the current through resistor R1
What’s the current through resistor R2?
What’s the current through each battery?
1
1
2
Non-Ideal Batteries
• You have two ideal identical batteries, and a
resistor. Do you connect the batteries in series or
in parallel to get maximum current through R?
• Does the answer change if you have non-ideal (but
still identical) batteries?
Apply loop and junction rules
until you have current in R.
More Light Bulbs
• If all batteries are ideal,
and all batteries and light
bulbs are identical, in
which arrangements will
the light bulbs as bright as
the one in circuit X?
• Does the answer change if
batteries are not ideal?
Calculate i and V across each bulb.
P = iV = “brightness”
or
Calculate each i with R’s the same:
P = i2R