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Algebra Test
Suppose x + y – 5 = 2 and 3x – 2y + 1 = 0. What is x?
A) zero
B) 13 / 5
C) 2 / 3
D) 7
E) 17 / 4
Algebra Test
Suppose x + y – 5 = 2 and 3x – 2y + 1 = 0. What is x?
A) zero
B) 13 / 5
C) 2 / 3
D) 7
Solve first equation for y:
x+y =7
y =7–x
Plug into second equation
3x – 2(7 – x) + 1 = 0
E) 17 / 4
3x – 14 + 2x + 1 = 0
5x – 13 = 0
5x = 13
x = 13 / 5
Circuits
Which of the equations is valid for the circuit shown below?
1
A) 2 – I1 – 2I2 = 0
B) 2 – 2I1 – 2I2 – 4I3 = 0
I2
C) 2 – I1 – 4 – 2I2 = 0
D) I3 – 2I2 – 4I2 = 0
E) 2I1 – I1 – 4I2 = 0
2
6V
22VV
4V
I1
1
3
I3
Circuits
Which of the equations is valid for the circuit shown below?
1
A) 2 – I1 – 2I2 = 0
B) 2 – 2I1 – 2I2 – 4I3 = 0
I2
C) 2 – I1 – 4 – 2I2 = 0
D) I3 – 2I2 – 4I2 = 0
E) 2I1 – I1 – 4I2 = 0
2
6V
22VV
4V
I1
1
3
I3
ConcepTest
R1
R2
=8V
In the circuit above, if all the resistors are identical,
what is the voltage drop across resistor R2?
A) 0 V
B) 2 V
C) 4 V
D) 6 V
E) 20 V
ConcepTest
Same current as R1 V = +6 V
and same resistance
V = +6 V
Loop rule 
(right loop and
top left loop)
V = +4 V
R1
R2
=8V
V = –4
? V
Loop rule
(outer loop)
Loop rule
(bottom left loop)
V = –8 V
In the circuit above, if all the resistors are identical, what is the
voltage drop across resistor R2?
A) 0 V
B) 2 V
C) 4 V
D) 6 V
E) 20 V
Circuits
The light bulbs in the circuit are identical. When the switch is
closed
A) both bulbs go out
B) the intensity of both bulbs increases
C) the intensity of both bulb decreases
D) nothing changes
Circuits
The light bulbs in the circuit are identical. When the switch is
closed
A) both bulbs go out
B) the intensity of both bulbs increases
C) the intensity of both bulb decreases
D) nothing changes
Circuits
The light bulbs in the circuit are identical. When the switch is
closed,
a) the intensity of bulb A increases
b) the intensity of bulb A decreases
c) the intensity of bulb B increases
d) the intensity of bulb B decreases
e) nothing changes
Circuits
The light bulbs in the circuit are identical. When the switch is
closed,
a) the intensity of bulb A increases
b) the intensity of bulb A decreases
c) the intensity of bulb B increases
d) the intensity of bulb B decreases
e) nothing changes
Answer
• Write a loop law for original loop:
+12V - I1R = 0
I1 = 12V/R
• Write a loop law for the new loop:
12 V
R
I
12 V
12 V
R
+12V +12V - I0R - I0R = 0
...or
I0 = 12V/R
• The key here is to determine the potential (Va–Vb)
before the switch is closed.
• From symmetry, (Va–Vb) = +12V.
• Therefore, when the switch is closed, NO additional
current will flow!
12 V
• Therefore, the current after the switch is closed is
equal to the current after the switch is closed.
12 V
R
a
I
12 V
R
b
Sample Problem
I2
V1 = 9 V
Determine the magnitudes and
directions of the currents
through R1 in the figure at right.
2
R2 = 15 
I1
Label Currents
1
Junction Rule
I1 = I2 + I3
Loop Rule (1)
+V2 – I1R2 = 0
Loop Rule (2)
+I1R2 + I2R1 + V1 = 0
Algebra
Solve Loop Rule (1) for I1
Plug into Loop Rule (2)
Comments
R1 = 25 
I3
V2 = 6 V
I2 =
–V1 – V2
R1
I2 = –3/5 Amp What does the minus sign mean?
Current goes in opposite direction of arrow