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...
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pose 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 Algebra Test
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Transcript of 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...
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 TestAlgebra 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
Solve first equation for y:
x + y = 7
y = 7 – x
Plug into second equation
3x – 2(7 – x) + 1 = 0
3x – 14 + 2x + 1 = 0
5x – 13 = 0
5x = 13
x = 13 / 5
Algebra TestAlgebra Test
CircuitsCircuits
Which of the equations is valid for the circuit shown below? Which of the equations is valid for the circuit shown below?
A) 2 – I2 – I11 – 2I – 2I22 = 0 = 0
B) 2 – 2I2 – 2I11 – 2I – 2I2 2 – 4I– 4I3 3 = 0= 0
C) 2 – I2 – I11 – 4 – 2I – 4 – 2I2 2 = 0= 0
D) II33 – 2I – 2I2 2 – 4I– 4I2 2 = 0= 0
E) 2I2I11 – I – I11 – 4I – 4I2 2 = 0= 0
2 V
2
2 V 6 V
4 V
3 1
1
I1 I3
I2
CircuitsCircuits
Which of the equations is valid for the circuit shown below? Which of the equations is valid for the circuit shown below?
A) 2 – I2 – I11 – 2I – 2I22 = 0 = 0
B) 2 – 2I2 – 2I11 – 2I – 2I2 2 – 4I– 4I3 3 = 0= 0
C) 2 – I2 – I11 – 4 – 2I – 4 – 2I2 2 = 0= 0
D) II33 – 2I – 2I2 2 – 4I– 4I2 2 = 0= 0
E) 2I2I11 – I – I11 – 4I – 4I2 2 = 0= 0
2 V
2
2 V 6 V
4 V
3 1
1
I1 I3
I2
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
ConcepTestConcepTest
= 8 V
R1
R2
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
ConcepTestConcepTest
V = ?
V = +6 V
V = +6 V
V = –8 V
–4 V
Same current as R1 and same resistance
Loop rule (bottom left loop)
Loop rule (outer loop)
= 8 V
V = +4 V
Loop rule (right loop andtop left loop)
R1
R2
CircuitsCircuitsThe light bulbs in the circuit are identical. When the switch is The light bulbs in the circuit are identical. When the switch is closedclosed
A) both bulbs go outboth bulbs go out
B) the intensity of both bulbs increasesthe intensity of both bulbs increases
C) the intensity of both bulb decreasesthe intensity of both bulb decreases
D)D) nothing changes nothing changes
CircuitsCircuitsThe light bulbs in the circuit are identical. When the switch is The light bulbs in the circuit are identical. When the switch is closedclosed
A) both bulbs go outboth bulbs go out
B) the intensity of both bulbs increasesthe intensity of both bulbs increases
C) the intensity of both bulb decreasesthe intensity of both bulb decreases
D)D) nothing changes nothing changes
CircuitsCircuitsThe light bulbs in the circuit are identical. When the switch is The light bulbs in the circuit are identical. When the switch is closed,closed,
a) the intensity of bulb A increasesthe intensity of bulb A increases
b) the intensity of bulb A decreasesthe intensity of bulb A decreases
c) the intensity of bulb B increasesthe intensity of bulb B increases
d)d) the intensity of bulb B decreases the intensity of bulb B decreases
e)e) nothing changes nothing changes
CircuitsCircuitsThe light bulbs in the circuit are identical. When the switch is The light bulbs in the circuit are identical. When the switch is closed,closed,
a) the intensity of bulb A increasesthe intensity of bulb A increases
b) the intensity of bulb A decreasesthe intensity of bulb A decreases
c) the intensity of bulb B increasesthe intensity of bulb B increases
d)d) the intensity of bulb B decreases the intensity of bulb B decreases
e)e) nothing changes nothing changes
R
12 V
12 V
R
12 VI
+12V +12V - I0R - I0R = 0
I0 = 12V/R
• Write a loop law for original loop:
• Write a loop law for the new loop:
+12V - I1R = 0
I1 = 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!
• Therefore, the current after the switch is closed is equal to the current after the switch is closed.
...or...or
R
12 V
12 V
R
12 VI
a
b
AnswerAnswer
Determine the magnitudes and Determine the magnitudes and directions of the currents directions of the currents through Rthrough R11 in the figure at right. in the figure at right.
V2 = 6 V
R2 = 15
R1 = 25 V1 = 9 V
Label CurrentsI1
I3
I2
Junction Rule
Loop Rule (1)
Loop Rule (2)
I1 = I2 + I3
+V2 – I1R2 = 0
+I1R2 + I2R1 + V1 = 0
Algebra
1
Sample ProblemSample Problem
Solve Loop Rule (1) for I1
Plug into Loop Rule (2) I2 = –V1 – V2
R1
2
Comments I2 = –3/5 Amp What does the minus sign mean?Current goes in opposite direction of arrow
