Circuits Which of the equations is valid for the circuit shown below? 2 – I 1 – 2I 2 = 0 A) 2...
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Transcript of Circuits Which of the equations is valid for the circuit shown below? 2 – I 1 – 2I 2 = 0 A) 2...
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
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
In the circuit below, if all the resistors are identical, what is the voltage drop across resistor R2?
= 8 V
R1
R2
V = –8 VLoop rule
(bottom left loop)
V
V Same current as R1 and same resistance V =V
V = ?
V
right loopV V = 0
How do the currentscompare here and here?
Give me an expression that describes the current I1 through R1
In the circuit below all the resistors are identical.
= 8 V
R1
R2
V = –8 VLoop rule
(bottom left loop)
V
V Same current as R1 and same resistance V =V
V = ?
V
right loopV V = 0
Both same current, I2
I1 = 2I2
What equation best describes the CW full outermost loop?
1) 8v + I2R1 + I2R I2R2 = 02) -8v + I2R1 + I2R + I2R2 = 03) -8v + 2I2R1 + 2I2R + I2R2 = 04) 8v 2I2R1 2I2R I2R2 = 0
In the circuit below all the resistors are identical.
= 8 V
R1
R2
V = –8 VLoop rule
(bottom left loop)
V
V Same current as R1 and same resistance V =V
V = ?
V
right loopV V = 0
Both same current, I2
I1 = 2I2
What is the voltage drop across resistor R2?
1) 0.2 volts 2) 1.0 volts3) 1.2 volts 4) 1.6 volts5) 2.0 volts 6) 4.0 volts
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