1. 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

2. 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 Test

3. Circuits Which of the equations is valid for the circuit shown below? 2 – I 1 – 2I 2 = 0 A) 2 – I 1 – 2I 2 = 0 2 – 2I 1 – 2I 2 – 4I 3 = 0 B) 2 – 2I 1 – 2I 2 – 4I 3 = 0 2 – I 1 – 4 – 2I 2 = 0 C) 2 – I 1 – 4 – 2I 2 = 0 I 3 – 2I 2 – 4I 2 = 0 D) I 3 – 2I 2 – 4I 2 = 0 2I 1 – I 1 – 4I 2 = 0 E) 2I 1 – I 1 – 4I 2 = 0 2 V 2  2 V 6 V 4 V 3  1  I1I1 I3I3 I2I2

4. Circuits Which of the equations is valid for the circuit shown below? 2 – I 1 – 2I 2 = 0 A) 2 – I 1 – 2I 2 = 0 2 – 2I 1 – 2I 2 – 4I 3 = 0 B) 2 – 2I 1 – 2I 2 – 4I 3 = 0 2 – I 1 – 4 – 2I 2 = 0 C) 2 – I 1 – 4 – 2I 2 = 0 I 3 – 2I 2 – 4I 2 = 0 D) I 3 – 2I 2 – 4I 2 = 0 2I 1 – I 1 – 4I 2 = 0 E) 2I 1 – I 1 – 4I 2 = 0 2 V 2  2 V 6 V 4 V 3  1  I1I1 I3I3 I2I2

5. In the circuit above, if all the resistors are identical, what is the voltage drop across resistor R 2 ? A) 0 VB) 2 VC) 4 VD) 6 VE) 20 VConcepTest  = 8 V R1R1 R2R2

6. In the circuit above, if all the resistors are identical, what is the voltage drop across resistor R 2 ? A) 0 VB) 2 VC) 4 VD) 6 VE) 20 VConcepTest  V = ?  V = +6 V  V = –8 V –4 V Same current as R 1 and same resistance Loop rule (bottom left loop) Loop rule (outer loop)  = 8 V  V = +4 V Loop rule (right loop and top left loop) R1R1 R2R2

7. Circuits The light bulbs in the circuit are identical. When the switch is closed both bulbs go out A) both bulbs go out the intensity of both bulbs increases B) the intensity of both bulbs increases the intensity of both bulb decreases C) the intensity of both bulb decreases D) nothing changes

8. Circuits The light bulbs in the circuit are identical. When the switch is closed both bulbs go out A) both bulbs go out the intensity of both bulbs increases B) the intensity of both bulbs increases the intensity of both bulb decreases C) the intensity of both bulb decreases D) nothing changes

9. Circuits The light bulbs in the circuit are identical. When the switch is closed, the intensity of bulb A increases a) the intensity of bulb A increases the intensity of bulb A decreases b) the intensity of bulb A decreases the intensity of bulb B increases c) the intensity of bulb B increases d) the intensity of bulb B decreases e) nothing changes

10. Circuits The light bulbs in the circuit are identical. When the switch is closed, the intensity of bulb A increases a) the intensity of bulb A increases the intensity of bulb A decreases b) the intensity of bulb A decreases the intensity of bulb B increases c) the intensity of bulb B increases d) the intensity of bulb B decreases e) nothing changes

11. R 12 V R I +12V +12V - I 0 R - I 0 R = 0 I 0 = 12V/R Write a loop law for original loop: Write a loop law for the new loop: +12V - I 1 R = 0 I 1 = 12V/R The key here is to determine the potential (V a –V b ) before the switch is closed. From symmetry, (V a –V b ) = +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 R 12 V R I a bAnswer

12. Determine the magnitudes and directions of the currents through R 1 in the figure at right. V 2 = 6 V R 2 = 15  R 1 = 25  V 1 = 9 V Label Currents I1I1 I3I3 I2I2 Junction Rule Loop Rule (1) Loop Rule (2) I 1 = I 2 + I 3 +V 2 – I 1 R 2 = 0 +I 1 R 2 + I 2 R 1 + V 1 = 0 Algebra 1 Sample Problem Solve Loop Rule (1) for I 1 Plug into Loop Rule (2) I 2 = –V 1 – V 2 R1R1 2 Comments I 2 = –3/5 AmpWhat does the minus sign mean? Current goes in opposite direction of arrow