1. 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

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. In the circuit below, if all the resistors are identical, what is the voltage drop across resistor R 2 ?  = 8 V R1R1 R2R2  V = –8 V Loop rule (bottom left loop) VV VV Same current as R 1 and same resistance  V =  V  V = ? VV right loop  V  V = 0 How do the currents compare here and here? Give me an expression that describes the current I 1 through R 1

4. In the circuit below all the resistors are identical.  = 8 V R1R1 R2R2  V = –8 V Loop rule (bottom left loop) VV VV Same current as R 1 and same resistance  V =  V  V = ? VV right loop  V  V = 0 Both same current, I 2 I 1 = 2 I 2 What equation best describes the CW full outermost loop? 1) 8v + I 2 R 1 + I 2 R  I 2 R 2 = 0 2)-8v + I 2 R 1 + I 2 R + I 2 R 2 = 0 3)-8v + 2I 2 R 1 + 2I 2 R + I 2 R 2 = 0 4) 8v  2I 2 R 1  2I 2 R  I 2 R 2 = 0

5. In the circuit below all the resistors are identical.  = 8 V R1R1 R2R2  V = –8 V Loop rule (bottom left loop) VV VV Same current as R 1 and same resistance  V =  V  V = ? VV right loop  V  V = 0 Both same current, I 2 I 1 = 2 I 2 What is the voltage drop across resistor R 2 ? 1)0.2 volts2) 1.0 volts 3)1.2 volts4) 1.6 volts 5) 2.0 volts6) 4.0 volts

6. 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