1. 1. Write the three rules governing voltage, current, and resistance relationships in a series circuit containing a battery and three resistors. Some Electricity Problems Answers R eq = R 1 + R 2 + R 3 V = V 1 + V 2 + V 3 I = I 1 = I 2 = I 3 (1) (2) (3)

2. 2. Write the three rules governing voltage, current, and resistance relationships in a parallel circuit containing a battery and three resistors. V = V 1 = V 2 = V 3 I = I 1 + I 2 + I 3 1 R eq = 1 R1R1 + 1 R2R2 1 R3R3 + (1) (2) (3)

3. 3. Draw a schematic diagram of a circuit containing a 6.0-V battery hooked in series with three resistors of 2-Ω, 2-Ω, and 4-Ω, respectively. 6 V2 Ω 4 Ω 2 Ω

4. 6 VR eq R eq = R 1 + R 2 + R 3 = (2 Ω) + (2 Ω) + (4 Ω) R eq = 8 Ω 6 V2 Ω 4 Ω 2 Ω 3. Draw a schematic diagram of a circuit containing a 6.0-V battery hooked in series with three resistors of 2-Ω, 2-Ω, and 4-Ω, respectively. (a) Determine the equivalent resistance.

5. V = I RI = V R eq = 8 Ω 6 V I = 0.75 A Use simpler circuit 3. Draw a schematic diagram of a circuit containing a 6.0-V battery hooked in series with three resistors of 2-Ω, 2-Ω, and 4-Ω, respectively. (b) Determine the current I emerging from the battery. 6 V8 Ω6 V2 Ω 4 Ω 2 Ω ITIT ITIT

6. 6 V V 1 V 3 0.75 A I1I1 I2I2 I3I3 2 Ω 4 Ω 2 Ω V2V2 (c) Determine the voltage drops V 1, V 2, and V 3. 0.75 A V 1 = I 1 R 1 = (0.75 A)(2 Ω) V 1 = 1.5 V V 2 = I 2 R 2 = (0.75 A)(2 Ω) V 2 = 1.5 V V 3 = I 3 R 3 = (0.75 A)(4 Ω) V 3 = 3.0 V Current is the same everywhere = I R 1 = I R 2 = I R 3 Note that V 1 + V 2 + V 3 = 6 V

7. 4. Draw a schematic diagram of a circuit containing a 6.0-V battery hooked in parallel with three resistors of resistances 2.0-Ω, 2.0-Ω, and 4.0-Ω, respectively. 6 V 2 Ω 4 Ω

8. 1 R eq = 1 R1R1 + 1 R2R2 1 R3R3 += 1 2 + 1 2 1 4 + = 2 4 + 2 4 + 1 4 = 5 4 = 1 = 0.80 Ω 6 V 2 Ω 4 Ω6 VR eq 4. Draw a schematic diagram of a circuit containing a 6.0-V battery hooked in parallel with three resistors of resistances 2.0-Ω, 2.0-Ω, and 4.0-Ω, respectively. (a) Determine the equivalent resistance R eq.

9. V = I R V R eq = 0.80 Ω 6 V I = 7.5 A I = 6 V 2 Ω 4 Ω6 V0.80 Ω 4. Draw a schematic diagram of a circuit containing a 6.0-V battery hooked in parallel with three resistors of resistances 2.0-Ω, 2.0-Ω, and 4.0-Ω, respectively. (b) Determine the current emerging from the battery. ITIT ITIT

10. (c) Determine the branch currents I 1, I 2, and I 3. 6 V I1I1 I2I2 I3I3 ITIT I 1 = V1V1 R1R1 = V R1R1 = 6 V 2 Ω = I 1 = 3.0 A I 2 = V2V2 R2R2 = V R2R2 = 6 V 2 Ω = I 2 = 3.0 A Voltage is the same everywhere I 3 = V3V3 R3R3 = V R3R3 = 6 V 4 Ω = I 3 = 1.5 A Note that I 1 + I 2 + I 3 = 7.5 A, the total current from (b)

11. 120 V 5. Consider the complex circuit below. 12 Ω 30 Ω60 Ω 48 Ω 24 Ω (a) How many resistors does each electron pass through on its way through the circuit? Each electron goes through 3 resistors example of one way through circuit

12. 12 Ω 30 Ω60 Ω 48 Ω 24 Ω 12 R’ R’’ 5. (b) Determine the equivalent resistance of the circuit. The 30-Ω and 60-Ω resistor are in parallel, and can be combined into a single resistor, call it R’ 120 V Likewise, the 48-Ω and 24-Ω resistor are in parallel, and can be combined into a single resistor, call it R’’

13. 5. (b) Determine the equivalent resistance of the circuit. 120 V 12 Ω 30 Ω60 Ω 48 Ω 24 Ω 12 R’ R’’ 1 R’ = 1 30 + 1 60 = 2 + 1 3 =R’ = 20 Ω 1 R’’ = 1 24 + 1 48 = 2 + 1 3 =R’’ = 16 Ω

14. 5. (b) Determine the equivalent resistance of the circuit. 120 V 12 Ω 30 Ω60 Ω 48 Ω 24 Ω 12 20 16 R eq R eq = 12 + 20 + 16 Ω R eq = 48 Ω 120 V Resistors is series add

15. 5. (c) Determine the current emerging from the power source. 120 V 12 Ω 30 Ω60 Ω 48 Ω 24 Ω 12 20 16 48 Ω120 V I I V R eq I = 120 V 48 Ω = I = 2.5 A Use simpler circuit that has only one resistor

16. 5. (d) Determine V 1 and I 2. 120 V 12 Ω 30 Ω60 Ω 48 Ω 24 Ω 12 20 16 48 Ω120 V 2.5 A V1V1 I2I2 V 1 = I 1 R 1 I 1 = ? Every electron that comes out of battery goes through R 1 (see part (a)), so I 1 = 2.5 A V 1 = I 1 R 1 = (2.5 A)(12 Ω) V 1 = 30 V

17. 5. (d) Determine V 1 and I 2. 120 V 12 Ω 30 Ω 48 Ω 24 Ω 12 20 16 2.5 A V2V2 I2I2 V2V2 V2V2 R2R2 I 2 = V 2 = ?Use first equivalent circuit that is in series V 2 = I R’ = (2.5 A)(20 Ω) = 50 V = 50 V 30 Ω I 2 = 1.7 A

18. 6. Consider the complex circuit below. (a) How many resistors does each electron go through on its way around this circuit? 24 V 40 Ω 60 Ω 10 Ω 20 Ω 10 Ω 12 Ω 20 Ω An example of one way around this circuit Each electron goes through four resistors

19. 6. Consider the complex circuit below. (b) Determine the equivalent resistance of the circuit. 24 V 40 Ω 60 Ω 10 Ω 20 Ω 10 Ω 12 Ω 20 Ω 12R’ R’’ 20 The 10-Ω, 20-Ω, and 10-Ω resistors are in parallel, and can be combined into a single resistor, call it R’ Likewise, the 60-Ω and 40-Ω resistor are in parallel, and can be combined into a single resistor, call it R’’

20. 6. Consider the complex circuit below. (b) Determine the equivalent resistance of the circuit. 24 V 40 Ω 60 Ω 10 Ω 20 Ω 10 Ω 12 Ω 20 Ω 12R’ R’’ 20 1 R’ = 1 10 + 1 20 = 2 + 1 + 2 20 R’ = 4 Ω 1 R’’ = 1 60 + 1 40 = 2 + 3 120 5 =R’’ = 24 Ω + 1 10 = 5 20

21. 6. Consider the complex circuit below. (b) Determine the equivalent resistance of the circuit. 24 V 40 Ω 60 Ω 10 Ω 20 Ω 10 Ω 12 Ω 20 Ω 124 24 20 R eq 24 V R eq = 4 + 12 + 20 + 24 Ω R eq = 60 Ω

22. (b) Determine the current emerging from the power source. 24 V 40 Ω 60 Ω 10 Ω 20 Ω 10 Ω 12 Ω 20 Ω 124 24 20 60 Ω24 V I I V R eq I = 24 V 60 Ω = I = 0.40 A

23. 24 V 40 Ω 60 Ω 10 Ω 20 Ω 10 Ω 12 Ω 20 Ω 124 24 20 0.40 A (c) Determine V 1 and I 2. 60 Ω24 V I V1V1 I2I2 V 1 = I 1 R 1 I 1 = ? Every electron that comes out of battery goes through R 1 (see part (a)), so I 1 = 0.40 A V 1 = I 1 R 1 = (0.40 A)(12 Ω) V 1 = 4.8 V

24. 24 V 40 Ω 60 Ω 10 Ω 20 Ω 10 Ω 12 Ω 20 Ω 124 24 20 0.40 A (c) Determine V 1 and I 2. V1V1 I2I2 V2V2 R2R2 I 2 = V 2 = ? Use first equivalent circuit that is in series V 2 = I R’ = (0.40 A)(24 Ω) = 9.6 V = 9.6 V 60 Ω I 2 = 0.16 A V2V2 V2V2