2.  Combination or compound circuits contain both series and parallel connections.  Crucial that we understand the differences between series and parallel circuits.

4.  With the R total = 2Ω + 2Ω + 6Ω R total = 10Ω  Determine the total current in the circuit by using Ohm’s law. ∆V total = 12V R eq = 10Ω ∆V =IR I tot = ∆V/R I tot = 12V/10Ω I tot = 1.2A

5.  Resistor 1 and 4 are in Series, so I tot = I 1 = I 4 = 1.2 A  For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. So, I 2 = I 3 = 0.6 A

6.  Now that the current at each individual resistor location is known, use Ohm's law to determine the voltage drop across each resistor. V 1 = I 1 R 1 = (1.2A) (2Ω) = 2.4V V 2 = I 2 R 2 = (0.6A) (4Ω) = 2.4V V 3 = I 3 R 3 = (0.6A) (4Ω) = 2.4V V 4 = I 4 R 4 = (1.2A) (6Ω) = 7.2V

8. For the parallel part: 1 / R eq = 1/(4Ω) + 1/(12Ω) R eq = 3.00Ω For the total circuit. R tot = R 1 + R eq + R 4 = 5Ω + 3Ω+ 8Ω R tot = 16 Ω

9. I tot = V tot / R tot = (24V) / (16Ω) I tot = 1.5 Amp RResistors R 1 and R 4 are in series so, I tot = I 1 = I 4 = 1.5 Amp RResistors R 2 and R 3 are parallel, this means I 2 + I 3 = 1.5 Amp.

10. IIn our last example, R 2 and R 3 were the same. TTo determine the voltage drop R 2 and R 3, the voltage drop across the two series-connected resistors (R 1 and R 4 ) must first be determined. V 1 = I 1 R 1 = (1.5A) (5 ):V 1 = 7.5 V V 4 = I 4 R 4 = (1.5A) (8 ):V 4 = 12 V

11. WWe know, V tot = 24V and the 19.5V was consumed by R 1 and R 4. So, V 2 = V 3 = 4.5 V AApply Ohm’s law to determine the current at R 2 and R 3. I 2 = V 2 /R 2 = (4.5V)/(4Ω):I 2 = 1.125 A I 3 = V 3 /R 3 = (4.5V)/(12Ω):I 3 = 0.375 A

13.  Pg. 728 # 36 & 37