Procedure This figure represents two impedances connected in series with an ac voltage source. **Step 1**--Determine the total impedance of the circuit. This is a series circuit, so: Z_{T} = Z_{1} + Z_{2} Since this is a summation operation, the impedances are shown in rectangular form. The next step, however, uses Z_{T} in a division operation, so you should convert it to polar form as part of this step. **Step 2**--Use Ohm's law to determine the total current: **Step 3**--Determine the currents for Z_{1} and Z_{2}. These impedances are connected in series with the source voltage, so it follows that the currents are going to be equal to the source, or total, current: I_{Z1} = I_{T} I_{Z2} = I_{T} **Step 4**--Use Ohm's law to determine the voltage drops across Z_{1} and Z_{2}: V_{Z1} = I_{Z1}Z_{1} V_{Z2} = I_{Z2}Z_{2} This completes the basic analysis of this ac circuit. To check your results, convert the voltages from Step 4 into rectangular form and show that: V_{Z1} + V_{Z2} = V_{T} | Example Given: Z_{1} = 50Ð0° W, Z_{2} = 10Ð20° W, V_{T} = 12 V Find: Z_{T}, I_{T} ,V_{Z1}, and V_{Z2} Solution: **Step 1**--Determine the total impedance of the circuit Z_{1} = 50Ð0° W = 50 +j0 W Z_{2} = 10Ð20° W = 9.4 +j3.42 W Z_{T} = Z_{1} + Z_{2} Z_{T} = 50 +j0 + 9.4 +j3.42 Z_{T} = 59.4 +j3.42 W **Z**_{T} =59.5Ð3.3° W **Step 2**--Use Ohm's law to determine the total current I_{T} = 12Ð0° / 59.5Ð3.3° **I**_{T} = 0.20Ð-3.3° A **Step 3**--Determine the currents for Z_{1} and Z_{2} **I**_{Z1} = 0.20Ð-3.3° A I_{Z2} = 0.20Ð-3.3° A **Step 4**--Use Ohm's law to determine the voltage drops across Z_{1} and Z_{2} V_{Z1} = (0.20Ð-3.3°)(50Ð0°) **V**_{Z1} = 10Ð-3.3° V V_{Z2} = (0.20Ð-3.3°)(10Ð20°) **V**_{Z2} = 2Ð16.7° V **Check** V_{Z1} = 10Ð-3.3° = 9.98 -j0.58 V_{Z2} = 2Ð16.7° = 1.92 +j0.57 (9.98 -j0.58) + (1.92 +j0.57) = **11.9 -j0.01** **(close enough to V**_{T} = 12 +j0 V) |