1. Voltage Divider and Current Divider Rules
Week 10
Voltage Divider and Current Divider Rules

2. After completing this lesson, you should be able to apply:
voltage divider rule
current divider rule, and
in steady-state ac circuit analysis.
Learning Outcomes

3. Voltage Divider Rule
The voltage divider rule is a shortcut method for determining the voltage drop across an impedance within a series circuit.
To derive the voltage divider rule, consider the following phasor circuit, where N impedances are connected in series across an ac voltage source, V. KVL gives us
Now, Ohm’s law gives us

4. Voltage Divider Rule (continued)
Hence, we can write
The above equation can be rearranged into the more familiar form as
Similarly, it can be shown that

5. Thus, the general equation for finding the voltage drop across impedance Zi is
Note
The voltage divider rule is very useful in circuit analysis because it allows us to determine the voltage drop across an impedance within a series circuit without having to find the circuit current first.

6. The Voltage Division for two impedances in series is:

7. Worked Example Determine the voltage drop across ZR. Note
This is a badly formulated question I found on the internet. The circuit drawn mixes quantities in the time domain with those in the phasor domain. One must be careful not to apply KVL directly to the circuit.

8. Solution Obtain a phasor representation for the supply voltage.
Find the complex impedance of the capacitor.
We are given the value Xc = 5 Ω. Hence, the complex impedance of the capacitor is
Draw the phasor domain circuit.

9. Solution Use the voltage divider rule to determine VR. Hence,
vR(t)= P-1{vR(t)} = 8.94cos(6280t – 63.43o) V

10. Worked Example
Determine vL(t), and i(t). L
i(t)

11. Solution Obtain a phasor representation for the supply voltage.
Find the complex impedance of the inductor.
We are given the value L = 0.2 H. Hence, the complex impedance of the inductor is
Draw the phasor domain circuit.

12. Solution Use the voltage divider rule to determine VL.
Use Ohm’s law to find phasor current I.
Hence,
vL(t)= P-1{VL} = 2.236cos(10t – 26.56o) V
i(t)= P-1{I)} = 1.118cos(10t – o) A

13. Worked Example
Calculate vo in the following circuit.

14. Solution Obtain a phasor representation for the supply voltage.
Find the complex impedance of the inductor.
We are given the value L = 0.5 H. Hence, the complex impedance of the inductor is
Find the complex impedance of the capacitor.

15. Draw the phasor domain circuit.
Replace the parallel-connected inductor and resistor with an equivalent impedance.

16. The simplified phasor domain circuit is shown below.
Use the voltage divider rule to determine Vo.
Hence,
vo(t)= P-1{Vo} = 7.07cos(10t – 60o) V

17. Worked Example
Find the magnitude and phase of Vo at 5 kHz.

18. Solution Calculate the complex impedances of the two inductors.
Draw the phasor domain circuit.

19. Simplify the circuit by replacing the series-connected inductor and resistor with an equivalent impedance. VX
Note here that the voltage across Z1 is VX, not VO.

20. Further simplify the circuit by replacing the parallel-connected elements with an equivalent impedance. VX

21. Use voltage divider rule to determine VX.
Use voltage divider rule to determine VO.

22. Therefore,
Hence,
vo(t)= P-1{Vo} = 1.716cos(t – o) V

23. Worked Example
Find the voltage vc(t) in steady state. Find the phasor current through each element, and construct a phasor diagram showing the currents and source voltage. 23 23

24. Solution Write a phasor representation for the source voltage. Given
Therefore
Find the complex impedances of the inductor and the capacitor.
Given: L = 0.1 H; C = 10 µF. Hence,

25. Draw the phasor domain circuit.
Replace the parallel-connected elements with an equivalent impedance.

26. The simplified phasor domain circuit is shown below.
ZA= 50 – j50
Use voltage divider rule to determine VC.
Therefore,
and

27. Label the circuit currents.
Use Ohm’s law to determine IR, IC.
KCL gives

28. Therefore,
and
Phasor diagram

29. Worked Example
Find the output voltage vo if 10 V (peak) sinusoidal voltage is applied.

30. Solution
Define the source voltage as the reference phasor. With this definition we can then write
Replace the parallel-connected elements with an equivalent impedance.
Let ;
Therefore

31. Draw the simplified circuit.
Use voltage divider rule to determine V1.
Therefore,

32. Use voltage divider rule again to determine Vo.
Therefore, V1

33. Worked Example
Find v(t) in the RLC circuit shown below.

34. Solution
Define the source voltage as the reference phasor. With this definition we can then write
Replace the parallel-connected elements with an equivalent impedance.
Let ;
Therefore

35. Draw the simplified circuit.
1 Ω
Use voltage divider rule to determine .
Therefore,
giving

36. Worked Example
Find the steady-state expression for vo(t) if vS(t) = 64cos(8000t).
vS(t)

37. Solution
Draw the phasor domain circuit.
Let VO

38. Solution
Replace the parallel-connected elements with an equivalent impedance.
Let
Therefore,
Draw the simplified circuit.
1 Ω

39. Use voltage divider rule to determine .
1 Ω
Therefore,
giving

40. Current Divider Rule
The current divider rule is useful for determining the current flowing through an impedance within a parallel circuit.
To derive the current divider rule, consider the following circuit, where N impedances are connected in parallel across an ac voltage source, V. KCL gives us
Now, Ohm’s law gives us ; ;
…….

41. Therefore, we can write
Thus,

42. Current Divider Rule for Circuit with Two Parallel Impedances
For the simple case where there are only two parallel impedances across the voltage source V, we have

43. Current Divider Rule for Circuit with Two Parallel Impedances
Another useful current divider relationship can be obtained by taking the ratio of I1 to I2, as follows:
Therefore,
and

44. Worked Example
Find the steady-state expression for io(t) and i1(t) if is(t) = 125cos(500t) mA.
iS(t)
i1(t)
io(t)

45. Solution Write a phasor representation for the source current. Given
Therefore
Find the complex impedances of the inductor and the capacitor.
Given: L = 1 H; C = 20 µF. Hence,

46. Draw the phasor domain circuit.
Replace the series-connected elements with their equivalent impedances. ;

47. The simplified phasor domain circuit is shown below.Z1 Zo
Use current divider rule to determine Io.

48. Therefore,
and

49. Worked Example
If the voltage vo across the 2-Ω resistor in the circuit is 10cos2t V, obtain iS. iS

50. Solution
Write a phasor representation for the voltage across the 2 Ω resistor.
Given
Therefore
Find the complex impedances of the inductor and the capacitor.
Given: L = 0.5 H; C = 0.1 µF. Hence,

51. Draw the phasor domain circuit.
Replace the series-connected capacitor, inductor and resistor with an equivalent impedance.

52. The simplified phasor domain circuit is shown below.
Use Ohm’s law to determine Io. io

53. Use current divider rule to determine Is.
Therefore, plugging in the known values into the above expression, we obtain
Thus,
and

54. Summary In this lesson we have looked at The voltage divider rule,
The current divider rule.