Voltage Divider and Current Divider Rules Week 10 Voltage Divider and Current Divider Rules

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

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

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

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.

The Voltage Division for two impedances in series is:

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.

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.

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

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

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.

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 – 116.56o) A

Worked Example Calculate vo in the following circuit.

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.

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

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

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

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

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.

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

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

Therefore, Hence, vo(t)= P-1{Vo} = 1.716cos(t – 120.387o) V

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

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,

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

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

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

Therefore, and Phasor diagram

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

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

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

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

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

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

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

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

Solution Draw the phasor domain circuit. Let VO

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

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

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 ; ; …….

Therefore, we can write Thus,

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

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

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)

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,

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

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

Therefore, and

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

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,

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

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

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

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