RC and RL Circuits

Series and Parallel considerations

RC and RL Circuits Rules to remember

• ELI the ICE man: Voltage (E) leads Current (I) in an Inductive (L) circuit , whereas Current (I) leads Voltage (E) in a Capacitive (C) circuit

– This is only true for SERIES circuits. When it goes into a parallel configuration, the opposite occurs

• Current leads Voltage in a Parallel Inductive circuit

• Voltage leads Current in a Parallel Capacitive circuit

• This makes the parallel statement something like ILE get ECI stuff (maybe?)

RC and RL Circuits Example 1 Circuit

VT

5 Vrms

500 Hz

0°

C

0.1µF

R2.2kΩ

XC = ? ZT = ? VC = ? IT = ? VR = ? θZ = ?

RC and RL Circuits Example 1 Solution

• XC = 1

2𝜋𝑓𝐶 =

1

6.28 ×500 × 0.1×10−6 = 1

314.159×10−6 =

3.183kΩ

• ZT = 𝑋𝐶2 + 𝑅2 =

3.183 × 103 2 + 2.2 × 103 2 =

10.132 × 106 + 4.84 × 106 =

14.972 × 106 = 3.869kΩ

RC and RL Circuits

• IT = 𝑉

𝑍𝑇 =

5

3.869𝑘Ω = 1.292mA Since this is a

series circuit, all of the values of I should be equal

• VR = IR = 1.292mA × 2.2kΩ = 2.843V

• VC = IXC = 1.292mA × 3.183kΩ = 4.113V

RC and RL Circuits

Quick check:

• VT = 𝑉𝐶2 + 𝑉𝑅

2 = 2.483 2 + 4.113 2 =

8.083 + 16.917 = 24.999 = 4.999 ≅ 5V

RC and RL Circuits Example 1 Phasor diagram

VR = 2.843V(Adj)

VC= 4.113V(Opp)

RC and RL Circuits Phasor Triangle to solve for θ

• tan θ = 𝑂𝑝𝑝

𝐴𝑑𝑗 =

𝑉𝐶

𝑉𝑅 =

4.113𝑉

2.843𝑉 ∴ θ = tan−1 4.113𝑉

2.843𝑉 =

tan−1 1.447 = 55.347°

Quick check:

• cos θ = 𝐴𝑑𝑗

𝐻𝑦𝑝 =

𝑉𝑅

𝑉𝑇 =

2.843𝑉

5𝑉 ∴ θ = cos−1 2.843𝑉

5𝑉 =

cos−1 0.569 = 55.347°

RC and RL Circuits Example 2 Circuit

VA

5 Vrms

500 Hz

0°

C0.47µF

R680Ω

XC = ? IT = ? IC = ? Zeq = ? IR = ? θ = ?

RC and RL Circuits Example 2 Solution

• XC = 1

2𝜋𝑓𝐶 =

1

6.28 × 500 × 0.47×10−6 = 1

1.477×10−3 =

677.255Ω

• IC = 𝑉𝐴

𝑋𝐶 =

5

677.255 = 7.383mA

• IR = 𝑉𝐴

𝑅 =

5

680 = 7.353mA

RC and RL Circuits

• IT = 𝐼𝐶2 + 𝐼𝑅

2 =

7.383 × 10−3 2 + 7.353 × 10−3 2 =

54.504 × 10−6 + 54.065 × 10−6 =

108.57 × 10−6 = 10.42mA

• Zeq = 𝑉𝐴

𝐼𝑇 =

5

10.42𝑚𝐴 = 479.846Ω

RC and RL Circuits Example 2 Phasor diagram

IR = 7.353mA(Adj)

IC = 7.383mA(Opp)

RC and RL Circuits Phasor Triangle to solve for θ

• tan θ = 𝑂𝑝𝑝

𝐴𝑑𝑗 =

𝐼𝐶

𝐼𝑅 =

7.383𝑚𝐴

7.353𝑚𝐴 ∴ θ = tan−1 7.383𝑚𝐴

7.353𝑚𝐴 =

tan−1 1.004 = 45.117°

Quick check:

• cos θ = 𝐴𝑑𝑗

𝐻𝑦𝑝 =

𝐼𝑅

𝐼𝑇 =

7.353𝑚𝐴

10.42𝑚𝐴 ∴ θ = cos−1 7.353𝑚𝐴

10.42𝑚𝐴

= cos−1 0.706 = 45.117°

RC and RL Circuits Example 3 Circuit

VT

5 Vrms

500 Hz

0°

L

100mH

R1kΩ

XL = ? VL = ? ZT = ? VR = ? I = ? θ = ?

RC and RL Circuits Example 3 Solution

• XL = 2𝜋𝑓𝐿 = 6.28 × 500 × 100mH = 314.159Ω

• ZT = 𝑋𝐿2 + 𝑅2 =

314.159 2 + 1 × 103 2 =

98.696 × 103 + 1 × 106 = 1.099 × 106 = 1.048kΩ

RC and RL Circuits

• IT = 𝑉

𝑍𝑇 =

5

1.048𝑘Ω = 4.77mA Since this is a

series circuit, all of the values of I should be equal

• VR = IR = 4.77mA × 1kΩ = 4.77V

• VL = IXL = 4.77mA × 314.159Ω = 1.499V

RC and RL Circuits

Quick check:

• VT = 𝑉𝐿2 + 𝑉𝑅

2 = 1.499 2 + 4.77 2 =

2.247 + 22.753 = 24.999 = 4.999 ≅ 5V

RC and RL Circuits Example 3 Phasor diagram

y

x

VR = 4.77V(Adj)

VL = 1.499V(Adj)

RC and RL Circuits Phasor Triangle to solve for θ

• tan θ = 𝑂𝑝𝑝

𝐴𝑑𝑗 =

𝑉𝐿

𝑉𝑅 =

1.499

4.77 ∴ θ = tan−1 1.499

4.77 =

tan−1 0.314 = 17.446°

Quick check:

• sin θ = 𝑂𝑝𝑝

𝐻𝑦𝑝 =

𝑉𝐿

𝑉𝑇 =

1.499

5 ∴ θ = sin−1 1.499

5 =

sin−1 0.299 = 17.446°

RC and RL Circuits Example 4 Circuit

XL = ? IT = ? IL = ? Zeq = ? IR = ? θ1 = ?

L100mH

R1kΩ

VA

5 Vrms

2kHz

0°

RC and RL Circuits Example 4 Solution

• XL = 2𝜋𝑓𝐿 = 6.28 × 2k Hz × 100mH = 1.257kΩ

• IL = 𝑉𝐴

𝑋𝐿 =

5

1.257𝑘 = 3.979mA

• IR = 𝑉𝐴

𝑅 =

5

1𝑘 = 5mA

RC and RL Circuits

• IT = 𝐼𝐿2 + 𝐼𝑅

2 =

3.979 × 10−3 2 + 5 × 10−3 2 =

15.831 × 10−6 + 25 × 10−6 =

40.831 × 10−6 = 6.39mA

• Zeq = 𝑉𝐴

𝐼𝑇 =

5

6.39𝑚𝐴 = 782.479Ω

RC and RL Circuits Example 4 Phasor diagram

IR = 5.00 mA(Adj)

IL= 3.979 mA(Opp)

RC and RL Circuits Phasor Triangle to solve for θ

• tan θ = 𝑂𝑝𝑝

𝐴𝑑𝑗 =

𝐼𝐿

𝐼𝑅 =

3.979𝑚𝐴

5𝑚𝐴 ∴ θ = tan−1 3.979𝑚𝐴

5𝑚𝐴 =

tan−1 0.796 = 38.512°

Quick check:

• cos θ = 𝐴𝑑𝑗

𝐻𝑦𝑝 =

𝐼𝑅

𝐼𝑇 =

5𝑚𝐴

6.39𝑚𝐴 ∴ θ = cos−1 5𝑚𝐴

6.39𝑚𝐴 =

cos−1 0.782 = 38.512°

The End