rc rl circuits examples, Essays (university) of Circuit Theory

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Series and Parallel considerations

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?)

Example 1 Solution

• XC = 1 2𝜋𝑓𝐶 =^

1 6.28 ×500 × 0.1×10−6^ =^

1 314.159×10−6^ = 3.183kΩ

• ZT = 𝑋𝐶^2 + 𝑅^2 =

3.183 × 10^3 2 + 2.2 × 10^3 2 =

10.132 × 10^6 + 4.84 × 10^6 =

14.972 × 10^6 = 3.869kΩ

• 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

Example 1 Phasor diagram VR = (Adj) 2.843V

VC= (Opp) 4.113V

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°

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

• IT = 𝐼𝐶^2 + 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

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°

Example 3 Circuit

VT 5 Vrms 500 Hz 0°

L 100mH

R 1kΩ

XL =? VL =?

ZT =? VR =? I =? θ =?

• 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

Quick check:

• VT = 𝑉𝐿^2 + 𝑉𝑅^2 = 1.499 2 + 4.77 2 =

2.247 + 22.753 = 24.999 = 4.999 ≅ 5V

Phasor Triangle to solve for θ

• tan θ = 𝑂𝑝𝑝 𝐴𝑑𝑗 =^

𝑉𝐿 𝑉𝑅^ =^

4.77 ∴^ θ =^ tan

−1 1. 4.77 = tan−1^ 0.314 = 17.446°

Quick check:

• sin θ = 𝑂𝑝𝑝𝐻𝑦𝑝 = (^) 𝑉𝑉𝐿 𝑇

= 1.499 5 ∴ θ = sin−1 1.499 5 = sin−1^ 0.299 = 17.446°

Example 4 Circuit

XL =? IT =?

IL =? Zeq =? IR =? θ 1 =?

L 100mH

R 1kΩ

VA

5 Vrms 2kHz 0°