rc and rl circuits - electronics – online distance learning
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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