RLC Circuits elements Phasor Representation
- 2. Resistor • In dealing with RLC, what we need to do is to transform the voltage- current relationship from the time domain to the frequency domain for each element. Again, we will assume the passive sign convention. • We begin with the resistor. If the current through a resistor R is 𝑖 = 𝐼𝑚 cos 𝜔𝑡 + 𝜑 , the voltage across it is given by Ohm’s law as 𝑣 = 𝑖𝑅 = 𝑅𝐼𝑚 cos 𝜔𝑡 + 𝜑 • The phasor form of this voltage is 𝑉 = 𝑅𝐼𝑚∠φ But the phasor representation of the current is 𝐼 = 𝐼𝑚∠φ. Hence, 𝑉 = 𝑅𝐼
- 3. • Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain.
- 4. Inductor • For the inductor L, assume the current through it is 𝑖 = 𝐼𝑚 cos 𝜔𝑡 + 𝜑 . The voltage across the inductor is 𝑣 = 𝐿𝑑𝑖 𝑑𝑡 = −𝜔𝐿𝐼𝑚 s𝑖𝑛 𝜔𝑡 + 𝜑 Since – 𝑠𝑖𝑛𝐴 = cos 𝐴 + 90° , 𝑤𝑒 𝑐𝑎𝑛 𝑟𝑒𝑤𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑠 𝑣 = 𝜔𝐿𝐼𝑚 𝑐𝑜s 𝜔𝑡 + 𝜑 + 90° which transforms to the phasor 𝑉 = 𝜔𝐿𝐼𝑚𝑒𝑗(𝜑+90°) = 𝜔𝐿𝐼𝑚𝑒𝑗𝜑𝑒𝑗90° = 𝜔𝐿𝐼𝑚∠𝜑 + 90° But 𝐼𝑚 ∠𝜑 = I, and 𝑒𝑗90° = 𝑗. 𝑇ℎ𝑢𝑠 𝑉 = 𝑗𝜔𝐿𝐼
- 5. • Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.
- 6. Capacitor • For the capacitor C, assume the voltage across it is 𝑣 = 𝑉 𝑚 cos 𝜔𝑡 + 𝜑 . The current through the capacitor is 𝑖 = 𝐶𝑑𝑣 𝑑𝑡 𝐼 = 𝑗𝜔𝐶𝑽 ⇾ 𝑽 = 𝐼 𝑗𝜔𝐶
- 8. Example: The voltage 𝑣 = 12 cos 60𝑡 + 45° 𝑉 𝑖𝑠 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑎 0.1 𝐻 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑑𝑦 − 𝑠𝑡𝑎𝑡𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟. Solution: For the inductor, 𝑉 = 𝑗𝜔𝐿𝐼, 𝑤ℎ𝑒𝑟𝑒 𝜔 = 60 𝑟𝑎𝑑 𝑠 , 𝑎𝑛𝑑 𝑉 = 12∠45° V. 𝐼 = 𝑉 𝑗𝜔𝐿 = 12∠45° 𝑗60(0.1) = 12∠45° 6∠90° Converting this to time domain, 𝑖 𝑡 = 2 cos 60𝑡 − 45° 𝐴 𝐼 = 2∠ − 45° A
- 9. Example 2: If voltage 𝑣 = 10cos(100𝑡 + 30°) is applied to a 50μF capacitor, calculate the current through the capacitor. Solution: 𝐼 = 𝑗𝜔𝐶𝑽 = j(100)(50x10−6)(10∠30°) 𝐼 = (5𝑥10−3∠90°)(10∠30°) 𝐼 = 50∠120°)mA In time domain, 𝐼 = 50cos 100t + 120° mA
- 10. Impedance and Admittance • Ohm’s Law in Phasor Form for any element: • where Z is a frequency-dependent quantity known as impedance, measured in ohms.
- 11. Impedance • The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms (Ω). • The impedance represents the opposition that the circuit exhibits to the flow of sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity. • When (i.e. 𝜔 = 0,for dc sources),𝑍𝐿 = 0 and 𝑍𝐶 = ∞ confirming what we already know—that the inductor acts like a short circuit, while the capacitor acts like an open circuit. When (i.e. 𝜔 = ∞, for high frequencies), 𝑍𝐿 = ∞ and 𝑍𝐶 = 0 indicating that the inductor is an open circuit to high frequencies, while the capacitor is a short circuit.
- 12. Impedance • As a complex quantity, the impedance may be expressed in rectangular form as: 𝑍 = 𝑅 + 𝑗𝑋 • where R = Re Z is the resistance and X = Im Z is the reactance. • The reactance X may be positive or negative. We say that the impedance is inductive when X is positive or capacitive when X is negative. Thus, impedance is said to be inductive or lagging since current lags voltage, while impedance is capacitive or leading because current leads voltage. The impedance, resistance, and reactance are all measured in ohms.
- 13. Polar form of Impedance: • It is sometimes convenient to work with the reciprocal of impedance, known as admittance measured in siemens. The admittance of an element (or a circuit) is the ratio of the phasor current through it to the phasor voltage across it, or 𝑍 = 𝑍 ∠𝜃
- 14. Admittance • As a complex quantity, we may write Y as where G = Re Y is called the conductance and B = Im Y is called the susceptance. Admittance, conductance, and susceptance are all expressed in the unit of siemens (or mhos). • By rationalization,
- 15. Sample Problem: Find v(t) and i(t) in the circuit shown Solution:
- 16. • Sample 2: Find v(t) and i(t) in the circuit shown
- 17. Questions?