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Series/Parallel Inductors and Capacitors
Chapter 6 – Inductance, Capacitance and Mutual Inductance Terminology Inductors Capacitors Series/Parallel Inductors and Capacitors Mutual Inductance
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Terminology Inductance– Relates the induced voltage to the current
Capacitance – Relates the displacement current to the voltage Passive Elements – cannot generate energy Mutual Inductance– The voltage induced in one circuit that is related to The time varying current in another circuit.
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The Inductor 𝑣=𝐿 𝑑𝑖 𝑑𝑡 v = volts L = inductance (in Henrys) i = amperes t =in seconds Inductors store energy in terms of an electrical field
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The Inductor If current is constant, the voltage across the ideal inductor is zero, making The inductor behave like a short in the presence of a constant or DC current. Current cannot change instantaneously in an inductor, it would require an infinite voltage which is not possible. Inductors block AC and allow DC to flow
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The Inductor 𝑖=10𝑡 𝑒 −5𝑡 𝑑𝑖 𝑑𝑡 =10𝑡 𝑒 −5𝑡 −5 + 𝑒 −5𝑡 (10)
Example: The independent current source shown generates zero current for t < 0 and a pulse 10𝑡 𝑒 −5𝑡 A for t > 0. Sketch the current waveform. At what instant of time is the current Maximum 𝑖=10𝑡 𝑒 −5𝑡 Maximum occurs when the derivative is zero 𝑑𝑖 𝑑𝑡 =10𝑡 𝑒 −5𝑡 −5 + 𝑒 −5𝑡 (10) 𝑖 𝑚𝑎𝑥 = 𝑒 −5∗0.2 =0.736 0=−50𝑡 𝑒 −5𝑡 +10 𝑒 −5𝑡 50𝑡 𝑒 −5𝑡 =10 𝑒 −5𝑡 50𝑡=10 𝑡= =0.2
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The Inductor 100 mH inductor as a function of time.
C) Express the voltage across the terminals of the 100 mH inductor as a function of time. D) Sketch the voltage waveform 𝑣=𝐿 𝑑𝑖 𝑑𝑡 =(0.1)(−50𝑡 𝑒 −5𝑡 +10 𝑒 −5𝑡 ) 𝑣 0.2 =0𝑉 𝑣 0 = 𝑒 − −5 0 =1𝑉 𝑣=−5𝑡 𝑒 −5𝑡 +1 𝑒 −5𝑡 𝑣= 𝑒 −5𝑡 1−5𝑡 0= 𝑒 −5𝑡 1−5𝑡 0=1−5𝑡 𝑡=0.2
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The Inductor Current in terms of voltage: 𝑣=𝐿 𝑑𝑖 𝑑𝑡
Integrate the left side only 𝑣=𝐿 𝑑𝑖 𝑑𝑡 𝑖 𝑡 −𝑖 𝑡 0 = 1 𝐿 𝑡 0 𝑡 𝑣𝑑𝜏 𝑣 𝑑𝑡=𝐿 𝑑𝑖 𝑑𝑡 𝑑𝑡 Multiply both sides by the differential of time 𝑖 𝑡 = 1 𝐿 𝑡 0 𝑡 𝑣𝑑𝜏 +𝑖 𝑡 0 𝑣 𝑑𝑡=𝐿 𝑑𝑖 𝑡 0 𝑡 𝑣𝑑𝜏 = 𝑖 𝑡 0 𝑖(𝑡) 𝐿𝑑𝑥 In most practical applications 𝑡 0 =0 𝑖 𝑡 = 1 𝐿 0 𝑡 𝑣𝑑𝜏 +𝑖 0 𝐿 𝑖 𝑡 0 𝑖(𝑡) 𝑑𝑥 = 𝑡 0 𝑡 𝑣𝑑𝜏 Re-Write so solving Is more convenient
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The Inductor 𝑣 𝑡 =20𝑡 𝑒 −10𝑡 𝑡= 20 200 =0.1
Determine current given voltage Example: The voltage pulse applied to the 100mH inductor is 0 for t<0 and 𝑣 𝑡 =20𝑡 𝑒 −10𝑡 𝑉 for t >0. Sketch the voltages as a function of time. Find the inductor current as a function of time. Sketch the current as a function of time. 𝑣 𝑡 =20𝑡 𝑒 −10𝑡 𝑡= =0.1 𝑣 ′ 𝑡 =20𝑡 −10∗ 𝑒 −10𝑡 + 𝑒 −10𝑡 (20) 𝑣 0.1 = 𝑒 − =0.736 𝑣 ′ 𝑡 =−200𝑡 𝑒 −10𝑡 +20 𝑒 −10𝑡 𝑣 ′ 𝑡 = 𝑒 −10𝑡 (−200𝑡+20) Max when 𝑣 ′ 𝑡 =0 0= 𝑒 −10𝑡 (−200𝑡+20) 0=(−200𝑡+20)
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The Inductor 𝑖(𝑡)= 1 𝐿 0 𝑡 𝑣 𝑑𝜏 +𝑖(0) 𝑖= 1 0.1 0 𝑡 20𝜏 𝑒 −10𝜏 𝑑𝜏 +0
b) Find the inductor current as a function of time. 𝑖(𝑡)= 1 𝐿 0 𝑡 𝑣 𝑑𝜏 +𝑖(0) 𝑖= 𝑡 20𝜏 𝑒 −10𝜏 𝑑𝜏 +0 i=2 1−10𝑡 𝑒 −10𝑡 − 𝑒 −10𝑡 𝐴 c) Sketch the current as a function of time. 𝑖=200 0 𝑡 𝜏 𝑒 −10𝜏 𝑑𝜏 +0 Place the equation into your calculator. Use the integral table on pg 759 Appendix G 𝑖=200 − 𝑒 −10𝜏 𝜏+1 | 𝑡 0
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The Inductor 𝑝=𝑣𝑖 𝑝=𝑣 1 𝐿 𝑡 0 𝑡 𝑣𝑑𝜏+𝑖( 𝑡 0 ) 𝑝=𝐿𝑖 𝑑𝑖 𝑑𝑡
Power and energy in the inductor: 𝑝=𝑣𝑖 In terms of voltage In terms of current 𝑝=𝑣 1 𝐿 𝑡 0 𝑡 𝑣𝑑𝜏+𝑖( 𝑡 0 ) 𝑝=𝐿𝑖 𝑑𝑖 𝑑𝑡 In terms of energy 𝑝= 𝑑𝑤 𝑑𝑡 =𝐿𝑖 𝑑𝑖 𝑑𝑡 𝑑𝑤=𝐿𝑖 𝑑𝑖 0 𝑤 𝑑𝑥 =𝐿 0 𝑖 𝑦𝑑𝑦 𝑤= 1 2 𝐿 𝑖 2
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The Inductor 𝑝=𝑣𝑖 𝑝= 𝑒 −5𝑡 1−5𝑡 ∗10𝑡 𝑒 −5𝑡
Example: Determine plots of I, v, p, and w versus time. Done in previous example 𝑝=𝑣𝑖 𝑝= 𝑒 −5𝑡 1−5𝑡 ∗10𝑡 𝑒 −5𝑡 Done in previous example 𝑣= 𝑒 −5𝑡 (1−5𝑡) 𝑝=10𝑡 𝑒 −10𝑡 −50 𝑡 2 𝑒 −10𝑡 𝑊𝑎𝑡𝑡𝑠 𝑤= 1 2 𝐿 𝑖 2 𝑤= 1 2 ∗.1∗ 10𝑡 𝑒 −5𝑡 2 𝑤=5 𝑡 2 𝑒 −10𝑡 Joules
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The Inductor Example: In what time interval is energy being stored in the inductor? From 0 to 0.2s, also corresponds to p>0 In what time interval is energy being extracted from the inductor? From 0.2s to ∞, also corresponds to p<0 What is the maximum energy stored in the inductor? From previous example maximum current is 0.736A 𝑤= 1 2 𝐿 𝑖 2 𝑤= ∗ =27.08𝑚𝐽
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The Inductor Example: Evaluate the integrals of power. From before:
𝑤= 𝑝𝑑𝑡 𝑤= 0.2 ∞ 𝑝𝑑𝑡 𝑝=10𝑡 𝑒 −10𝑡 −50 𝑡 2 𝑒 −10𝑡 𝑊𝑎𝑡𝑡𝑠 Use a larger number like 100 to evaluate In your calculator 𝑤= 𝑡 𝑒 −10𝑡 −50 𝑡 2 𝑒 −10𝑡 𝑑𝑡 0.2 ∞ 10𝑡 𝑒 −10𝑡 −50 𝑡 2 𝑒 −10𝑡 𝑑𝑡 Use your calculator to save time Use your calculator to save time 𝑝=27.1𝑚𝐽 𝑝=−27.1𝑚𝐽 Energy Stored Energy Extracted After current peaks, no energy is store in the inductor
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The Capacitor 𝑖=𝐶 𝑑𝑣 𝑑𝑡 Capacitors store electric charge i in Amperes
C in Farads V in Volts t in seconds Capacitors store electric charge Passive sign convention
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The Capacitor Voltage cannot change instantaneously across the terminals of a capacitor or it would produce infinite current If voltage across the terminals is constant, the capacitor current is zero, because a conduction current cannot be established in the dielectric material. It must have a time varying voltage to operate Therefore a capacitor acts like an open circuit in the presence of a constant voltage AC can pass through better, but a capacitor does not block DC.
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The Capacitor 𝑖=𝐶 𝑑𝑣 𝑑𝑡 𝑡 0 𝑡 𝑑𝑣 = 1 𝐶 𝑡 0 𝑡 𝑖 𝑑𝜏 𝑖𝑑𝑡=𝐶𝑑𝑣
In terms of voltage 𝑖=𝐶 𝑑𝑣 𝑑𝑡 𝑡 0 𝑡 𝑑𝑣 = 1 𝐶 𝑡 0 𝑡 𝑖 𝑑𝜏 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑖𝑑𝑡=𝐶𝑑𝑣 𝑣 𝑡 −𝑣 𝑡 0 = 1 𝐶 𝑡 0 𝑡 𝑖 𝑑𝜏 𝑖 𝐶 𝑑𝑡=𝑑𝑣 For simplicity 𝑣 𝑡 = 1 𝐶 𝑡 0 𝑡 𝑖 𝑑𝜏 +𝑣 𝑡 0 1 𝐶 𝑡 0 𝑡 𝑖 𝑑𝜏 = 𝑡 0 𝑡 𝑑𝑣 𝑡 0 =0 𝑣 𝑡 = 1 𝐶 𝑡 0 𝑡 𝑖 𝑑𝜏 +𝑣 0
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The Capacitor In terms of power 𝑖=𝐶 𝑑𝑣 𝑑𝑡 𝑝=𝑣𝑖=𝐶𝑣 𝑑𝑣 𝑑𝑡 Alternatively:
Substitute Alternatively: V calculated previously 𝑝=𝑖 1 𝐶 𝑡 0 𝑡 𝑖 𝑑𝜏 +𝑣 0
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The Capacitor In terms of energy 𝑝= 𝑑𝑤 𝑑𝑡 𝑥| 𝑤 0 =𝐶 𝑦 2 2 | 𝑣 0
𝑝= 𝑑𝑤 𝑑𝑡 𝑥| 𝑤 0 =𝐶 𝑦 2 2 | 𝑣 0 𝑝=𝑣𝑖=𝐶𝑣 𝑑𝑣 𝑑𝑡 𝑑𝑤=𝑝𝑑𝑡 𝑑𝑤=𝐶𝑣 𝑑𝑣 𝑑𝑡 ∗𝑑𝑡 𝑤−0=𝐶 𝑣 2 2 − Integrate 𝑑𝑤=𝐶𝑣𝑑𝑣 𝑤= 1 2 𝐶 𝑣 2 0 𝑤 𝑑𝑥 =𝐶 0 𝑣 𝑦𝑑𝑦 Energy in Joules
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The Capacitor 𝑣 𝑡 = 0, 4𝑡 𝑉 4 𝑒 − 𝑡−1 𝑉 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠
Example: The voltage pulse described by the following equations is impressed across the terminals of a 0.5𝜇𝐹 Capacitor 𝑣 𝑡 = 0, 4𝑡 𝑉 4 𝑒 − 𝑡−1 𝑉 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 Derive the expressions for the capacitor current: 𝑖=𝐶 𝑑𝑣 𝑑𝑡 =𝐶∗𝑣′(𝑡) Take the derivative of each voltage with respect to time 𝑖= 0, 2𝜇𝐴 −2 𝑒 − 𝑡−1 𝜇𝐴 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 𝑣′ 𝑡 = 0, 4 𝑉 −4 𝑒 − 𝑡−1 𝑉 𝑖= 0.5× 10 −6 (0), 0.5× 10 −6 (4), 0.5× 10 −6 (−4 𝑒 − 𝑡−1 )
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The Capacitor 𝑣 𝑡 = 0, 4𝑡 𝑉 4 𝑒 − 𝑡−1 𝑉 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠
Example: The voltage pulse described by the following equations is impressed across the terminals of a 0.5𝜇𝐹 Capacitor 𝑣 𝑡 = 0, 4𝑡 𝑉 4 𝑒 − 𝑡−1 𝑉 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 𝑖= 0, 2𝜇𝐴 −2 𝑒 − 𝑡−1 𝜇𝐴 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 Previously calculated Derive the expressions for the power: 𝑝=𝑣𝑖 Substitute in for v and i 𝑝= 0, 8𝑡 𝜇𝑊 −8 𝑒 −2 𝑡−1 𝜇𝑊 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 𝑝= 0∗0, 4𝑡 ∗2𝜇 4 𝑒 − 𝑡−1 ∗−2 𝑒 − 𝑡−1 𝜇
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The Capacitor 𝑣 𝑡 = 0, 4𝑡 𝑉 4 𝑒 − 𝑡−1 𝑉 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠
Example: The voltage pulse described by the following equations is impressed across the terminals of a 0.5𝜇𝐹 Capacitor 𝑣 𝑡 = 0, 4𝑡 𝑉 4 𝑒 − 𝑡−1 𝑉 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 Derive the expressions for the energy: 𝑤= 1 2 𝐶 𝑣 2 Substitute in for C and v 𝑤= 0, 4 𝑡 2 𝜇𝐽 4 𝑒 −2 𝑡−1 𝜇𝐽 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 𝑤= 0, 𝑡 (0.5)16 𝑒 −2 𝑡−1
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The Capacitor 𝑣 𝑡 = 0, 4𝑡 𝑉 4 𝑒 − 𝑡−1 𝑉 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 Example:
The voltage pulse described by the following equations is impressed across the terminals of a 0.5𝜇𝐹 Capacitor 𝑣 𝑡 = 0, 4𝑡 𝑉 4 𝑒 − 𝑡−1 𝑉 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 Sketch the voltage, current, power, and energy as functions of time: 𝑖= 0, 2𝜇𝐴 −2 𝑒 − 𝑡−1 𝜇𝐴 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 𝑤= 0, 4 𝑡 2 𝜇𝐽 4 𝑒 −2 𝑡−1 𝜇𝐽 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 𝑝= 0, 8𝑡 𝜇𝑊 −8 𝑒 −2 𝑡−1 𝜇𝑊 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠
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The Capacitor Example: Specify the interval of time when energy is being stored in the capacitor. 𝑊ℎ𝑒𝑛 0≤𝑡≤1, 𝑤ℎ𝑒𝑛𝑒𝑣𝑒𝑟 𝑝𝑜𝑤𝑒𝑟 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 Specify the interval of time when energy is delivered by the capacitor. 𝑊ℎ𝑒𝑛 𝑡>1, 𝑤ℎ𝑒𝑛𝑒𝑣𝑒𝑟 𝑝𝑜𝑤𝑒𝑟 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒
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The Capacitor Evaluate the integrals for the capacitors energy
Example: Evaluate the integrals for the capacitors energy 𝑝= 0, 8𝑡 𝜇𝑊 −8 𝑒 −2 𝑡−1 𝜇𝑊 𝑡≤0𝑠 0𝑠≤𝑡≤1𝑠 𝑡≥1𝑠 𝑤= 1 ∞ 𝑝 𝑑𝑡 𝑤= 0 1 𝑝 𝑑𝑡 Use a larger value to replace infinity 𝑤= 0 1 8𝑡 𝑑𝑡 𝑤= 1 ∞ −8 𝑒 −2 𝑡−1 𝑑𝑡 Use your calculator for the derivatives 𝑤=4𝜇𝐽 𝑤=−4𝜇𝐽
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Series-Parallel Combinations of Inductance and Capacitance
Inductors in Series Inductors in Parallel 𝑇𝑜𝑡𝑎𝑙 𝐼𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒=𝐿1+𝐿2+𝐿3 𝑇𝑜𝑡𝑎𝑙 𝐼𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒= 𝐿1 −1 + 𝐿2 −1 + 𝐿3 −1 −1 𝐿 𝑇 =60𝑚𝐻+120𝑚𝐻+240𝑚𝐻=420𝑚𝐻 𝐿 𝑇 = 60 − − −1 −1 =34.28𝑚𝐻
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Series-Parallel Combinations of Inductance and Capacitance
Capacitors in Series Capacitors in Parallel 𝑇𝑜𝑡𝑎𝑙 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒= 𝐶1 −1 + 𝐶2 −1 + 𝐶3 −1 −1 𝑇𝑜𝑡𝑎𝑙 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒=𝐶1+𝐶2+𝐶3 𝐶 𝑇 = 20 − − −1 −1 𝐶 𝑇 =𝐶1+𝐶2+𝐶3 𝐶 𝑇 =12.5𝑚𝐻 𝐶 𝑇 = =170𝜇𝐹
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