1. Series/Parallel Inductors and Capacitors
Chapter 6 – Inductance, Capacitance and Mutual Inductance
Terminology
Inductors
Capacitors
Series/Parallel Inductors and Capacitors
Mutual Inductance

2. 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.

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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)

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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.

16. 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

17. 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

18. 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

19. 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 )

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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