1. Capacitors in a circuit 1

2. Example Let’s find the current through the following capacitor with v(t)=5 cos(2000  t)V. C = 30  F What if v(t) = 5A? = 0 (A capacitor “looks like” an open circuit to DC) 2

3. Inductors Inductor is a passive circuit element which stores energy in its magnetic field. It consists of a coil of wire wrapped around a core. 3

4. Inductance Inductors are defined by their inductance (L) measured in henrys (H), where  is the magnetic flux through the core of the inductor. 4

5. Inductors in a circuit  = L i 5

6. Example Let’s find the inductor voltage assuming 1)i(t) = 5 A, then 2)i(t) = 5 cos(2000  t) A. L = 3 mH 1) = 0 (An inductor “looks like” a short circuit to DC) 2) 6

7. 20 mH 10  50 cos (100 t) V Let’s find v in this circuit. Apply KVL: You can either solve the differential equation, or transform the problem from the time domain to the frequency domain. Then the differential equation becomes a linear algebraic equation. 7

8. Time Domain Representation Frequency Domain (Phasor Representation) Circuit Component Sources M cos (  t +  ) V or AM/  V or A Resistors R  R/0   InductorsL H j  L =  L/90   CapacitorsC F 1/(j  C) = 1/(  C)/-90   8 Time Domain: Frequency Domain: Re Im Argand Diagram   M Transforming to the frequency domain is done using phasors. Start by writing each circuit quantity as a phasor.

9. In the frequency domain, R, L, and C values are expressed in ohms. These ohmic values are called “impedances.” Impedance (Z) = Resistance (R) + j Reactance (X) Z = R + jX Ohm’s Law becomes: V = I Z 9

10. 20 mH 10  50 cos (100 t) V Let’s find i. V max  Apply KVL: 50/0  V 10/0   j  L = j (100)(20  10 -3 ) j 2 = 2/90   50/0  = (10/0  ) i + (2/90  ) i Solve: 50/0  = (10/0  + 2/90  ) i 50/0  = (10.2/11.3  ) i i = = 4.9/-11.3  Transform back to time domain: i (t) = 4.9 cos (100 t – 11.3  ) A 10 Remember: j = 1/90 

11. 20 mH 10  50 cos (100 t) V Now let’s find v. Apply Ohm’s Law: 50/0  V 10/0   j  L = j (100)(20  10 -3 ) j 2 = 2/90   i (t) = 4.9 cos (100 t – 11.3  ) A i = 4.9/-11.3  v = i Z = (4.9/-11.3  ) (2/90  ) = 9.8/78.7  V Transform back to time domain: v (t) = 9.8 cos (100 t + 78.7  ) V 11

12. Let’s find v and i. 10 k  10  F 50 cos (10t + 30  ) V 50/30  V 10 4 /0   1/(j  C) = 1/[(10)(10  10 -6 )/-90  ] = 10 4 /-90   Apply KVL: 50/30  = 10 4 /0  i + 10 4 /-90  i Solve: 50/30  = (10 4 /0  + 10 4 /-90  ) i 50/30  = (1.414  10 4 /-45  ) i i = 3.54/75  mA Ohm’s Law: v = (3.54  10 -3 /75  ) (10 4 /-90  ) = 35.4/-15  V i (t) = 3.54 cos (10 t + 75  ) mA v (t) = 35.4 cos (10 t – 15  ) V 12

13. 13 v s (t) = 50 cos (10 t + 30  ) V i (t) = 3.54 cos (10 t + 75  ) mA v c (t) = 35.4 cos (10 t – 15  ) V v r (t) = 35.4 cos (10 t + 75  ) V Time Domain v s = 50/30  V i = 3.54/75  mA v c = 35.4/– 15  V v r = 35.4/75  V Frequency Domain Re Im 