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CIRCUIT ANALYSIS: PORT POINT OF VIEW
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[CHAP. 1
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Example 1.11. Since the average value of a sinusoidal function of time is zero, the half-cycle average value, which is nonzero, is often useful. Find the half-cycle average value of the current through a resistance R connected directly across a periodic (ac) voltage source v t Vm sin !t. By Ohm s law, i t v t Vm sin !t R R
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and from (1.20), applied over the half cycle from t0 0 to T=2 , 1  Vm 1 Vm 2 Vm I0 sin !t d !t cos !t  !t 0  0 R  R  R
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Example 1.12. Consider a resistance R connected directly across a dc voltage source Vdc . The power absorbed by R is Pdc
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2 Vdc R
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1:23 The instantaneous power is now given by 1:24
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Now replace Vdc with an ac voltage source, v t Vm sin !t.
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p t
v t sin2 !t R R
2 Vm
Hence, the average power over one period is, by (1.20), P0 1 2 2
0 2 Vm V2 sin2 !t d !t m R 2R
1:25
Comparing (1.23) and (1.25), we see that, insofar as power dissipation is concerned, an ac source of amplitude Vm is equivalent to a dc source of magnitude s Vm 1 T 2 p v t dt  V 1:26 T 0 2 p For this reason, the rms value of a sinusoid, V Vm = 2, is also called its e ective value. From this point on, unless an explicit statement is made to the contrary, all currents and voltages in the frequency domain (phasors) will re ect rms rather than maximum values. Thus, the p time-domain voltage " v t Vm cos !t  will be indicated in the frequency domain as V Vj, where V Vm = 2. Example 1.13. A sinusoidal source, a dc source, and a 10  resistor are connected as shown by Fig. 1-12. If vs 10 sin !t 308 V and VB 20 V, use SPICE methods to determine the average value of i I0 , the rms value of i I , and the average value of power P0 supplied to R.
Fig. 1-12 The netlist code below describes the circuit. Notice that the two sources have been combined as a 10 V sinusoidal source with a 20-V dc bias. The frequency has been arbitrarily chosen as 100 Hz as the solution is independent of frequency.
CHAP. 1]
CIRCUIT ANALYSIS: PORT POINT OF VIEW
Ex1_13.CIR - Avg & rms current, avg power vsVB 1 0 SIN(20V 10V 100Hz 0 0 -30deg) R 1 0 10ohm .PROBE .TRAN 5us 10ms .END
The Probe feature of PSpice is used to display the instantaneous values of i t and pR t . The running average and running RMS features of PSpice have been implemented as appropriate. Both features give the correct fullperiod values at the end of each period of the source waveform. Figure 1-13 shows the marked values as I0 2:0 A, I 2:1213 A, and P0 45:0 W.
Fig. 1-13
Solved Problems
1.1 Prove that the inductor element of Fig. 1-1(b) is a linear element by showing that (1.2) satis es the converse of the superposition theorem.
Let i1 and i2 be two currents that ow through the inductors. inductor for these currents are, respectively, v1 L di1 dt and v2 L Then by (1.2) the voltages across the
di2 dt
CIRCUIT ANALYSIS: PORT POINT OF VIEW
[CHAP. 1
Now suppose i k1 i1 k2 i2 , where k1 and k2 are distinct arbitrary constants. v L
Then by (1.2) and (1), 2
d di di k i k2 i2 k1 L 1 k2 L 2 k1 v1 k2 v2 dt dt dt 1 1
Since (2) holds for any pair of constants k1 ; k2 , superposition is satis ed and the element is linear.
If R1 5 , R2 10 , Vs 10 V, and Is 3 A in the circuit of Fig. 1-14, nd the current i by using the superposition theorem.
R1 + Vs _ b i R2 Is a
Fig. 1-14 With Is deactivated (open-circuited), KVL and Ohm s law give the component of i due to Vs as i0 Vs 10 0:667 A R1 R2 5 10
With Vs deactivated (short-circuited), current division determines the component of i due to Is : i 00 By superposition, the total current is i i 0 i 00 0:667 1 1:667 A R1 5 I 3 1A R1 R2 s 5 10
In Fig. 1-14, assume all circuit values as in Problem 1.2 except that R2 0:25i . Determine the current i using the method of node voltages.
By (1.1), the voltage-current relationship for R2 is vab R2 i 0:25i i 0:25i2 p i 2 vab
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