rdlc data matrix

rdlc data matrix

rdlc data matrix

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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.

By the distance formula (2.1), c= Hence, c2 = (x a)2 + y2 = (b cos a)2 + (b sin )2 (x a)2 + (y 0)2 = (x a)2 + y2

Fig. 1-13

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.

The letters should be in proper sequence, beginning with the accounts at the top of the page 3 Extend the trial balance gures and the adjustment gures to ei ther the income statement or balance sheet columns The process of ex tending the balances horizontally should begin with the account at the top of the sheet The revenue and expense accounts should be extended to the income statements columns; the assets, liabilities, and capital to the bal ance sheet columns Each gure is extended to only one of the columns After the adjusted trial balance column totals have been proved out, then the income statement columns and the balance sheet columns should also prove out 4 Total the income statement columns and the balance sheet col umns.

= b2 cos2 2ab cos + a2 + b2 sin2 = a2 + b2 (cos2 + sin2 ) 2ab cos = a2 + b2 2ab cos

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

Then by (1.2) and (1), 2

Since (2) holds for any pair of constants k1 ; k2 , superposition is satis ed and the element is linear.

(a) By the algebra of Problem 26.3, (sin cos )2 = sin2 2 sin cos + cos2 = 1 2 sin cos = 1 sin 2 1 (cos2 sin2 ) 1 cos 2 = sin 2 2 sin cos = = sin2 + sin2 2 sin cos sin (2 sin )(sin ) = (2 sin )(cos ) cos [by Theorem 26.1] [by Theorem 26.8]

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.

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

[by Theorem 26.8]

The difference between the debit and credit totals in both sets of columns should be the same amount, which represents net income or net loss for the period 5 Enter the net income or net loss Net income increases capital, so the net income gure should go on the credit side of the balance sheet Conversely, a net loss decreases the capital, so the net loss gure should go on the debit side of the balance sheet..

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 Theorem 26.1]

By (1.1), the voltage-current relationship for R2 is vab R2 i 0:25i i 0:25i2 p i 2 vab

Fig. 9-14 The principle of superposition is applicable to this linear circuit. With vS2 0 (shorted), the voltage appearing at the noninverting terminal is found by voltage division to be v2 Let vo1 be the value of vo with vS2 0. vo1 Similarly, with vS1 0, R v v S1 R R S1 2 1

26.5 Given that is in the fourth quadrant and that cos = 2 , nd: 3