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EXAMPLE 9.7 A practical coil is connected in series between two voltage sources v1 5 cos !1 t and v2 10 cos !2 t 608 such that the sources share the same reference node. See Fig. 9-54. The voltage di erence across the terminals of the coil is therefore v1 v2 . The coil is modeled by a 5-mH inductor in series with a 10- resistor. Find the current i t in the coil for (a) !1 !2 2000 rad/s and (b) !1 2000 rad/s, !2 2!1 . p (a) The impedance of the coil is R jL! 10 j10 10 2 458 . The phasor voltage between its terminals is p V V1 V2 5 10 608 j5 3 V. The current is p V j5 3 j8:66 % 0:61 1358 A I p Z 10 2 458 14:14 458 i 0:61 cos 2000t 1358

/4, AB > BD > AD. Since /2

The instantaneous power ow into the port is given by p t v t i t Vm Im cos !t cos !t 1 Vm Im cos 2!t cos 2 By (1.20), P0 1 2 2

(b) Because the coil has di erent impedances at !1 2000 and !2 4000 rad/s, the current may be represented in the time domain only. By applying superposition, we get i i1 i2 , where i1 and i2 are currents due to v1 and v2 , respectively. V1 5 0:35 458 A; i1 t 0:35 cos 2000t 458 Z1 10 j10 V 10 608 0:45 3:48 A; i2 t 0:45 cos 4000t 3:48 I2 2 Z2 10 j20 i i1 i2 0:35 cos 2000t 458 0:45 cos 4000t 3:48 I1

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3.10 Applying isosceles triangle principles In each of Fig. 3-18(a) and (b), nI can be proved congruent to nII. Make a diagram showing the congruent parts of both triangles and state the congruency principle involved.

2

From Table 9-1, vL !LI cos !t 908 100 cos 2000t 908 (V). vL 100 sin 2000t 1808 100 sin 2000t As a sine function, V

Fig. 3-18

A series circuit, with R 10 and L 20 mH, has current i 2:0 sin 500t (A). Obtain total voltage v and the angle by which i lags v.

After the integration is performed and its limits evaluated, the result is P0 V m Im V Im cos pm p cos VI cos 2 2 2

(a) Since AB > BC, /A (b) Since AB > AC, /B /C. nI /C. nI nII by SAS [see Fig. 3-19(a)]. nII by ASA [see Fig. 3-19(b)].

CHAP. 9]

Fig. 3-19

CHAP. 1]

By the methods of Example 9.1, 500 20 10 3 458 arctan 10 q v I R2 !L 2 sin !t 28:3 sin 500t 458 It is seen that i lags v by 458.

AB > BC AC is trisected at D and E. To Prove: /1 /2 Plan: Prove nI nII to obtain BD > BE. Given:

Find the two elements in a series circuit, given that the current and total voltage are i 10 cos 5000t 23:138 A v 50 cos 5000t 308 V

1.23 Prove that the capacitor element of Fig. 1-1(c) is a linear element by showing that it satis es the converse of the superposition theorem. (Hint: See Problem 1.1.) Use the superposition theorem to nd the current i in Fig. 1-14 if R1 5 ; R2 10 ; Vs 10 cos 2t V, and Is 3 cos 3t =4 A. Ans: i 0:667 cos 2t cos 3t =4 A In Fig. 1-23, (a) nd the Thevenin equivalent voltage and impedance for the network to the left of terminals a; b, and (b) use the Thevenin equivalent circuit to determine the current IL . Ans: a VTh V1 I2 R2 ; ZTh R1 R2 ; b IL V1 I2 R2 = R1 R2 RL

PROOF:

Hence,

Fig. 1-23

Statements 1. AC is trisected at D and E 2. AD > EC 3. AB > BC 4. /A 5. nI 7. /1 /C nII /2 1. Given

Since i lags v (by 53.138), the elements are R and L. The ratio of Vmax to Imax is 50/10. q 50 5000L and tan 53:138 1:33 R2 5000L 2 10 R Solving, R 3:0 , L 0:8 mH.

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