EECE 210 Homework 3 with Solution, Fall 2013

Homework 3 P3.1.2 Determine Req in Figure P3.1.2. Solution: On the RHS, (0.5 k||2 k) = 1/2.5 = 0.4 k; in series with 0.1 k, this becomes 0.5 k; in parallel with 2 k, this gives again 0.4 k. On the LHS, (0.6 k||1.2 k) = 0.72/1.8 = 0.4 k; in series with 0.2 k, this becomes 0.6 k; in parallel with 0.3 k, this gives 0.18/0.9 = 0.2 k. This resistance appears in series with the0.4 k from the RHS, giving a total of 0.6 k. Req = (1.8 k||0.6 k) = 0.45 k. P3.1.9 Determine between terminals ab in Figure P3.1.9. Solution: G in series with G is 0.6 G. The G resistor connected to node b may be split into two G resistors, each of which is in series with 0.6 G. This gives ; Gab  . P3.1.14 Determine each of the resistors of the equivalent delta between terminals ‘a’, ‘b’, and ‘c in Figure P3.1.14, expressed in S. Solution: The inner  is transformed to three 2/3  resistors 1-3 a b G G 2 G Figure P3.1.9 3G/2 3G/2 3G/2 2  3 S 3 S 1/6 S a 1/6 S 0.5 k 0.1 k 2 k 1.2 k 2 k 0.3 k 0.2 k 0.6 k 1.8 k Req a b Figure P3.1.2 connected in Y. To each is added 1/3  in series to give three 1  resistors connected in Y. These are transformed to three 1/3 S resistors connected in . Each of the resistors of the equivalent  will be (1/3 + 1/6) = 1/2 S. P3.2.3 Determine Ix and VO in Figure P3.2.3 by transforming: (a) the current source to a voltage source; (b) the voltage source to a current source. Note that Ix and VO can be identified with respect to terminals ‘ab’. Solution: (a) The current source is transformed to a voltage source 0.5Ix2 = Ix A in series with 2  (Figure P3.2.3a). It follows that , which gives Ix = 1 A.; from KVL, VO = 4 – 11 = 3 V. 2-3 – + Ix Ix 2  1  4 V – + VO a b + – (a) Ix + a 1.5Ix Ix 0.5Ix 2  1  4 V Figure P3.2.3 – + VO a b + – (b) The voltage source is transformed to a current source 4/1 = 4 V in parallel with 1  (Figure P3.2.3b). The two current sources are combined to (4 + 0.5Ix) and the two resistances to (1||2) = 2/3 . Hence, , with VO = 21.5Ix. = 3Ix. This gives Ix = 1 A and VO = 3 V, as before. P3.2.12 Determine Vx and Iy in Figure P2.3.12. Solution: , , . V. A, A. 3-3 4  40  60  120  30  80  – + 25 V IY + – Vx Figure P3.2.12 a b c Ia 4  30  16  25 V IY Vx Vy – + – + – + Ia a b c P3.2.15 Determine IO in Figure P3.2.15 using source transformations. Solution: The voltage of node ‘a’ with respect to the lower reference node is 3Ix V, and the voltage of node ‘b’ is (3Ix – 2IO) V. The 10 V source in series with 2  is transformed to a 5 A source in parallel with 2 . This resistance is combined with the 3  resistance to give a resistance of 23/5 = 1.2 . The current in this resistance is 3Ix/1.2 = 2.5 Ix. The 3Ix dependent source in series with 5  is transformed to a 0.6 A source in parallel with 2 . This resistance is combined with the 5  resistance to give a resistance of 25/7 = 10/7 . The current in this resistance is 7(3Ix – 2IO)/10 = 2.1Ix – 1.4IO. KCL at node ‘a’ gives, 2.5Ix + IO + 5 = 0, or Ix = -2 – 0.4IO. KCL at node ‘b’ gives: IO + 0.6Ix = 2.1Ix – 1.4IO, or 2.4IO = 1.5Ix, or Ix = 1.6IO. Substituting the first equation, 1.6IO = -2 – 0.4IO, gives IO = -1 A. 4-3 + – 10 V 2  2  3  2  – + 5  3Ix Ix IO Figure P3.2.15 5 A 2  1.2  10/7  0.6Ix 2.5Ix IO 3Ix V (3Ix – 2IO) V 2.1Ix – 1.4IO a b P3.3.1 Determine ISRC in Figure P3.3.1 so that no current flows in Solution: When no current flows in RL, Vad is, from voltage division, V, and ISRC flows in the 3 k resistor only. From KVL around the mesh ‘abcd’, Vad + Vba + Vbc + Vcd = 0, or 3 + 6 +0 + Vcd = 0. This gives Vcd = -9 = 3Isrc. so that ISRC  -3 mA. 5-3 3 k RL 1 k 3 k 6 V 12 V ISRC Figure P3.3.1 + – + – – + – + 3 k RL 1 k 3 k 6 V 12 V ISRC a b c d 0 P3.3.7 Determine Vx in Figure P3.3.7. Transform the voltage source to a current source and apply KCL and KVL. Solution: The 30 V source in series with 2  transforms to a 15 A source in parallel with 2 ; the parallel combination of this resistor and the 4  resistor is a resistor of , which, from KCL, carries a current of 24 A and therefore has a voltage drop across it of V. From KCL, the current in the 3  resistor is 6 A, so that the voltage across it is 18 V. It follows that Vx  32 + 18  50 V. 6-3 9 A 3 A 30 V 2  4  3  + – Vx Figure P3.3.7 + – 9 A 3 A 15 A 4/3  3  Vx 24 A 9 A 6 A 18 V – + – +