Question

Problem 1 (10 marks): Find the Thevenin equivalent at terminals a-b of the following circuit: +- 10 kΩ 20 kΩ 70ν (Ξ)

Answer 1

Denote redraw the loop current and the venin voltage the circuit as shown in Figure 1. + - I /ь 2012 log.n + 4V 7ov ☺ from Figure I value of voltage No is as follows Vo = (10x103) 1 - 0 Apply kirchhoff's voltage low in loop abcdef. -70 + (10x103) I + (20 % 103) I +4 Vo =D 130 X 103) I +4V. = 70 substitute (10x109 I) for vo. (30 x103) I + 4 (10x103)I = 70 40x103)I = 70 I - 70 70x103 FOA I mA law in loop abef... Apply Kirchhoff's voltage vo + (ox to I - 5 substitute 1 mA for I.

o VE 70 - (10x103) (1x103) = 70-10 Hence, the value of Thevenin bov] voltage, von is since the circuit has 4V depen dent source, thus, to find the Thevenin equivalent resistance, place. a IV test source across the open terminals a and b. determine the current supplied the test source; Thereafter, find the Thevenin equivalent resistance as follows: 1. Ron V - Draw the modified circuit . as shown in Figure 2. . 4VO vo & loke 9 P figure, 2, the value of voltage. No is as follows From the Vo+1 =0 Vo = -1V

calculate the value of current is ie -V. 10x103 substitute -Iv for V 10x103 loop mnopqr in e 0.1mA apply kirchhoff's voltage law in fig. 2. - 60x103) ia + (20x103) 6 + 4V. - O substitute olma for å and IV for vo... - (10x103) (0.1X103) + (20x103) 16 + 4(-1)=0 -1 + (20 %10) 1 - 420 (20x103) *= 5 delivered by the IV io = 5 x 103 0.25 MA calculate the total current test source o = latib substitute oil ma for ia and L = (0.1X10-3) + (0.25 x103) = 0.35 x 10-3 A = 0.35 mA 0.25 mA for ib.

substitute 0.35 mA for ia in equation (2) Rin = 35x 103 = 2857 · 143 12 = 2.86 Hence, the value of the venin resistance, Riho is (2.86 kg The Thewenin equivalent circuit is the series connection of Thevenin voltage source (VW) and Thevenin resistance (Riw Draw the Thevenin equivalent circuit across the terminals a-b as shown in Figure -3. 2.86 ks whoa Gov -ob Hence, the Thevenin equivalent circuit is drawn.