B.Tech./EE-304 (Syllabus - 2017) Class Test – II (February 24, 2023) (Electronics and Communication Engineering) EE – 304 Electrical Network Theory (3rd Semester) Questions & Answers ✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿ Duration: 1 hr. Total Marks: 20 1. (a) The circuit in Fig. 1(a) models a common-emitter transistor amplifier. Find ix using source transformation. (b) Find the Thévenin equivalent at the terminals, a−b, in the circuit shown in Fig. 1(b). + − 2V A 6Ω b βix + − − Ro 5Ω A 23 V vs + − 2Ω 2V Rs + − ix 4Ω − a + 18 V 10 V a b Fig. 1 Soln .: (a) In the given circuit of Fig. 1(a), we transform the current source, βix into a voltage as shown in Fig. 1(c) resulting ix being the only current flowing in the circuit. Applying KVL ix − vs + Rs ix + Ro ix + βRo ix = 0 ⇒ (Rs + Ro + βRo ) ix = vs vx ∴ ix = Rs + (1 + β) Ro Rs Ro + − + vs − βRo ix Fig. 1(c) (b) First of, we shall find the Thévenin voltage, VT H , across the terminals, a − b, from the given circuit of Fig. 1(b). For this, we will use mesh analysis and identify only 3 (three) possible meshes in the circuit as shown Fig. 1(d). Accordingly, all the mesh currents, i1 , i2 and i3 , have also been labeled in clock-wise directions. Because of 2 A and 3 A, we can’t apply KVL on any of the mesh. Therefore, we combine meshes 2 and 3 to form supermesh and apply KVL on it as 18 + 4(i2 − i1 ) + 6(i3 − i1 ) + 5i3 − 10 = 0 ⇒ 4i2 + 11i3 − 10i1 + 8 = 0 20230224 North-Eastern Hill University (1.1) Page 1/6 [5+5] EE–304 ENT (Class Test - II) Electronics and Communication Engineering but we know that + − 2V A i1 = 2 A and i2 − i3 = 3 A i1 Using above expressions in Eq. (1.1), we have 18 V + 4Ω 6Ω − a b + − i3 A 23 V 5Ω 4(i3 + 3) + 11i3 = 12 ∴ i3 = 0 A 2Ω i2 2 V + − − and hence 10 V i2 = 3 A Fig. 1(d) We now can find VT H from either VT H = vab = 18 + v4 Ω + v6 Ω = 18 + 4(i2 − i1 ) + 6(i3 − i1 ) = 18 + 4(3 − 2) + 6(0 − 2) = 18 + 4 − 12 = 10 V or VT H = vab = 10 + v5 Ω = 10 + 0 = 10 V Next, we find RT H by suppressing all independent sources in the given circuit of Fig. 1(b) resulting a resistive network as shown in Fig. 1(e). We know + − 2V RT H = Rab = (R4 Ω + R6 Ω ) ||R5 Ω = (4 + 6)||5 10 × 5 = 10 + 5 ∴ RT H = 3.3333 Ω 4Ω 6Ω a b + − We now draw the Thévenin equivalent of the given circuit as below 2Ω 2V 5Ω + − 2V Fig. 1(e) RT H a 3.3333 Ω + − VT H 10 V b Fig. 1(f): Thévenin equivalent network of Fig. 1(b) Alternative Solution Above solution/equivalency can also be obtained by ✿✿✿✿✿✿✿✿ source ✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿ transformation as shown below: 2A − + 12 V + 18 V − a + 10 V 20230224 − (4 + 6)Ω + 30 V a 10 Ω − 2A b 5Ω b 2A 5Ω ⇒ North-Eastern Hill University Page 2/6 EE–304 ENT (Class Test - II) Electronics and Communication Engineering 3A 3A 10 Ω a a b 5Ω b 3.3333 Ω ⇒ RT H a 3.3333 Ω + − VT H + a 3.3333 Ω − 10 V b 9.9999 V ≈ 10 V b ⇒ Fig. 1(g): Thévenin equivalent network of Fig. 1(b) obtained using source transformation 2. Find voltage drop across the 4 Ω resistor in the circuit of Fig. 1(b) using superposition principles. Soln .: The given circuit of Fig. 1(b) redrawn as shown in Fig. 2(a)(i) by assigning the required voltage, v4Ω to be determined. We have also drawn below other circuits in Fig. 2(a)(ii)2(a)(v), for applying superposition principles, wherein only 1 (one) independent source is active at a time: 2A + + 10 V v4Ω 3A 6Ω 4Ω b − a + 1 v4Ω 5Ω 18 V 6Ω a b − + − + 4Ω 4Ω − 18 V a 2A + 5Ω − (i) (iii) + 3 v4Ω 3A 4Ω 6Ω b − a + 4 v4Ω 10 V 5Ω + 4Ω b 5Ω (ii) a 2 v4Ω 6Ω − 6Ω b − 5Ω − (iv) (v) Fig. 2(a) From Fig. 2(a)(ii) using the current division and Ohm’s law, we get   5 5 1 × −2 = −8 × = −2.6667 V v4Ω = 4 (4 + 6) + 5 15 Again from Fig. 2(a)(iii) using the voltage divider, we have 2 v4Ω = 4 4 × −18 = −18 × = −4.8 V 4 + (6 + 5) 15 And again from Fig. 2(a)(iv) using the current division and Ohm’s law, we get   6+5 11 3 v4Ω = 4 × 3 = 12 × = 8.8 V 4 + (6 + 5) 15 And lastly from Fig. 2(a)(v) using the voltage divider, we have 4 v4Ω = 20230224 4 4 × 10 = 10 × = 2.6667 V 4 + (6 + 5) 15 North-Eastern Hill University Page 3/6 [10] EE–304 ENT (Class Test - II) Electronics and Communication Engineering Therefore, the required voltage is given by superposition principles as 1 2 3 4 v4Ω = v4Ω + v4Ω + v4Ω + v4Ω = −2.6667 − 4.8 + 8.8 + 2.6667 = 4 V (2.1) Verification We will use Thévenin theorem to find the value of v4Ω . For this, we redraw the given circuit of Fig. 1(b) as shown in Fig. 2(b)(iii). From this circuit, we remove the resistor, 4 Ω, and find its equivalent Thévenin for the remaining network. Corresponding circuits to find VT H and RT H are shown in Fig. 2(b)(iv) and Fig. 2(b)(v) respectively. a + − 10 V + − 18 V 18 V a + 18 V + − a 4Ω + b − v4Ω v4Ω 3A 6Ω 4Ω 2A + + 5Ω b b (i) 4Ω − 5Ω − v4Ω 2A 3A 6Ω 3A − 10 V + − − 5Ω 10 V + 2A 6Ω (iii) (ii) a + 18 V − a RT H + + − 10 V 11 Ω 2A 3A VT H 5Ω VT H + − v4Ω −5 V 5Ω b b 6Ω 6Ω (iv) 4Ω − RT H − + Network - A (v) Network - B (vi) Fig. 2(b) To find VT H from Fig. 2(b)(iv), we shall use nodal analysis. Assigning node a as a reference node, we apply KCL at node b as vb − (−10) −2+3=0 5 vb + 10 +1=0 ⇒ 5 ⇒ vb = −15 V ∵ i6Ω = 3 A Also vb being a node voltage with respect to the reference node a, we know − vb = v5Ω + 10 ⇒ v5Ω = −vb − 10 = 15 − 10 = 5 V Now, we can find VT H as VT H = −18 + 10 + v5Ω + v6Ω = −8 + 5 + 6 × 3 = −3 + 18 = 15 V From Fig. 2(b)(v), we get RT H = 5 + 6 = 11 Ω Finally, the required voltage, v4Ω , is calculated using voltage division from the circuit of Fig. 2(b)(vi) as v4Ω = 4 × 15 = 4 V 4 + 11 which ✿✿✿✿✿✿✿✿ matches✿✿✿✿✿ with✿✿✿✿✿ that✿✿✿ of ✿✿✿✿ Eq.✿✿✿✿✿✿ (2.1).✿✿✿✿✿✿✿✿ Hence, ✿✿✿✿ the ✿✿✿✿✿✿ result ✿✿ is✿✿✿✿✿✿✿✿ verified. 20230224 North-Eastern Hill University Page 4/6 EE–304 ENT (Class Test - II) Electronics and Communication Engineering [5+5] 2 V11′ + 4Ω 1 − 3. For the circuit of Fig. 2, obtain (a) V11′ using superposition principles, and + (b) The Norton equivalent at terminals, 1 − 1′ . 5V + − 3A V11′ − 1′ Fig. 2 Soln .: (a) The given circuit of Fig. 2 is redrawn by making only one independent source active at a time as shown below. Let V11′ = V111 ′ + V112 ′ where V111 ′ and V112 ′ are voltages due to the 5V and 3A sources when acted one at a time respectively. + 5V + − 1 1 + V111 ′ − 2 V112 ′ + − + 4Ω − 2 V111 ′ 4Ω V112 ′ 3A − 1′ 1′ (ii) (i) Fig. 3(a) From Fig. 3(a)(i) by KVL, we have − 5 + 4 × 0 + 2V111 ′ + V111 ′ = 0 5 ⇒ V111 ′ = = 1.6667 V 3 ∵ i4 Ω = 0 A Applying KVL in From Fig. 3(a)(ii), we have 4 × (−3) + 2V112 ′ + V112 ′ = 0 12 =4V ⇒ V112 ′ = 3 Therefore, the required voltage is V11′ = V111 ′ + V112 ′ = 1.6667 + 4 = 5.6667 V (b) We first find Norton current, IN , by shorting the terminals, 1 − 1′ , in the circuit of Fig. 3(b)(i), which in turn makes the 3A source shorted. Therefore, V11′ = 0 V. + − + 5V + − 3A V11′ 1 2 V11′ 1V − − 1′ (ii) (i) + − V11′ V11′ 1′ 1 + itest + IN − IN I1 4Ω 2 V11′ 4Ω + 1 − + 2 V11′ − 4Ω 1′ (iii) Fig. 3(b) Applying KVL in the perimeter (outer loop) of the circuit, we have − 5 + 4I1 + 2V11′ = 0 5 ⇒ I1 = = 1.25 A ∵ V11′ = 0 V as both terminals are shorted 4 We also know IN − I1 = 3 ⇒ IN = 3 + I1 = 3 + 1.25 = 4.25 A 20230224 North-Eastern Hill University Page 5/6 EE–304 ENT (Class Test - II) Electronics and Communication Engineering Now, for RT H , we use an external 1V voltage source as shown in Fig. 3(b)(iii). Applying KVL in the circuit, we have a 4 × −itest + 2V11′ + 1 = 0 ⇒ − 4itest + 2 + 1 = 0 ∵ V11′ = 1 V 3 ∴ itest = = 0.75 A 4 IN 4.25 A RN 1.3333 Ω b Fig. 3(c): Norton equivalent network of Fig. 2 Hence, the Norton resistance, RN , is given by RN = 20230224 1 1V = = 1.3333 Ω itest 0.75 North-Eastern Hill University Page 6/6