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Unformatted text preview: ENGR 210 – Spring 2008 Homework #5 Solutions Problem 1 Use the mesh‐current method to determine the currents Ia, Ib , Ic and Ic in the circuit shown below. Solution First we draw our mesh current loops. I always begin each loop in the bottom left corner and go around in a clockwise direction. i3 i1 i2 Now that we have our mesh current loops, we can write out equations to describe each of them. i1: i3: i2: −110 + 10(i1 − i3 ) + 3(i1 − i2 ) − 12V + 4i1 = 0 12V + 3(i2 − i1 ) + 12(i2 − i3 ) + 70 + 2i2 = 0 6i3 + 12(i3 − i2 ) + 10(i3 − i1 ) = 0 Now that we have the mesh equations, we put them in standard form and solve them using matrix math. i1 :17i1 − 3i2 − 10i3 = 122 i2 : −3i1 + 17i2 − 12i3 = −82 i3 : −10i1 − 12i2 + 28i3 = 0 Solving for the three variables, we get: i1 = 8 A And lastly we find the labeled currents from the problem: i2 = −2 A i3 = 2 A I a = i1 = 8 A I b = i3 = 2 A I c = i2 = −2 A I d = i1 − i2 = 10 A Problem 2 Use the mesh current method to find the power developed in the dependent voltage source in the circuit shown below. Be sure to indicate if it is generating or absorbing power. Solution First we draw our mesh current loops: i2 i1 i3 Next, we write out the loop equations: i1 :125V + 15(i1 − i2 ) + 100(i1 − i3 ) + 35i1 = 0 i2 : 2.65VΔ + 25(i2 − i3 ) + 15(i2 − i1 ) = 0 i3 :100(i3 − i1 ) + 25(i3 − i2 ) − 125V + 85(i3 ) = 0 And we know we have the constraint: VΔ = 100(i1 − i3 ) By substituting in the above constraint (VΔ) into the equation for i2 and placing the equations in standard form, we can solve using matrix math. 150i1 − 15i2 − 100i3 = −125 250i1 + 40i2 − 290i3 = 0 −100i1 − 25i2 + 210i3 = 125 Solving, we get the following: i1 = 1.2 A i3 = 2 A i2 = 7 A The problem asks us to find the power developed by the dependent source. P = IV I = i2 = 7 A V = 2.65 × [100(i1 − i3 ) ] = −212V P= ‐ 1484W and negative indicates that is generating power. Problem 3 Use the mesh‐current method to determine the currents Ia, Ib, and Ic associated with each source in the circuit shown below. Solution First we draw our mesh loops and label them. i3 i1 i2 We look at this circuit and notice that loops i1 and i3 are normal. Loop i2 has a current source in line with it. That means that the current i2 must be the same as the current through the source. We write out the loop equations for i1 and i3 like normal: i1 : −90V + 2(i1 − i3 ) + 3(i1 − i2 ) + 165V = 0 i3 : 7(i3 ) + 1(i3 − i2 ) + 2(i3 − i1 ) = 0 We know the current through i2 is the same as through the source: i2 = 0.5VΔ And we can write the following relationship for VΔ VΔ = 2(i2 − i3 ) Solve the above system of equations. The easiest Finally, solve for the labeled currents: way is to plug VΔ into the i2 equation and then plug I a = i1 = −33 A the new equation for i2 into the equations for i1 I b = i2 − i1 = 9 A and i3. After that you’ll have 2 equations and 2 I c = .5VΔ = −24 A unknowns and you can solve for all 4 variables: i1 = −33 A i3 − 9 A i2 = −24 A VΔ = −48V Problem 4 (a) Use the mesh‐current method to find the branch currents ia, ib, ic, id, and ie for the circuit shown above. (b) Show that the total power generated equals the total power dissipated for this circuit. Solution First we notice we have a current source that is shared by 2 meshes. So we must define a super mesh: VA i2 VB i1 i3 supermesh i4 Now we need to write the equation for the super mesh: 40(i3 − i1 ) + 10(i3 − i2 ) + 35(i4 − i2 ) + 150 = 0 We also can write the equation for the mesh loop i2: i2 :15id + 35(i2 − i4 ) + 10(i2 − i5 ) = 0 There are a few constraints defined in the circuit aswell: Constraint for current source (again we don’t need to make a loop equation for i1 because there is a current source in line with it: i1 = 30 A Constraint for id: id = i4 Constraint for ia: ia = i1 − i3 Constraint for dependent current source (the supermesh constraint): 3ia = i3 − i4 Solving using the above equations, we find the following: i1 = 30 A i2 = 8 A i3 = 24 A i4 = 6 A We can now solve for each ia, ib, ic, id, and ie. ia = i1 − i3 = 6 A ib = i2 − i3 = −16 A ic = i2 − i4 = 2 A id = i4 = 6 A ie = i2 = 8 A In part b, we are to show that the total power generated equals the total power disipated. First we find the voltage at VA and VB (we need voltages if we’re going to find power): VA = 40Ωiia = 240V VB = 150V − 35i1 = 80V Now we can solve for each of the powers: P30 A = −30VA = −7200( gen ) P id = 15id ie = 720W ( diss ) 15 P3ia = 3iaVb = 1440W ( diss ) P V = 150id = 900W (diss ) 150 P40Ω = (6) 2 (40) = 1440W ( diss ) P Ω = ( −16) 2 (10) = 2560W (diss ) 10 P35Ω = (2) 2 (35) = 140W ( diss )
Finally we check that the generating and dissipating values are the same by summing the positive and negative values: ∑P ∑P gen = 7200W = 7200W diss Problem 5 The circuit shown below is a direct‐current version of a typical three‐wire electric power distribution system. The resistors Ra, Rb, and Rc represent the resistances of the three conductors that connect the three loads R1, R2, and R3 to the 110/220 V voltage supply (the power lines up on the poles). The resistors R1 and R2 represent loads connected to the 110 V circuits, and R3 represents a load connected to the 220 V circuit. (a) What circuit analysis method will you use and why? (b) Calculate v1, v2, and v3. (c) Calculate the power delivered to R1, R2, and R3. (d) What percentage of the total power developed by the sources is delivered to the loads? You will probably need to calculate the powers delivered to Ra, Rb and Rc to answer this. (e) The Rb branch represents what is called the neutral conductor. What adverse effect occurs if the neutral conductor is opened, i.e., Rb goes to infinity? (Hint: Calculate v1 and v2 for this situation and note that appliances or loads designed for use in this circuit would have a nominal voltage rating of 110 V.) Solution a) If we use mesh current we have 3 meshes. If we use node voltage we have 3 nodes. Neither method is clearly superior. I’ll solve this using mesh current since last time we solved this problem we used node voltage. To do the rest of the problem, we must define mesh current loops. P1 i1 i3 P2 i2 b) Next we find the equation for each loop: i1 : −110 + .1(i1 ) + 18(i1 − i3 ) + .2(i1 − i2 ) = 0 i2 : −110 + .2(i2 − i1 ) + 110.5(i2 − i3 ) + .1(i2 ) = 0 i3 :110.5(i3 − i2 ) + 18(i3 − i1 ) + 54.625(i3 ) = 0 By solving in standard form or with a calculator, we get: i1 = 10 A i3 = 4 A i2 = 5 A We can now solve for V1, V2 and V3. V1 = 18(i1 − i3 ) = 108V V2 = 110.5(i2 − i3 ) = 110.5V V3 = 54.625i3 = 218.5V
c) Next, we’re asked to calculate the power dissipated in each resistor representing an appliance: P = I 2R PR1 = (i1 − i3 ) 2 (18) = 648W PR2 = (i2 − i3 ) 2 (110.5) = 110.5W PR3 = i32 (54.625) = 874W
Summing the power, the total load power is 1632.5W. d) To find the percentage of power delivered to the load, we find the power generated by each of the sources, sum them, and then find the percentage: P = 110i1 = 1100W 1 P2 = 110i2 = 550W % Delivered = 1632.5 × 100% = 98.9% 1650 e) Finally, the problem asks us to analyze what happens if the neutral conductor is opened. We will redraw the circuit to analyze this scenario: i1 i2 We can now re‐write the loop equations: i1 : −220 + 0.1i1 + 18(i1 − i2 ) + 110.5(i1 − i2 ) + 0.1i1 = 0 i2 :110.5(i2 − i1 ) + 18(i2 − i1 ) + 54.625(i2 ) = 0 After converting to standard form and solving, we get: i1 = 5.71A i2 = 4.01 A i1 − i2 = 1.7 A We can now find the voltage across the two loads: V1 = (1.7 A)(18) = 30.6V V2 = (1.7)(110.5) = 187.85V V1 drops low and V2 goes high. In this scenario, the appliance at load 1 would probably not turn on, and the appliance at load 2 would likely fail (explode/melt). ...
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 Fall '09

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