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- 1. Sample Problems and their solutions… Solution Nodal The reference node was selected and the three other nodes of the circuit were labeled V1, V2, and V3. Note that the node voltages are all with respect to the reference. So for example, V1 is the voltage with the "+" at the label V1 and the "-" at the reference node. The following equations can then be written: Current leaving V1 (v1 - v4)/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0 Current leaving V2 (V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0 Current leaving V3
- 2. We don't know the current through the 12V nor through the 18V, and we cannot use Ohm's law to find it, since they are not resistors. So add unknown current labels to the circuit, as shown: Now we can write the KCL equations: iy + (V3 - V2)/5 - ix = 0 Current leaving V4 -iy + (V4 - V1)/2 = 0 Current leaving V5 ix + (V5 - 0)/1 = 0 Extra Equations We now have five equations, but we have seven unknowns (the 5 unknown voltages plus 2 more unknown currents). We can get two extra equations -- just look at the voltage sources that caused the problems in the first place, where we had to add ix and iy. We'll write the voltage relationships between the node voltages: V4 - V3 = 12V V3 - V5 = 18V Now we have 7 equations and 7 unknowns. To solve these equations, let's simplify it down to fewer equations. Solve for V4 and solve for V5 from those last two we added: V4 = V3 + 12V V5 = V3 - 18V Substitute these two equations into the previous five, so that we eliminate V4 and V5. We now have 5 equations and 5 unknowns: (v1 - [v3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0 (V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0 iy + (V3 - V2)/5 - ix = 0 -iy + ([V3 + 12] - V1)/2 = 0 ix + ([V3 - 18] - 0)/1 = 0 Continue using substitution (Gaussian elimination) to narrow it down. Next, solve for ix in the last equation: ix = 18 - V3 and substitute back to get 4 equations and 4 unknowns: (v1 - [v3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0 (V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0 iy + (V3 - V2)/5 - [18 - V3] = 0
- 3. -iy + ([v3 + 12] - V1)/2 = 0 Next, solve for iy in the last equation: iy = ([v3 + 12] - V1)/2 and substitute back to get 3 equations and 3 unknowns: (V1 - [V3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0 (V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0 ([V3 + 12] - V1)/2 + (V3 - V2)/5 - [18 - V3] = 0 Now multiply each equation by a constant to clear out fractions. That is multiply by the Least Common Denominator (multiply first by 6, second by 60, third by 10). 3 * (V1 - [V3 + 12]) + 2 * (V1 - V2) + 3 * (V1 - 0) = 0 12 * (V2 - V3) + 15 * (V2 - 0) + 20 * (V2 - V1) = 0 5 * ([V3 + 12] - V1) + 2 * (V3 - V2) - 10 * [18 - V3] = 0 Next, multiply through 3*V1 - 3*V3 - 36 + 2*V1 - 2*V2 + 3*V1 = 0 12*V2 - 12*V3 + 15*V2 + 20*V2 - 20*V1 = 0 5*V3 + 60 - 5*V1 + 2*V3 - 2*V2 -180 + 10*V3 = 0 Now gather like terms: 8*V1 -2*V2 -3*V3 = 36 -20*V1 +37*V2 = 12*V3 = 0 -5*V1 -2*V2 + 15*V3 = 120 Continue with substitution until you have it solved.
- 4. Solution Mesh First, label each mesh (window pane) with a mesh current. For consistency, make each mesh in a clock-wise direction. Now write KVL equations for each loop. KVL for i1: -18V + 5(i1-i2) + 4(i1-i3) + 1(i1) = 0 then gather terms: 10i1 - 5i2 - 4 i3 -18V = 0 Note that the i1 term is positive, and all other current terms are negative (because they are all clockwise, all other panes will contribute a negative term). Let's do the other two panes with terms gathered up directly (write the total resistance of the loop multiplied by the mesh current that goes through that total resistance): KVL for i2: -5i1 + 10i2 - 3i3 - 12 = 0 KVL for i3: -4i1 -3i2 +9i3 = 0 Now solve the three equations in three unknowns: i1 = 7.02A i2 = 6.28A i3 = 5.21A
- 5. The current through the 5 ohm resistor is just i1 - i2, or 0.74A. The current through the 18V is i1, or 7.02A. All the branch currents are shown below from a Pspice simulation: Solution Nodal First, select a reference node and label the other nodes. Since each node has the same number of connected branches (4), we'll simply choose the bottom node as the reference. There are only two other nodes, which we will label V1 and V2.
- 6. Now write KCL at each node (except the reference): KCL at V1: -5A + V1/5 + (V1-V2)/10 + [V1-(V2+4)]/10 = 0 Note that there are four terms in the equation, one for each branch leaving the node. The terms list the current leaving right, down, left, and up. KCL at V2: (V2-V1)/10 + V2/2 - 2A + [V2-(V1-4)]/10 = 0 Note that there are four terms in the equation, one for each branch leaving the node. The terms list the current leaving right, down, left, and up. Now gather terms (multiplying through by 10 to clear up the fractions): 4V1 - 2V2 = 54 -2V1 + 7V2 = 16 Now solve the set of 2 equations with 2 unknowns. V1 = 17.08V V2 = 7.17V We can now determine the current through the 5 ohm by Ohm's law: I = V1/5 = 3.41A The current through the 4V source can be found as: I = [V1-(V2+4)]/10 = 0.59A
- 7. Solution Mesh First, define a mesh current arround each mesh (window pane) of the circuit. Define each one in a clockwise direction. Now write KVL equations for each loop. On the first loop, we run into a problem. We do not know the voltage across the 5A source. (This is one of the difficulties you can run into when using mesh current.) We add a new unknown to handle this problem. We run into the same problem with the 2A source, so two new unknowns (V5 and V2) are labeled:
- 8. Now we are ready to write the KVL equations. KVL loop 1: -V5 + 5(i1-i2) = 0 KVL loop 2: 5(i2-i1)+10(i2-i4)+2(i2-i3) = 0 or -5i1 + 17i2 - 2i3 = 0 KVL loop 3: 2(i3-i2) + V2 = 0 KVL loop 4: 10i4 + 4V + 10(i4-i2) = 0 Or -10i2 + 20i4 = -4 We have two extra unknowns, so we need two more equations. These can be found by examining the loops with the sources: i1= 5A i3 = -2A We can then solve for the remaining unknowns: i2 = 1.58A i4 = 0.59A We can then find the current through the 5 ohm resistor as: I = i1 - i2 = 3.42A The current through the 4V source is simply i4, or 0.59A.
- 9. Solution Nodal First, select a reference node and label the other nodes. All the nodes have three connected branches, so we will simply choose the bottom node as the reference.
- 10. Now write KCL equations for each node except the reference, in terms of the node voltages: KCL at V1: -3A + (V1-V2)/5 + (V1-V3)/1 = 0 KCL at V2: (V2-V1)/5 + V2/3 + (V2-V3)/2 =0 KCL at V3: (V3-V2)/2 + (V3-V1)/1 - 8A = 0 Now gather terms and clear up the fractions: 6V1 - V2 - 5V3 = 15 -6V1 + 31V2 - 15V3 =0 -2V1 -V2 +3 V3 = 16 Finally, solve the 3 equations in 3 unknowns. V1 = 48.625V V2 = 33 V V3 = 48.75V The current through the 5 ohm resistor can be found by Ohm's law: I = (V1 - V2)/5 = 3.125A The voltage over the 3A source is simply V1, or 48.625V.
- 11. Solution First, select a reference node and label the other nodes. All the nodes have three connected branches, so we will simply choose the bottom node as the reference. Now write KCL equations for each node except the reference, in terms of the node voltages: KCL at V1: -3A + (V1-V2)/5 + (V1-V3)/1 = 0 KCL at V2: (V2-V1)/5 + V2/3 + (V2-V3)/2 =0 KCL at V3: (V3-V2)/2 + (V3-V1)/1 - 8A = 0 Now gather terms and clear up the fractions: 6V1 - V2 - 5V3 = 15 -6V1 + 31V2 - 15V3 =0
- 12. -2V1 -V2 +3 V3 = 16 Finally, solve the 3 equations in 3 unknowns. V1 = 48.625V V2 = 33 V V3 = 48.75V The current through the 5 ohm resistor can be found by Ohm's law: I = (V1 - V2)/5 = 3.125A The voltage over the 3A source is simply V1, or 48.625V.
- 13. Solve for the current ix flowing right through the 4 ohm resistor using Mesh-Current Analysis. Solution Label each mesh (window pane) with a mesh current. Then write the KVL equations for each pane. Note that we were forced to label the voltage over the current source (Vx) in order to write the voltage term there: We now have an extra unknown (Vx), so we need another equation. It is found be relating the two mesh currents to the current source. Note that i1 is positive because it is in the same direction of the source. I2 is negative because it is in the opposite direction as the source.
- 14. Now solve the three equations in three unknowns. I1 is found to be -320mA. Since ix is in the opposite direction of i1, then ix = 320mA. Write the node-voltage equations for the circuit shown below, but do not solve. Write the equation for the power absorbed by the 3V supply in terms of the node voltages. Solution Label Label each spot where two or more wires come together (this is every spot that has an independent voltage). Let's label the bottom node where the ground symbol as REF. Then label the lower left corner as V1, upper left as V2, top middle as V3, middle as V4, and upper right as V5. Equations KCL @ V1 -ix + V1/4 = 0
- 15. KCL @ V2 ix + (V2 - V3)/8 = 0 Also, V2 - V1 = 3V KCL @ V3 (V3-V2)/8 + 200mA + (V3-V5)/6 = 0 KCL @ V4 -200mA + V4/5 = 0 KCL @ V5 (V5-V3)/6 + iz = 0 Also, V5 = 7V Power P = V * i = (3V) * (ix) = (3) * (V3-V2)/8 a) Label the node-voltages for the circuit shown above. b) Write the Node-Voltage equations, but do not solve. c) Write the symbolic equation (in terms of the node voltages) for the power in source Va. d) Label the mesh-currents for the circuit shown above. e) Write the Mesh-Current equations, but do not solve. f) Write the symbolic equation (in terms of the mesh currents) for the power in source Va. a) Label Nodes
- 16. Label each spot where two or more wires come together (this is every spot that has an independent node voltage). Here's how I labeled them: Notice that node V2 is all the wire over the top and extends down to a dot on the right and in the middle of the circuit. That's all one node because it is all connected with wire (which means it is at the same potential). b) Node Voltage equations The equations are Kirchoff Current Law (KCL) equations. I'll write them as the sum of the currents leaving a node, so currents entering will be written negative. The sum should be zero. Node V1 V1 = 8V. This is not a KCL -- I can just directly to the solution here because of the voltage source connected to the reference node. KCL @ V2 (V2 - V1)/1K + ia + (V2 - V4)/6K + (V2 - V4)/7K - 12mA + (V2 - V5)/4k = 0 Notice that node V2 always shows up positive and all the other node voltages always show up negative (because we are writing all the currents leaving). Also, I needed to label the current leaving node V2 down through the voltage source Va as a new unknown, ia. I could not write this as 9V because that's not a current and I need to write all currents for KCL.
- 17. KCL @ V3 -ia + V3/3K = 0 Notice that I must be consistent with my direction of ia. Since it was heading down through Va, now I need to write it negative because it is entering node V3 (and I'm writing all currents in terms of current leaving, so entering is negative). KCL @ V4 (V4 - V2)/(6K || 7K) + 12mA + V2/5K = 0 Since the 6K and 7K have one end connected to V4 and the other end connected to V2 (their head and tail are connected), then they are in parallel (i.e., they share voltage). Thus I can write them both into the first current term in the equation above. KCL @ V5 (V5 - V2)/4K + V5/2K = 0 Extra Equation Because I added an extra unknown, ia, I need an extra equation. We can get it from the voltage source Va: V2 - V3 = 9V c) Power in Va To find the power dissipated by source Va, we cannot use Ohm's law because that only works for resistors. So we use the direct definition: P = V * I P = 9V * ia d) Label mesh currents
- 18. Label each "window pane" that traverses a complete inner loop of the circuit. Here's how I labeled them: e) Mesh current equations For mesh current, we write Kirchoff Voltge Law (KVL) around each loop. I'll write them all in a clockwise direction. KV around i1 (i1 - i4)*6K + (i1 - i2)*7K = 0 KV around i2 (i2 - i1)*7K - Vc = 0 Note that I don't know the voltage over the current source, so I need to add an extra unknown, Vc. I labeled it with the + on the left side and the - on the right side. Since we are going the i2 loop in a clock-wise direction, we hit the minus side first, so we write Vc as negative. KV around i3 i3 * 1K + 9V + (i3 - i4)*3K - 8V = 0 Notice that we hit the 9V source on the plus side and the 8V source on the minus side as we traverse in a clockwise direction. KV around i4 (i4 - i3)*3K -9V + (i4 - i1)*6K + (i4 - i5)*5K = 0
- 19. KV around i5 (i5 - i4) * 5K + Vc + i5 * (4K + 2K) = 0 Since the 4K and 2K are in series (they share current), we can combine them in the last term above. Extra equation Since I added a new unknown, Vc, I also need an extra equation. I get it from the current source: i5 - i2 = 12mA Note that i5 is in the same direction as the current source, so it is written positive. i2 is in the opposite direction, so it is negative. f) Power in Va P = V * I P = 9V * (i3 - i4) Write KCL equations for each node. For consistency, we will always write the currents leaving a node as positive, entering as negative. KCL at Node V1 Instead of writing the normal current equation for this node, we notice that there is a voltage source connected to the node (for which we don't know the current, off-hand). Because the reference node is on the other side of that voltage, we can then simply solve for V1 directly:
- 20. V1 = 8V KCL at Node V2 Here we write the currents leaving left, down, and right: KCL at Node V3 Here we write the currents leaving left, down, and right again. However, in this case, there is a voltage source to the right, so we don't know its current. We thus add an unknown current, ix. KCL at Node V4 Here we write the currents leaving left, up, and right. KCL at Node V5 Here we write the currents leaving left, and up. However, we notice that leaving up we find a voltage source. To be consistent, we use the same unknown current ix , as we did for V3, but in the opposite direction in this case. Extra Equation
- 21. We now have 5 equations, but 6 unknowns, because we added ix. The extra equation to make this set solvable comes from the voltage source that forced us to add the unknown current in the first place. V3 - V5 = 7V Notice that we use V3 as positive because it is at the "+" end of the 7 volt source and V5 as negative because it is at the "-" end. You can think of the 7 volt source as analagous to an elevator that goes up 7 floors, starting from V5 to get to V3.