Lecture 4 Circuit Analysis Techniques PDF

EE 401 Electrical Circuits I Lecture 4 Circuit Analysis Techniques Superposition Theorem The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to ea...

EE 401 Electrical Circuits I Lecture 4 Circuit Analysis Techniques Superposition Theorem The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone. A linear circuit is one whose output is linearly related (or directly proportional) to its input. Keep in mind that superposition is based on linearity. For this reason, it is not applicable to the effect on power due to each source, because the power absorbed by a resistor depends on the square of the voltage or current. If the power value is needed, the current through (or voltage across) the element must be calculated first using superposition. The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately. However, to apply superposition principle, we must keep two things in mind: 1. We consider one independent source at a time while all other independent sources are turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and every current source by 0 A (or an open circuit). This way, we obtain a simpler and more manageable circuit. 2. Dependent sources are left intact because they are controlled by circuit variables. Steps to apply superposition principle: 1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using the techniques covered in the previous topics. 2. Repeat step 1 for each of the other independent sources. 3. Find the total contribution by adding algebraically all the contributions due to the independent sources. Example 4.1. Use the superposition principle to find v in the given circuit below. Page 1 of 8 EE 401 Electrical Circuits I Lecture 4 Circuit Analysis Techniques Example 4.2 Find 𝑖𝑜 in the given circuit using superposition principle. Example 4.3 Use the superposition principle to find i. Source Transformation A source transformation is the process of replacing a voltage source v s in series with a resistor R by a current source is in parallel with a resistor R, or vice versa. Shown in figure 1 below is the transformation of independent sources. Fig. 1 Source transformation requires that 𝑣𝑠 𝑣𝑠 = 𝑖𝑠 𝑅 𝑜𝑟 𝑖𝑠 = (𝑒𝑞. 1) 𝑅 Source transformation also applies to dependent sources, provided we carefully handled the dependent variable. As shown in figure below, a dependent voltage source in series with a resistor can be transformed to a dependent current source in parallel with the resistor or vice versa where we make sure that equation (1) is satisfied. Page 2 of 8 EE 401 Electrical Circuits I Lecture 4 Circuit Analysis Techniques Fig. 2 Like the wye-delta transformation, a source transformation does not affect the remaining part of the circuit. When applicable, source transformation is a powerful tool that allows circuit manipulations to ease circuit analysis. However, we should keep in mind the following points in mind when dealing with source transformation. 1. Note from figure 1 (or figure 2) that the arrow of the current source is directed toward the positive terminal of the voltage source. 2. Note from equation 1 that source transformation is not possible when R = 0, which is the case with an ideal voltage source. However, for a practical, non-ideal voltage source, R ≠ 0. Similarly, an ideal current source with R = ∞ cannot be replaced by a finite voltage source. Example 4.4 Find the power associated with the 6-V source for the circuit given below. Example 4.5 Use source transformation to find 𝑣𝑜 in the given circuit below. Thevenin’s Theorem Thevenin’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source V th in series with a resistor Rth where Vth is the open circuit voltage at the terminals and R th is the input or equivalent resistance at the terminals when the independent sources are turned off. The Thevenin equivalent circuit (TEC) is shown in figure 3. Page 3 of 8 EE 401 Electrical Circuits I Lecture 4 Circuit Analysis Techniques Fig. 3 TEC Thevenn’s theorem is very important in circuit analysis. It helps simplify a circuit. A large circuit may be replaced by a single independent voltage source and a single resistor. This replacement technique is a powerful tool in circuit design. Our major concern right now is how to find the Thevenin equivalent voltage V th and resistance Rth. The Vth is the open-circuit voltage across the terminals a-b when the load is removed as shown in figure 4a; that is, 𝑉𝑡ℎ = 𝑣𝑜𝑐 (𝑒𝑞 2) Again, with the load disconnected and terminals a-b open-circuited, we turn off all independent sources. Rth is the input resistance at the terminals when the independent sources are turned off, as shown in figure 4b; that is 𝑅𝑡ℎ = 𝑅𝑖𝑛 (𝑒𝑞 3) Fig. 4 To apply this idea in finding the Thevenin resistance Rth, we need to consider two cases. Case 1. If the network has no dependent sources, we turn off all independent sources. R th is the input resistance of the network looking between terminals a and b. A linear circuit with a variable load can be replaced by the Thevenin equivalent, exclusive of the load. The equivalent network behaves the same way externally as the original circuit. Consider a linear circuit terminated by a load RL, as shown in figure 5. Fig 5 Fig 3 Page 4 of 8 EE 401 Electrical Circuits I Lecture 4 Circuit Analysis Techniques The current IL through the load and the voltage VL across the load are easily determined once the Thevenin equivalent of the circuit at the load’s terminals is obtained, as shown in figure 3. From this figure, we obtain 𝑉𝑡ℎ 𝐼𝐿 = (𝑒𝑞 4) 𝑅𝑡ℎ + 𝑅𝐿 𝑅𝐿 𝑉𝐿 = 𝑅𝐿 𝐼𝐿 = 𝑉 (𝑒𝑞 5) 𝑅𝑡ℎ + 𝑅𝐿 𝑡ℎ Example 4.6 Find the Thevenin equivalent circuit of the circuit shown in figure 4.16, to the left of the terminals a-b. Then find the current through Rl = 6 Ω, 16 Ω, and 36 Ω. Case 2. If the network has dependent sources, we turn off all independent sources. As with superposition, dependent sources are not to be turned off because they are controlled by circuit variables. We apply a voltage source v0 at terminals a and b and determine the resulting current i0. Then Rth = v0/i0, as shown in figure 6a. Alternatively, we may insert a current source i0 at terminals a-b as shown in figure 6b and find the terminal voltage v 0. Again Rth = v0/i0. Either the two approaches will give the same result. In either approach we may assume any value of v 0 and i0. For example, we may use v0 = 1 V or i0 = 1 A, or even unspecified values of v0 or i0. It often occurs that Rth takes a negative value. In this case, the negative resistance implies that the circuit is supplying power. This is possible in a circuit with dependent sources. Example 4.7. Find the Thevenin Equivalent Circuit. Page 5 of 8 EE 401 Electrical Circuits I Lecture 4 Circuit Analysis Techniques Norton’s Theorem Norton’s Theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source I N in parallel with resistor RN, where IN is the short circuit current through the terminals and R N is the input or equivalent resistance at the terminals when the independent sources are turned off. The Norton’s Equivalent Circuit (NEC) is shown in figure 6b. Fig 6a. Original circuit Fig 6b. Norton Equivalent Circuit We are mainly concerned with how to get RN and IN. We find RN in the same way we find Rth. In fact, from what we know about source transformation, the Thevenin and Norton resistances are equal; that is 𝑅𝑁 = 𝑅𝑡ℎ (𝑒𝑞 6) To find the Norton current IN, we determine the short circuit current flowing from terminal a to b in both circuits in figure 6a and 6b. Thus, 𝐼𝑁 = 𝐼𝑠𝑐 (𝑒𝑞 7) Dependent and independent sources are treated the same way as in Thevenin’s theorem. Observe close relationship between Norton’s and Thevenin’s theorems: R N = Rth as in equation 6, and 𝑉𝑡ℎ 𝐼𝑁 = (𝑒𝑞 8) 𝑅𝑡ℎ This is essentially source transformation. For this reason, source transformation is often called Thevenin-Norton transformation. Since Vth, IN, and Rth are related according to eq 8, to determine the Thevenin or Norton equivalent circuit requires that we find: The open-circuit voltage voc across terminals a and b. The short-circuit current isc at terminals a and b. The equivalent or input resistance Rin at terminals a and b when all independent sources are turned off. We can calculate any two of the three using the method that takes the least effort and use them to get the third using Ohm’s law. Also, since 𝑉𝑡ℎ = 𝑣𝑜𝑐 𝐼𝑁 = 𝑖𝑠𝑐 𝑣𝑜𝑐 𝑅𝑡ℎ = = 𝑅𝑁 𝑖𝑠𝑐 Page 6 of 8 EE 401 Electrical Circuits I Lecture 4 Circuit Analysis Techniques the open-circuit and short-circuit tests are sufficient to find any Thevenin or Norton equivalent, of a circuit which contains at least one independent source. Example 4.8 Find the Norton equivalent circuit of the circuit in Fig. 4.39 at terminals a-b. Example 4.9. Find the Norton Equivalent Circuit (NEC). Maximum Power Transfer Maximum power is transferred to the load when the load resistance equals the Thevenin resistance as seen from the load, that is 𝑅𝐿 = 𝑅𝑡ℎ The maximum power transfer is given by the equation (𝑉𝑡ℎ )2 𝑃𝑚𝑎𝑥 = (𝑒𝑞 9) 4𝑅𝑡ℎ Equation 9 applies only when RL = Rth. Example 4.10. Find the value of RL for maximum power transfer. Also, find the maximum power. Page 7 of 8 EE 401 Electrical Circuits I Lecture 4 Circuit Analysis Techniques Example 4.11. Determine the value of RL that will draw the maximum power from the rest of the circuit. Calculate the maximum power. Page 8 of 8