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Download Analysis of a single-loop circuit using the KVL method
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Analysis of a single-loop circuit using the KVL method Figure 1 is our circuit to analyze. We shall attempt to determine the current through each element, the voltage across each element, and the power delivered to or absorbed by each element. You will note that the KVL method determines the unknown current in the loop by using a sum of voltages in the loop. R1 30 V2 30V R1 30 R2 15 V1 120V V1 120V I Figure 1: Single loopV2 circuit. R1 30V 30 The first step in the analysis is to assume a reference direction for the unknown current. We do I=2A not know apriori what the direction is, nor does it matter. If our assumption is wrong, the current Vr1 will simply have a minus sign associated with it. The circuit below is shown with 120Van arbitrary R2 direction for the current I. V1 Vr2 15 120V I voltage source with reference current V2 V2 R1 R1 start here 30V 30V direction30 as initially 30 assumed 120V voltage s V2 referenc 30V computin Vr1 R2 15 V1 120V R1 30 I I R1 30 Vr1 I R2 15 V1 120V Vr2 I I Veq Req 90 Figure 2: Single loop circuit with I direction. 15 V2 V2 30V 30V The second step in the analysis is to assign the voltage references (where for each element I=2A I =needed) −2A I in the circuit. We already know that the passive sign convention for resistors demands that the sense of the current and voltage be120V selected such that the current enters the more positive terminal. 120V V R2 The choice of direction in which we draw the current arrow is arbitrary, but once the direction is chosen, the Vr2 15 I withvoltages across the voltage sources and their polarity orientation across the resistorsvoltage is then source fixed. The voltage source with current polarities are taken as given and cannot be current altered as they are independent sources. Therefore, reference model for computing reference arrow placed for voltages Vr 1 and Vr 2 in this loop must be assigned as shown in figure 4. direction as initially power dissipated computing power dissipation assumed 1 R1 30 R2 15 start here V1 120V e V V2 30V R1 30 R2 15 I q V2 30V R1 30 V2 30V R1 30 Vr1 I R2 15 V1 120V Vr2 I R2 15 start here Figure 3: Single loop circuit, with V signs. Next, we apply Kirchhoffs voltage law to the single closed path. We may sum the voltages by I=2A I = −2A I traversing the circuit in either direction, but let’s do so in the clockwise direction, beginning at the lower left120V corner, (where it says start here) 120V V and write down each voltage first encountered at its positive reference and write down the negative of every voltage encountered at its negative voltage source with terminal. Thus we get, voltage source with current −120 +model Vr1 +for 30computing + Vr2 = 0 . reference current reference arrow placed for direction as initially power This is a single equation with two unknowns. Todissipated solve this equation, we must find some way to computing power dissipation assumed express Vr1 and Vr2 in one variable. Since this is a series circuit and the same current flows through all elements, we may express Vr1 and Vr2 in terms of the current I and the individual resistances using Ohms law. So we apply the Ohm’s law substitution step. We know that for R1 and R2 , Ohms law states that: I Vr1 = I × 30, and Vr2 = I × 15 . If we substitute these expressions for Vr 1 and Vr 2 into the first equation, we get: Req 15 −120 + 30I + 30 + 15I = 0 . Solving for I, we obtain: 45I = 90, thus I = 2A. Since the solution for current is a positive value, we know that the assumed direction for current is correct. If the answer had been -2A, we would know that the current would be actually flowing the opposite direction. Before we proceed to compute the power for the components in this circuit, let’s repeat the problem but reverse the assumed direction of current flow. Therefore we have the situation shown in figure 4. TITLE FILE: REVISION: PAGE OF 2 DRAWN BY: R2 15 V1 120V V1 120V I V2 30V R1 30 I=2A Vr1 V1 120V Vr2 I R2 15 start here 120V voltage source with reference current direction as initially assumed Figure 4: Single loop circuit, with V signs. R1 R2 30 I traversing We will now write the KVL equation by the 15 circuit in the clockwise direction as we did I write before. Again we write down each voltage first encountered at its positive reference and down the negative of every voltage encountered at its negative terminal. Thus we get, V1 120V −120 − Vr1 + 30 − Vr2 = 0 . Veq Req 90 15 Applying the Ohms law substitution step we get: V2 30V −120 − 30I + 30 − 15I = 0 . Solving for I, we obtain: −45I = 90, thus I = −2A. Since the resulting current is negative, we know that the assumed direction for current is incorrect. However, our answer is correct considering the direction of the arrow. To compute the power dissipation, we will assume that a positive current was obtained as in the first example. Knowing the equation for power dissipation; P = I 2 R, we can calculate the power dissipated by the resistors. For the 30Ω resistor: P = 2 × 2 × 30 = 120 Watts. For the 15Ω resistor: P = 2 × 2 × 15 = 60 Watts. Resistors always dissipate positive power. They can never generate power. How much power is generated or consumed by the voltage sources? Remember to compute power dissipated, we must take each element and redefine the voltage reference sign and the current reference arrow so that the arrow points into the positive terminal of the component such that it looks like the model for computing power dissipated shown below. In figure 5 we see the progression of taking the 120V source and transforming its current arrow so that the arrow points into the positive terminal just like the model for computing power. 3 120V voltage referen comput 15 120V 15 I 120V 15 I start here V2 30V R2 15 I=2A Vr2 R2 15 I = −2A I 120V 120V voltage source with reference current direction as initially assumed V voltage source with current reference arrow placed for computing power dissipation model for computing power dissipated Figure 5: Orient the voltage source to match the PSC model Note that the current I arrow for the 120V source is now drawn as the passive sign convention demands; it points into the positive terminal. The voltage references are identical to the model so they were not changed. Now we can compute the power dissipation directly as: Veq Req 90 15 P120V = −2 × 120 = −240 Watts. The negative power dissipation indicates that the source is actually delivering power, not dissipating it. For the 30V source, again the current arrow is drawn as the passive sign convention demands. However in this case, the current into the source is in the same direction as the model dictates, thus: P30V = 2 × 30 = 60 Watts. In this case, the 30 volt voltage source is dissipating power. A useful check of your calculations may be accomplished by using the power check. This method relies on the fact of the conservation of energy. In other words, the power dissipated by all the elements must equal the power supplied by all the elements. In our example: P120V + PR1 + P30V + PR2 = 0 −240 + 120 + 60 + 60 = 0 0 = 0. Thus, our calculations are correct. To summarize the KVL method: 1. Assign a reference direction for the unknown current 2. Assign voltage references to the elements 3. Apply KVL to the closed loop path 4. Substitute in Ohms law where needed to get an equation in I 5. Solve for the current I 4 TITLE FILE: PAGE RE OF DR Remember that: 1. The assumed direction for the current flowing in the loop does not matter. If the assumed direction is backwards to the actual flow, the magnitude will simply be negative. 2. The polarity across the resistors must conform to the passive sign convention. 3. The direction in which you sum voltages does not matter. Neither does it matter where in the loop you start the summation process. 5