Analysis of a singleloop circuit using the KVL method Below 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. The first step in the analysis is to assume a reference direction for the unknown current.we do not know apriori what the direction is, nor does it matter. f our assumption is wrong, the current will simply have a minus sign associated with it. The circuit below is shown with an arbitrary direction for the current. The second step in the analysis is to assign the voltage references (where needed) for each element in the circuit. We already know that the passive sign convention for resistors demands that the sense of the current and voltage be selected such that the current enters the more positive terminal. The choice of direction in which we draw the current arrow is arbitrary, but once the direction is chosen, the polarity orientation across the resistors is then fixed. The voltages across the voltage sources and their polarities are taken as given and cannot be altered Page 1
as they are independent sources. Therefore, voltages Vr1 and Vr2 in this loop must be assigned as shown below. start here Vr1 Vr2 Next, we apply Kirchoff s voltage law to the single closed path. We may sum the voltages by traversing the circuit in either direction, but let s do so in the clockwise direction, beginning at the lower left corner, (where it says start here ) and write down each voltage first encountered at its positive reference and write down the negative of every voltage encountered at its negative terminal. Thus we get, 120 V r1 V r2 = 0 This is a single equation with two unknowns. To solve this equation, we must find some way to 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 and the individual resistances using Ohm s law. So we apply the ohms law substitution step. We know that for and R2, Ohm s law states that: V r1 = *, and V r2 = * 15 f we substitute these expressions for V r1 and V r2 into the first equation, we get: Solving for, we obtain: 120 15 = 0 45 = 90, thus = 2A Since the solution for current is a positive value, we know that the assumed direction for current is correct. f the answer had been 2A, we would know that the current would be actually Page 2
flowing the opposite direction. Before we proceed to compute the power for the components in this circuit, lets repeat the problem but reverse the assumed direction of current flow. Therefore we have the situation below. start here Vr1 Vr2 We will now write the KVL equation by traversing the circuit in the clockwise direction as we did before. Again we write down each voltage first encountered at its positive reference and write down the negative of every voltage encountered at its negative terminal. Thus we get, 120 V r1 V r2 = 0 Applying the ohms law substitution step we get: Solving for, we obtain: 120 15 = 0 45 = 90, thus = 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 = 2 * R, we can calculate the power dissipated by the resistors. For the ohm resistor: P = 2 * 2 * = 120 watts, For the 15 ohm 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 Page 3
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. Below we see the progression of taking the source and transforming its current arrow so that the arrow points into the positive terminal just like the model for computing power. = 2A = 2A V Voltage source as in circuit Voltage source altered to look like power dissipation model Model for computing power dissipation Note that the current arrow for the 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: P 120v = 2 * 120 = 240 watts The negative power dissipation indicates that the source is actually delivering power, not dissipating it. For the 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: P = 2 * = 60 watts n this case, the volt voltage source is dissipating power. Page 4
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. n other words, the power dissipated by all the elements must equal the power supplied by all the elements. n our example: P 120 P r1 P P r2 = 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 5. Solve for Remember that: 1. The assumed direction for the current flowing in the loop does not matter. f 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. Page 5
Single Loop KVL Tricks For simple loops, there are a few simplifying steps we can take to quickly find the current. Since the loop is a series circuit, the same current flows through all elements. When we write the KVL equation and do the Ohm s law substitution we obtain a sum of voltages. Since addition is commutative we can add the elements in any order. This means we can rearrange the equation as well as the circuit. For example, our KVL equation was written: 120 15 = 0 t is also correct to write it this way: 120 15 = 0 Which means our circuit could be drawn as shown below in the schematic diagram on the left. After rearranging the circuit, we see that the two voltage sources are connected so as to oppose each other. Furthermore, the source is the bigger one and thus overcomes the source. This result can be seen in the schematic on the right where the series combination of the two sources is redrawn as single source of 90V with the polarity of the bigger source. Since we know that resistors in series add directly, they may be summed and result in one 45 ohm resistor. With this reduced circuit shown on the right, it is easy to compute the current. 15 R2 V eq 90V R eq 45 = 90 45 = 2A Page 6
Sometimes you need to fine the voltage between two points in the single loop KVL circuit that are unusual or don t use what would be considered a convenient reference point. There is a visual method that can be used to help determine the voltage. For example, suppose we have the circuit below and are supposed to find Vx. We have already found the value of. 2A Vx R2 15 The negative sign of Vx marks the point from which we are to find the voltage. n other words, we are to find the voltage with reference to where the negative sign is positioned. To visualize this, imagine a set of stairs. Going up the stairs indicates a voltage increase, going down is a voltage decrease relative to where we were. You begin on a staircase at the reference point which is 0 volts. 0V 60V Vx this step is from the source this step is from R2 As electrons flow through a circuit, they experience either a voltage increase or decrease as they pass through an element. Starting at the negative sign, we traverse the circuit first encountering Vr2 which is volts. As we pass through R2 we experience a increase in voltage potential as we approach its positive terminal With the staircase we can show a step up of volts. Next, we pass through the source experiencing another volt rise in potential as we approach the positive terminal. This is shown as another step up. Now we have reached the positive terminal of Vx, the point at which we are to determine the voltage. We are 60 volts above our reference, therefore Vx = 60V. Page 7