Since the wires have negligible resistance, the voltage remains constant as we cross the wires connecting the components. These may be currents, voltages, or resistances. Each time a rule is applied, it produces an equation. If there are as many independent equations as unknowns, then the problem can be solved. When locating the junctions in the circuit, do not be concerned about the direction of the currents. If the direction of current flow is not obvious, choosing any direction is sufficient as long as at least one current points into the junction and at least one current points out of the junction.
If the arrow is in the opposite direction of the conventional current flow, the result for the current in question will be negative but the answer will still be correct. The number of nodes depends on the circuit. Each current should be included in a node and thus included in at least one junction equation.
Do not include nodes that are not linearly independent, meaning nodes that contain the same information. Consider Figure 6. There are two junctions in this circuit: Junction and Junction. Points , , , and are not junctions, because a junction must have three or more connections.
The equation for Junction is , and the equation for Junction is. These are equivalent equations, so it is necessary to keep only one of them. When choosing the loops in the circuit, you need enough loops so that each component is covered once, without repeating loops. Option d reflects more loops than necessary to solve the circuit. Consider the circuit in Figure 6. Let us analyse this circuit to find the current through each resistor. First, label the circuit as shown in part b.
Next, determine the junctions. In this circuit, points and each have three wires connected, making them junctions. Junction shows that and Junction shows that. Since Junction gives the same information of Junction , it can be disregarded. This circuit has three unknowns, so we need three linearly independent equations to analyse it. Next we need to choose the loops.
In Figure 6. The loop starts at point , then travels through points , , and , and then back to point. The second loop, Loop , starts at point and includes resistors and and the voltage source.
Starting at point and moving to point , the resistor is crossed in the same direction as the current flow , so the potential drop is subtracted. Moving from point to point , the resistor is crossed in the same direction as the current flow so the potential drop is subtracted.
Moving from point to point , the voltage source is crossed from the negative terminal to the positive terminal, so is added. There are no components between points and. The sum of the voltage differences must equal zero:.
Finally, we check loop. We start at point and move to point , crossing in the opposite direction as the current flow. The potential drop is added. Next, we cross and in the same direction as the current flow and subtract the potential drops and. Note that the current is the same through resistors and , because they are connected in series.
Finally, the voltage source is crossed from the positive terminal to the negative terminal, and the voltage source is subtracted. The sum of these voltage differences equals zero and yields the loop equation.
To solve the three equations for the three unknown currents, start by eliminating current. First add Eq. The result is labeled as Eq. We can solve Eqs. Adding seven times Eq. Using Eq. Finally, Eq. One way to check that the solutions are consistent is to check the power supplied by the voltage sources and the power dissipated by the resistors:. Note that the solution for the current is negative.
This is the correct answer, but suggests that the arrow originally drawn in the junction analysis is the direction opposite of conventional current flow. The power supplied by the second voltage source is and not. Currents have been labeled , , and in the figure, and assumptions have been made about their directions. Locations on the diagram have been labeled with letters through.
In the solution, we apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents. Applying the junction and loop rules yields the following three equations. We have three unknowns, so three equations are required. Simplify the equations by placing the unknowns on one side of the equations. Skip to content 21 Circuits and DC Instruments.
The sum of all currents entering a junction must equal the sum of all currents leaving the junction. The algebraic sum of changes in potential around any closed circuit path loop must be zero. Figure 5.
This circuit is similar to that in Figure 1 , but the resistances and emfs are specified. Each emf is denoted by script E. The currents in each branch are labeled and assumed to move in the directions shown.
If a current is unknown, you must assign it a direction. This is necessary for determining the signs of potential changes. If you assign the direction incorrectly, the current will be found to have a negative value—no harm done. Apply the junction rule to any junction in the circuit.
Each time the junction rule is applied, you should get an equation with a current that does not appear in a previous application—if not, then the equation is redundant.
Apply the loop rule to as many loops as needed to solve for the unknowns in the problem. There must be as many independent equations as unknowns. To apply the loop rule, you must choose a direction to go around the loop. Then carefully and consistently determine the signs of the potential changes for each element using the four bulleted points discussed above in conjunction with Figure 4.
Solve the simultaneous equations for the unknowns. This may involve many algebraic steps, requiring careful checking and rechecking. Check to see whether the answers are reasonable and consistent. The numbers should be of the correct order of magnitude, neither exceedingly large nor vanishingly small. The signs should be reasonable—for example, no resistance should be negative. Check to see that the values obtained satisfy the various equations obtained from applying the rules.
The currents should satisfy the junction rule, for example. Conceptual Questions 1: Can all of the currents going into the junction in Figure 6 be positive? There is no risk here, for if you choose the wrong direction, the current will be of the correct magnitude but negative. For example, in Figure 3 the loop was traversed in the same direction as the current clockwise.
Again, there is no risk; going around the circuit in the opposite direction reverses the sign of every term in the equation, which is like multiplying both sides of the equation by —1.
See Figure 4. Example 1. If a current is unknown, you must assign it a direction. This is necessary for determining the signs of potential changes.
If you assign the direction incorrectly, the current will be found to have a negative value—no harm done. Apply the junction rule to any junction in the circuit. Each time the junction rule is applied, you should get an equation with a current that does not appear in a previous application—if not, then the equation is redundant. Apply the loop rule to as many loops as needed to solve for the unknowns in the problem.
There must be as many independent equations as unknowns. To apply the loop rule, you must choose a direction to go around the loop. Then carefully and consistently determine the signs of the potential changes for each element using the four bulleted points discussed above in conjunction with Figure 4. Solve the simultaneous equations for the unknowns. This may involve many algebraic steps, requiring careful checking and rechecking.
Check to see whether the answers are reasonable and consistent. The numbers should be of the correct order of magnitude, neither exceedingly large nor vanishingly small. The signs should be reasonable—for example, no resistance should be negative. Check to see that the values obtained satisfy the various equations obtained from applying the rules. The currents should satisfy the junction rule, for example.
Conceptual Questions 1. Can all of the currents going into the junction in Figure 6 be positive? Figure 6. Figure 7. Apply the loop rule to loop afedcba in Figure 7. Apply the loop rule to loops abgefa and cbgedc in Figure 7. Apply the loop rule to loop abcdefgha in Figure 5 shown again below. Apply the loop rule to loop aedcba in Figure 5.
Apply the junction rule at point a in Figure 8. Figure 8. Figure 9. Licenses and Attributions.
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