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Applying kirchhoff’s rules

By applying Kirchhoff’s rules, we generate equations that allow us to find the unknowns in circuits. The unknowns may be currents, emfs, or resistances. Each time a rule is applied, an equation is produced. If there are as many independent equations as unknowns, then the problem can be solved. There are two decisions you must make when applying Kirchhoff’s rules. These decisions determine the signs of various quantities in the equations you obtain from applying the rules.

  1. When applying Kirchhoff’s first rule, the junction rule, you must label the current in each branch and decide in what direction it is going. For example, in [link] , [link] , and [link] , currents are labeled I 1 size 12{I rSub { size 8{1} } } {} , I 2 size 12{I rSub { size 8{2} } } {} , I 3 size 12{I rSub { size 8{3} } } {} , and I size 12{I} {} , and arrows indicate their directions. There is no risk here, for if you choose the wrong direction, the current will be of the correct magnitude but negative.
  2. When applying Kirchhoff’s second rule, the loop rule, you must identify a closed loop and decide in which direction to go around it, clockwise or counterclockwise. For example, in [link] 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.

[link] and the following points will help you get the plus or minus signs right when applying the loop rule. Note that the resistors and emfs are traversed by going from a to b. In many circuits, it will be necessary to construct more than one loop. In traversing each loop, one needs to be consistent for the sign of the change in potential. (See [link] .)

This figure shows four situations where current flows through either a resistor or a source, and the calculation of the potential change across each. The first two diagrams show the potential drop across a resistor, with the current flowing from left to right or right to left. The other two diagrams show a potential drop across a voltage source, when the terminals are in one orientation and then another.
Each of these resistors and voltage sources is traversed from a to b. The potential changes are shown beneath each element and are explained in the text. (Note that the script E stands for emf.)
  • When a resistor is traversed in the same direction as the current, the change in potential is IR size 12{- ital "IR"} {} . (See [link] .)
  • When a resistor is traversed in the direction opposite to the current, the change in potential is + IR size 12{+ ital "IR"} {} . (See [link] .)
  • When an emf is traversed from to + (the same direction it moves positive charge), the change in potential is +emf. (See [link] .)
  • When an emf is traversed from + to (opposite to the direction it moves positive charge), the change in potential is size 12{ - {}} {} emf. (See [link] .)

Calculating current: using kirchhoff’s rules

Find the currents flowing in the circuit in [link] .

The diagram shows a complex circuit with two voltage sources E sub one and E sub two and several resistive loads, wired in two loops and two junctions. Several points on the diagram are marked with letters a through h. The current in each branch is labeled separately.
This circuit is similar to that in [link] , 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. This example uses Kirchhoff’s rules to find the currents.

Strategy

This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the series-parallel techniques—it is necessary to use Kirchhoff’s rules. Currents have been labeled I 1 size 12{I rSub { size 8{1} } } {} , I 2 size 12{I rSub { size 8{2} } } {} , and I 3 size 12{I rSub { size 8{3} } } {} in the figure and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h. In the solution we will apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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