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Kirchhoff’s second rule requires emf Ir IR 1 IR 2 = 0 size 12{"emf" - ital "Ir" - ital "IR" rSub { size 8{1} } - ital "IR" rSub { size 8{2} } =0} {} . Rearranged, this is emf = Ir + IR 1 + IR 2 size 12{"emf"= ital "Ir"+ ital "IR" rSub { size 8{1} } + ital "IR" rSub { size 8{2} } } {} , which means the emf equals the sum of the IR size 12{ ital "IR"} {} (voltage) drops in the loop.

Part a shows a schematic of a simple circuit that has a voltage source in series with two load resistors. The voltage source has an e m f, labeled script E, of eighteen volts. The voltage drops are one volt across the internal resistance and twelve volts and five volts across the two load resistances. Part b is a perspective drawing corresponding to the circuit in part a. The charge is raised in potential by the e m f and lowered by the resistances.
The loop rule. An example of Kirchhoff’s second rule where the sum of the changes in potential around a closed loop must be zero. (a) In this standard schematic of a simple series circuit, the emf supplies 18 V, which is reduced to zero by the resistances, with 1 V across the internal resistance, and 12 V and 5 V across the two load resistances, for a total of 18 V. (b) This perspective view represents the potential as something like a roller coaster, where charge is raised in potential by the emf and lowered by the resistances. (Note that the script E stands for emf.)

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] .)

Questions & Answers

state Faraday first law
aliyu Reply
what does the speedometer of a car measure ?
Jyoti Reply
Car speedometer measures the rate of change of distance per unit time.
Moses
describe how a Michelson interferometer can be used to measure the index of refraction of a gas (including air)
WILLIAM Reply
using the law of reflection explain how powder takes the shine off a person's nose. what is the name of the optical effect?
WILLIAM
is higher resolution of microscope using red or blue light?.explain
WILLIAM
can sound wave in air be polarized?
WILLIAM Reply
Unlike transverse waves such as electromagnetic waves, longitudinal waves such as sound waves cannot be polarized. ... Since sound waves vibrate along their direction of propagation, they cannot be polarized
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what is the difference between mass and weight
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Isru
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what is airflow
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derivative of first differential equation
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why static friction is greater than Kinetic friction
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draw magnetic field pattern for two wire carrying current in the same direction
Ven Reply
An American traveler in New Zealand carries a transformer to convert New Zealand’s standard 240 V to 120 V so that she can use some small appliances on her trip.
nkombo Reply
What is the ratio of turns in the primary and secondary coils of her transformer?
nkombo
what is energy
Yusuf
How electric lines and equipotential surface are mutually perpendicular?
Abid Reply
The potential difference between any two points on the surface is zero that implies È.Ŕ=0, Where R is the distance between two different points &E= Electric field intensity. From which we have cos þ =0, where þ is the angle between the directions of field and distance line, as E andR are zero. Thus
MAHADEV
sorry..E and R are non zero...
MAHADEV
By how much leeway (both percentage and mass) would you have in the selection of the mass of the object in the previous problem if you did not wish the new period to be greater than 2.01 s or less than 1.99 s?
Elene Reply
hello
Chichi
Hi
Matthew
hello
Sujan
Hi I'm Matthew, and the answer is Lee weighs in mass 0.008kg OR 0.009kg
Matthew
14 year old answers college physics and the crowd goes wild!
Matthew
Hlo
spread
Practice Key Terms 4

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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