<< Chapter < Page Chapter >> Page >

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.

Solution

We begin by applying Kirchhoff’s first or junction rule at point a. This gives

I 1 = I 2 + I 3 , size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } } {}

since I 1 size 12{I rSub { size 8{1} } } {} flows into the junction, while I 2 size 12{I rSub { size 8{2} } } {} and I 3 size 12{I rSub { size 8{3} } } {} flow out. Applying the junction rule at e produces exactly the same equation, so that no new information is obtained. This is a single equation with three unknowns—three independent equations are needed, and so the loop rule must be applied.

Now we consider the loop abcdea. Going from a to b, we traverse R 2 size 12{R rSub { size 8{2} } } {} in the same (assumed) direction of the current I 2 size 12{I rSub { size 8{2} } } {} , and so the change in potential is I 2 R 2 size 12{ - I rSub { size 8{2} } R rSub { size 8{2} } } {} . Then going from b to c, we go from to +, so that the change in potential is + emf 1 size 12{+"emf" rSub { size 8{1} } } {} . Traversing the internal resistance r 1 size 12{r rSub { size 8{1} } } {} from c to d gives I 2 r 1 size 12{ - I rSub { size 8{2} } r rSub { size 8{1} } } {} . Completing the loop by going from d to a again traverses a resistor in the same direction as its current, giving a change in potential of I 1 R 1 size 12{ - I rSub { size 8{1} } R rSub { size 8{1} } } {} .

The loop rule states that the changes in potential sum to zero. Thus,

I 2 R 2 + emf 1 I 2 r 1 I 1 R 1 = I 2 ( R 2 + r 1 ) + emf 1 I 1 R 1 = 0 . size 12{ - I rSub { size 8{2} } R rSub { size 8{2} } +"emf" rSub { size 8{1} } - I rSub { size 8{2} } r rSub { size 8{1} } - I rSub { size 8{1} } R rSub { size 8{1} } = - I rSub { size 8{2} } \( R rSub { size 8{2} } +r rSub { size 8{1} } \) +"emf" rSub { size 8{1} } - I rSub { size 8{1} } R rSub { size 8{1} } =0} {}

Substituting values from the circuit diagram for the resistances and emf, and canceling the ampere unit gives

3 I 2 + 18 6 I 1 = 0 . size 12{ - 3I rSub { size 8{2} } +"18" - 6I rSub { size 8{1} } =0} {}

Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives

+ I 1 R 1 + I 3 R 3 + I 3 r 2 emf 2 = + I 1 R 1 + I 3 R 3 + r 2 emf 2 = 0 . size 12{+I rSub { size 8{1} } R rSub { size 8{1} } +I rSub { size 8{3} } R rSub { size 8{3} } +I rSub { size 8{3} } r rSub { size 8{2} } - "emf" rSub { size 8{2} } "=+"I rSub { size 8{1} } R rSub { size 8{1} } +I rSub { size 8{3} } left (R rSub { size 8{3} } +r rSub { size 8{2} } right ) - "emf" rSub { size 8{2} } =0} {}

Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered, this becomes

+ 6 I 1 + 2 I 3 45 = 0 . size 12{+6I rSub { size 8{1} } +2I rSub { size 8{3} } - "45"=0} {}

These three equations are sufficient to solve for the three unknown currents. First, solve the second equation for I 2 size 12{I rSub { size 8{2} } } {} :

I 2 = 6 2 I 1 . size 12{I rSub { size 8{2} } =6 - 2I rSub { size 8{1} } } {}

Now solve the third equation for I 3 size 12{I rSub { size 8{3} } } {} :

I 3 = 22 . 5 3 I 1 . size 12{I rSub { size 8{3} } ="22" "." 5 - 3I rSub { size 8{1} } } {}

Substituting these two new equations into the first one allows us to find a value for I 1 size 12{I rSub { size 8{1} } } {} :

I 1 = I 2 + I 3 = ( 6 2 I 1 ) + ( 22 . 5 3 I 1 ) = 28 . 5 5 I 1 . size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } = \( 6 - 2I rSub { size 8{1} } \) + \( "22" "." 5 - 3I rSub { size 8{1} } \) ="28" "." 5 - 5I rSub { size 8{1} } } {}

Combining terms gives

6 I 1 = 28 . 5, and size 12{6I rSub { size 8{1} } ="28" "." 5} {}
I 1 = 4 . 75 A . size 12{I rSub { size 8{1} } =4 "." "75"" A"} {}

Substituting this value for I 1 size 12{I rSub { size 8{1} } } {} back into the fourth equation gives

I 2 = 6 2 I 1 = 6 9.50 size 12{I rSub { size 8{2} } =6 - 2I rSub { size 8{1} } =6 - 9 "." "50"} {}
I 2 = 3 . 50 A . size 12{I rSub { size 8{2} } = - 3 "." "50"" A"} {}

The minus sign means I 2 size 12{I rSub { size 8{2} } } {} flows in the direction opposite to that assumed in [link] .

Finally, substituting the value for I 1 size 12{I rSub { size 8{1} } } {} into the fifth equation gives

I 3 = 22.5 3 I 1 = 22.5 14 . 25 size 12{I rSub { size 8{3} } ="22" "." 5 - 3I rSub { size 8{1} } ="22" "." 5 - "14" "." "25"} {}
I 3 = 8 . 25 A . size 12{I rSub { size 8{3} } =8 "." "25"" A"} {}

Discussion

Just as a check, we note that indeed I 1 = I 2 + I 3 size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } } {} . The results could also have been checked by entering all of the values into the equation for the abcdefgha loop.

Questions & Answers

what is temperature
Adeleye Reply
temperature is the measure of degree of hotness or coldness of a body. measured in kelvin
Ahmad
a characteristic which tells hotness or coldness of a body
babar
Average kinetic energy of an object
Kym
average kinetic energy of the particles in an object
Kym
A measure of the quantity of heat contained in an object which arises from the average kinetic energy of the constituent particles of that object. It can be measured using thermometers. It has a unit of kelvin in the thermodynamic scale and degree Celsius in the Celsius scale.
ibrahim
Mass of air bubble in material medium is negative. why?
Hrithik Reply
a car move 6m. what is the acceleration?
Umaru Reply
depends how long
Peter
What is the simplest explanation on the difference of principle, law and a theory
Kym Reply
how did the value of gravitational constant came give me the explanation
Varun Reply
how did the value of gravitational constant 6.67×10°-11Nm2kg-2
Varun
A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor.
Kris Reply
9.8m/s?
Muhammad
Sqrt(2*1.5m*9.81m/s^2)
Richard
0.5m* mate.
Muhammad
0.05 I meant.
Muhammad
Guess your solution is correct considering the ball fall from 1.5m height initially.
Muhammad
Sqrt(2*1.5m*9.81m/s^2)
Deepak
How can we compare different combinations of capacitors?
Prakash Reply
find the dimension of acceleration if it's unit is ms-2
Happiness Reply
lt^-2
Ahmad
b=-2 ,a =1
Ahmad
M^0 L^1T^-2
Sneha
what is botany
Masha
it is a branch of science which deal with the study of plants animals and environment
Varun
what is work
Sunday Reply
a boy moving with an initial velocity of 2m\s and finally canes to rest with a velocity of 3m\s square at times 10se calculate it acceleration
Sunday
.
Abdul
6.6 lol 😁😁
Abdul
show ur work
Sunday
sorry..the answer is -10
Abdul
your question is wrong
Abdul
If the boy is coming to rest then how the hell will his final velocity be 3 it'll be zero
Abdul
re-write the question
Nicolas
men i -10 isn't correct.
Stephen
using v=u + at
Stephen
1/10
Happy
ya..1/10 is very correct..
Stephen
hnn
Happy
how did the value 6.67×10°-11Nm2kg2 came tell me please
Varun
Work is the product of force and distance
Kym
physicist
Michael
what is longitudinal wave
Badmus Reply
A longitudinal wave is wave which moves parallel or along the direction of propagation.
sahil
longitudinal wave in liquid is square root of bulk of modulus by density of liquid
harishree
Is British mathematical units the same as the United States units?(like inches, cm, ext.)
Nina Reply
We use SI units: kg, m etc but the US sometimes refer to inches etc as British units even though we no longer use them.
Richard
Thanks, just what I needed to know.
Nina
What is the advantage of a diffraction grating over a double slit in dispersing light into a spectrum?
Uditha Reply
can I ask questions?
Boniface Reply
yes.
Abdul
Yes
Albert
sure
Ajali
yeap
Sani
yesssss
bilal
hello guys
Ibitayo
when you will ask the question
Ana
anybody can ask here
bichu
is free energy possible with magnets?
joel
no
Mr.
you could construct an aparatus that might have a slightly higher 'energy profit' than energy used, but you would havw to maintain the machine, and most likely keep it in a vacuum, for no air resistance, and cool it, so chances are quite slim.
Mr.
calculate the force, p, required to just make a 6kg object move along the horizontal surface where the coefficient of friction is 0.25
Gbolahan
Yes ask
Albert
if a man travel 7km 30degree east of North then 10km east find the resultant displacement
Ajali Reply
11km
Dohn
disagree. Displacement is the hypotenuse length of the final position to the starting position. Find x,y components of each leg of journey to determine final position, then use final components to calculate the displacement.
Daniel
1.The giant star Betelgeuse emits radiant energy at a rate of 10exponent4 times greater than our sun, where as it surface temperature is only half (2900k) that of our sun. Estimate the radius of Betelgeuse assuming e=1, the sun's radius is s=7*10exponent8metres
James Reply
2. A ceramic teapot (e=0.20) and a shiny one (e=0.10), each hold 0.25 l of at 95degrees. A. Estimate the temperature rate of heat loss from each B. Estimate the temperature drop after 30mins for each. Consider only radiation and assume the surrounding at 20degrees
James
Practice Key Terms 4

Get the best College physics course in your pocket!





Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask