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We conclude that if DOT and DOC are same then potential difference across resistor is negative and if they are opposite then the potential difference across resistor is positive.

The sign of power source is easier to decide. It merely depends on the direction of travel (DOT). Moving across a EMF source from negative to positive terminal is like moving from a point of lower to point of higher potential. Thus, if traveling across a source, we move from negative to positive terminal then potential difference is positive otherwise negative.

Combining above considerations, we write KVL equations for loops ABCFA and EDCFE as :

Voltage across circuit elements

Network of resistors

Loop EDCFE (Starting from E) :

V = - 10 - 5 I 1 + 4 I 2 + 8 = 0 5 I 1 4 I 2 = 2

Loop ABCFA (Starting from A) :

V = - 5 + 5 I 1 + I 2 + 4 I 2 + 8 = 0 5 I 1 + I 2 + 4 I 2 = - 3 5 I 1 + 9 I 2 = - 3

Subtracting first from second we eliminate I 1 and we have :

13 I 2 = - 1 I 2 = - 1 13 A

Current in ED,

I 1 = - 2 + 4 I 2 5 = - 2 + 4 X 1 13 5 = - 6 13 A

Current in BA,

I 1 + I 2 = - 1 13 6 13 = - 7 13 A

Clearly, direction of current in each of the branch are opposite to the ones assumed.

Problem : Consider the network of resistors as shown here :

Network of resistors

Network of resistors connected to EMF source

Each resistor in the network has resistance 2 Ω. The EMF of battery is 10 V having internal resistance 1/6 Ω. Determine the equivalent resistance of the network.

Solution : We have seen in the earlier example that if I be the current, then current is distributed in different branches of the network as shown in the figure.

Network of resistors

Network of resistors connected to EMF source

Clearly, we need to determine current I in order to calculate equivalent resistance of the network. For this, we consider the loop ABCMA in clockwise direction. Applying KVL :

V = I 3 I 6 I 3 I X 1 6 + 10 = 0 5 I 6 + I 6 = 10 I = 10 A

Let R e q be the equivalent resistance of the network. Reducing given circuit and applying KVL in clockwise direction, we have :

Equivalent resistance

Network of resistors connected to EMF source

V = - 10 X R e q 10 X 1 6 + 10 = 0 R e q = 50 60 = 5 6 Ω

Exercise

Consider the network of resistors as shown here :

Electrical network

Electrical Network

Determine the equivalent resistance of the network between A and C.

In order to determine equivalent resistance, we assume that given network is connected to an external source of EMF equal to E. Now, the external EMF is related to effective resistance as :

E = I R e q

Once this relation is known, we can determine equivalent resistance of the given network. It is important to note that current distribution is already given in the problem figure.

Electrical network

Electrical Network

Considering loop ABCEA in clockwise travel, we have KVL equation as :

V = - I 1 R 1 I 2 R 2 + E = 0 E = I 1 R 1 + I 2 R 2

Considering loop ABDA in clockwise travel, we have KVL equation as :

V = I 1 R 1 I 1 I 2 R 3 + I 2 R 2 = 0

We solve for I 2 to get an expression for it in terms of I 1 as :

I 2 = I 1 R 1 + R 3 R 2 + R 3

Substituting above expression of I 2 in the equation obtained earlier for E, we have :

E = I 1 R 1 + I 1 R 1 + R 3 R 2 R 2 + R 3 I 1 = R 2 + R 3 E [ R 3 R 1 + R 2 + 2 R 1 R 2 ]

Putting this expression for I 1 in the expression obtained earlier for I 2 , we have :

I 2 = R 1 + R 3 E [ R 3 R 1 + R 2 + 2 R 1 R 2 ]

But, we know that :

I = I 1 + I 2

I = R 1 + R 2 + 2 R 3 E [ R 3 R 1 + R 2 + 2 R 1 R 2 ]

Thus,

R e q = E I = [ R 3 R 1 + R 2 + 2 R 1 R 2 ] R 1 + R 2 + 2 R 3

Consider the network of resistors and batteries as shown here :

Electrical network

Electrical Network

Find the currents in different braches of the network.

We assign currents with directions in different branches as shown in the figure. We can, however, assign current directions in any other manner we wish. Here, starting from I 1 and I 2 in branches CA and AB respectively and applying KCL at A, the current in AD is I 1 - I 2 . Let the current in FE is I 3 . Applying KCL at B, current in BC is I 2 + I 3 . Again applying KCL at C, current in CD is I 2 + I 3 - I 1 .

Electrical network

Electrical Network

Considering loop ABCA in clockwise travel, we have KVL equation as :

V = - 2 I 2 1 I 2 + I 3 2 I 1 + 20 = 0 2 I 1 + 3 I 2 + I 3 = 20

Considering loop ADCA in anticlockwise travel, we have KVL equation as :

V = - 2 I 1 I 2 + I 2 + I 3 I 1 2 I 1 + 20 = 0 5 I 1 3 I 2 I 3 = 20

Considering loop BCDFEB in anticlockwise travel, we have KVL equation as :

V = I 2 + I 3 I 2 + I 3 I 1 + 10 2 I 3 = 0 - I 1 + 2 I 2 + 4 I 3 = 10

We have three equations with three variables. Solving, we have :

I 1 = 40 / 7 A ; I 2 = 13 / 7 A ; I 3 = 3 A

Currents in different branches are :

I C A = I 1 = 40 7 A ; I A B = I 2 = 13 7 A I A D = I 1 I 2 = 40 7 - 13 7 = 27 7 A I B C = I 2 + I 3 = 13 7 + 3 = 34 7 A I C D = I 2 + I 3 I 1 = 34 7 40 7 = 6 7 A I D F E B = I 3 = 3 A

Note that current in branch CD is negative. It means that current in the branch is opposite to the assumed direction.

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
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Sherica
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Sherica
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Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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J, combine like terms 7x-4y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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Ramkumar Reply
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what is system testing?
AMJAD
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
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Azam
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Electricity and magnetism. OpenStax CNX. Oct 20, 2009 Download for free at http://cnx.org/content/col10909/1.13
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