##
__Solved problems on Norton’s theorem__

__Solved problems on Norton’s theorem__

Q1) Find the Norton’s equivalent circuit across A-B
terminals for the circuit shown in figure 14.1

Answer:

First we remove the 10Ω resistor and short circuit the
terminals A&B. see figure 14.2

The current flowing through the short circuited terminals is
called the Norton’s curren I

_{N}.
To find the I

_{N}we apply nodal equation for point C
(V – 30)/5 + V/10 + V/3 = 0

**V = 180/19**

Ohms law to 3Ω resistance

I = V/R

**I**

_{N}= (180/19) / 3

__I___{N }= 60/19
This is the Norton’s current.

Now we are about to find the Norton’s resistance. Note that
this is equal to the thevenin’s resistance also.

See figure 14.3

R

_{N}= (10//5) + 3

__R___{N}= 19/3Ω
Q2) Find the Norton’s equivalent circuit across A-B
terminals for the circuit shown in figure 14.4

Answer:

First we short circuit the two terminals A&B and
calculate the current flowing

See figure 14.5

When we short circuit A-B terminals it results in shorting
out 12Ω resistor as shown in figure 14.6

Now we can simply find the current I

_{N}by applying Ohms law to 4Ω resistor,
I = V/R

I

_{N }= 24/4

__I___{N}= 6A
Then we find the R

_{N}. See figure 14.7
R

_{N}= 12//4

__R___{N}= 3Ω
Q3) Using Norton’s theorem calculates the current flowing
through 12Ω resistor. You may refer figure 14.8

Wait for my next post to see the solution. Till then, try
this yourself. It is very similar to Q1.

why the Q1 answer norton's current are not divided?..i mean y 180 can't divide in 19?

ReplyDeleteI'm sorry Can you please clarify your problem again?

ReplyDeletehow 2 solve the norton equation wchich ha dependent source

ReplyDeletethank u .......

ReplyDeleteI am clueless about this statement. First we remove the 10Ω resistance and short circuit the terminals A&B. see figure 14.2. Why isn't 12ohm resistor there?

ReplyDeleteI am just mentioning the 10 ohm resistor which is in between the points A&B

DeleteThis comment has been removed by a blog administrator.

ReplyDeleteanswers of the questions is;

ReplyDeleteIn=4amp & Rn=6ohm

is this right?

http://eefundamentals.blogspot.com/2012/09/nortons-theorem-solved-problems2.html

Deletecheck it over there. thank you for viewing

In=4/3amp,Rn=6ohm

Deletei got I = 3.24 A , R = 6 ohm

DeleteTHESE ARE TOP NOTCH INSTANCES

ReplyDeletei got current through 12 ohm resistor is 0.8866 ohm.Is it correct?

ReplyDeletehttp://eefundamentals.blogspot.com/2012/09/nortons-theorem-solved-problems2.html

Deleterefer the link

no current will flow there, current will choose flow where there is no resistance

ReplyDelete?how 12ohm get shorted? plsexplain wid reason

ReplyDeleteit's simply like this, when there is a highway to travel, no one uses the narrow road. electrons do so....

DeleteIs v thevinin =v Norton

ReplyDeleteIs v thevinin =v Norton

ReplyDeleteIs v thevinin =v Norton

ReplyDeleteHow 12 ohm get shorted? Plz say again

ReplyDeleteHow 12 ohm get shorted? Plz say again

ReplyDeleteOkay, this is how it happens, if you connect a resistor parallely with a conductor (say 12ohm resistor and 0 ohm resistor) current never flows through the resistor. it always choose the conductor. that is the meaning of "no current will flow in 12 ohms resistor as it is short circuited".

Deletethis is a rule you must apply on wherever you find a resistor is parallel connected with a conductor (or 0 ohm resistor). Hope you understood.

if still you have problems, please let me know.

How to solve the problem if load resistor is in the middle of the circuit?

ReplyDeleteIN Q1 fig 14.3 why 3 ohm resistance is consider ,even it is short

ReplyDeletesry not short but open . why?

DeleteIn= 4A, Rn= 6ohm, Current across the load resistor (i.e. Across 12ohm resistor) Il=1.33A, Voltage across 12 ohm = 16V.

ReplyDeleteLast one, I= 1.24V and R=5.4ohms

ReplyDeleteI found the answer of Q3 as follows

ReplyDeleteIL=1.33 A

I found the answer of Q3 as follows

ReplyDeleteIN=4A

IL=1.33 A

There is something wrong in the Q1.. We v and In.. In will be 90/19

ReplyDeleteNo. nothing is wrong with the question rishab. Please upload your work sheet. I will guide you and show u where u did wrong.

DeleteThank you.

The various sorts of ferrite materials are available yet ordinary kind of ferrites are nickel-zinc ferrite and manganese-zinc ferrite. Manganese Zinc is having maximal change thickness and its viable repeat broaden is under 5MHz yet banished ordinary mode inductors that impedance is awesome decision up to 70MHz. the Nickel zinc ferrite having unimportant movement thickness in the midst of manganese-zinc is higher.coil winding machine

ReplyDeleteCopper or aluminum wire utilized as the major current passing on transmitter in a transformer winding. Aluminum wire is irrelevant effort and lighter than the copper wire, which has a more prominent cross sectional region of transmitter used to pass on the liberal measure of current, so it is in a general sense utilized as a bit of expansive power transformer applications.automatic toroidal winding machine

ReplyDeleteIn the midst of each and every one of the six material may related in charge application in any case each material has solitary magnificent conditions. The key material of MPP has used for most irrelevant fixation bother, which affects interfacing with mollypermalloy powder focuses and this may fitting for flyback transformer, buckboost transformer. The doled out air opening of MPP focuses is cover stray interfacing with field and clears gap occasion botches up included with gapped ferrite focuses.automatic toroidal winding machine

ReplyDelete