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 IN.
To find the IN 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
IN = (180/19) / 3
IN = 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
RN = (10//5) + 3
RN = 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 IN by applying
Ohms law to 4Ω resistor,
I = V/R
IN = 24/4
IN = 6A
Then we find the RN . See figure 14.7
RN = 12//4
RN = 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.
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ReplyDeleteSir could you please solve the problem-1 in mesh analysis
ReplyDeletein problem 1st acc to ohms law v=ir
ReplyDeleteand In=6A it is wrong or right
Q.no.3
ReplyDeleteIN= 3.9A
RN=6ohm
Correct ans is
ReplyDeleteIN=_36
RN=7 ohm
Current -13.26 flowing through 12 ohm resistor