Electrical Engineering - Circuit Theorems and Conversions - Discussion
Discussion Forum : Circuit Theorems and Conversions - General Questions (Q.No. 19)
19.
Find the Norton circuit, that is, IIN and RN, for the circuit given below.
478 mA, 12.8 
750 mA, 12.8
478 mA, 6.8
750 mA, 6.8
Discussion:
5 comments Page 1 of 1.
M.V.KRISHNA/PALVONCHA said:
1 decade ago
From the circuit.
Vn=VS*R2/(R1+R2)
Vn=(12*12)/(12+12)=6V
Rn=R3+(R1||R2)
Rn=6.8+6=12.8
In=Vn/Rn
In=6/12.8
In=468mA(app. equal to 478mA)
So ans is option A
Vn=VS*R2/(R1+R2)
Vn=(12*12)/(12+12)=6V
Rn=R3+(R1||R2)
Rn=6.8+6=12.8
In=Vn/Rn
In=6/12.8
In=468mA(app. equal to 478mA)
So ans is option A
(2)
Sri said:
1 decade ago
Please explain me the basic and essential rules regarding norton theorem.
Prashant singh said:
1 decade ago
By applying superposition theorem ie. , 12v is kept on and 6v is off and then 12v is off and 6v is on.
Kumar said:
1 decade ago
Please explain me by keeping off and on the 12 and 6v source. We can calculate it?
Manidersingh said:
4 years ago
12*12/12+12=6 ohm.
6+6.8=12.8 ohm.
V-12/12+v/12 = 6v.
6/12.8= 468 mA.
6+6.8=12.8 ohm.
V-12/12+v/12 = 6v.
6/12.8= 468 mA.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers