Electrical Engineering - Branch, Loop and Node Analyses - Discussion
Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 3)
3.
Find branch current IR2.
Discussion:
14 comments Page 2 of 2.
BOSS said:
1 decade ago
@selva
ohms law V=IR | I=V/R
I= 4.1805/37 = 113mA
ohms law V=IR | I=V/R
I= 4.1805/37 = 113mA
Selva said:
1 decade ago
USING KIRCHOFF'S CURRENT LAW AT CENTRAL NODE,AND TAKE IT AS NODE A,AND GROUNDED POINT NODE B(Vb=0).THEN WE CAN GET,
((Va-12)/68)+((Va-0)/37)+((Va-4)/90)=0
(1665(Va-12)+3060Va+1258(Va-4))/113220=0
1665Va-19980+3060Va+1258Va-5032=0
5983Va=25012
so Va=25012/5983=4.1805 Volts
CURRENT THROUGH R2=37ohms,ie., IR2=4.1805*37=154.678amps
((Va-12)/68)+((Va-0)/37)+((Va-4)/90)=0
(1665(Va-12)+3060Va+1258(Va-4))/113220=0
1665Va-19980+3060Va+1258Va-5032=0
5983Va=25012
so Va=25012/5983=4.1805 Volts
CURRENT THROUGH R2=37ohms,ie., IR2=4.1805*37=154.678amps
Soham joshi said:
1 decade ago
First apply super position and than current divider rule.
Subasha Chandra Behera said:
1 decade ago
How to solve?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers