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.
SURYA said:
8 years ago
(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 Volt but the current through the r2 resistor is 4.
180/37 = 113mA.
(1665(Va-12)+3060Va+1258(Va-4))/113220 = 0.
1665Va-19980+3060Va+1258Va-5032 = 0.
5983Va = 25012.
So Va = 25012/5983 = 4. 1805 Volt but the current through the r2 resistor is 4.
180/37 = 113mA.
Nisar Ahmad Stanikzai said:
8 years ago
I=112.9mA.
Shadan said:
5 years ago
Can anyone please explain how came 1665, 3060, and 12558?
Divya N B said:
3 years ago
Multiply 68*37*90 = 226440
then 226440/68 = 3330 again 3330/2 = 1665.
226440/37 = 6120 again 6120/2 = 3060.
226440/90 = 2516 again 2516/2 = 1258.
then 226440/68 = 3330 again 3330/2 = 1665.
226440/37 = 6120 again 6120/2 = 3060.
226440/90 = 2516 again 2516/2 = 1258.
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