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 1 of 2.
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
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.
Rafiqulislam said:
9 years ago
Correct answer is:
(Va-12)/68-va/37-(va-4)/90 = 0.
=> 1665(va-12)-3060va-1258(va-4) = 0.
=> 5983va = 25012.
=> va = 4.180.
Hence,
IR2 = Va/37 = 4.180/37 = 112.97 = 113mA <=
(Va-12)/68-va/37-(va-4)/90 = 0.
=> 1665(va-12)-3060va-1258(va-4) = 0.
=> 5983va = 25012.
=> va = 4.180.
Hence,
IR2 = Va/37 = 4.180/37 = 112.97 = 113mA <=
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.
YAHYA said:
9 years ago
Answer of this question is 113 mA.
And by the way @Selva! how can we find the branch current using the formula which you used (IR2=4.1805*37=154.678amps)?
And by the way @Selva! how can we find the branch current using the formula which you used (IR2=4.1805*37=154.678amps)?
Shubham said:
10 years ago
Kirchoff law -net current at a junction is 0.
Drop across 37 ohm resistor comes around 5v.
I equals v/r.
Which gives answer as option C.
Drop across 37 ohm resistor comes around 5v.
I equals v/r.
Which gives answer as option C.
Soham joshi said:
1 decade ago
First apply super position and than current divider rule.
Shadan said:
5 years ago
Can anyone please explain how came 1665, 3060, and 12558?
Tamil said:
9 years ago
Correct answer is 113 mA. Thanks for giving me an idea.
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
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