# 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.

5.4 mA
–5.4 mA
113.0 mA
119.6 mA
Explanation:
No answer description is available. Let's discuss.
Discussion:
14 comments Page 1 of 2.

YAHYA said:   10 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)?
(1)

Subasha Chandra Behera said:   1 decade ago
How to solve?

Soham joshi said:   1 decade ago
First apply super position and than current divider rule.

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

@selva

ohms law V=IR | I=V/R

I= 4.1805/37 = 113mA

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.

Mohammad faisal said:   9 years ago
Correct answer is 113 mA. So question is wrong.

Rafiqulislam said:   9 years ago

(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 <=

Tamil said:   9 years ago
Correct answer is 113 mA. Thanks for giving me an idea.

Pardeep singh said:   9 years ago
How is this calculation please explain?