# Electrical Engineering - Branch, Loop and Node Analyses - Discussion

Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 3)

3.

Find branch current

*IR*_{2}.Discussion:

14 comments Page 1 of 2.
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.

Shadan said:
6 years ago

Can anyone please explain how came 1665, 3060, and 12558?

(1)

Nisar Ahmad Stanikzai said:
9 years ago

I=112.9mA.

SURYA said:
9 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.

Pardeep singh said:
9 years ago

How is this calculation please explain?

Tamil said:
9 years ago

Correct answer is 113 mA. Thanks for giving me an idea.

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 <=

Mohammad faisal said:
9 years ago

Correct answer is 113 mA. So question is wrong.

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)?

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)

Shubham said:
1 decade 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.

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