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

Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 2)

2.

What is the current through

*R*_{2}?Discussion:

16 comments Page 1 of 2.
Jaywonder said:
7 years ago

Can anyone find the answer using Thevenin's theorem?

Umair said:
7 years ago

@Somashekhar.

How you got that 30096? Explain.

How you got that 30096? Explain.

Krishna said:
8 years ago

The answer must me same irrespective of method we acquire but I did't get answer using mesh analysis.

Akinremi Akinola Musbau said:
8 years ago

@Shinghal.

Your working is right but take note that the current through 30 ohms is the sum of the current that emanate from the two cell and not the difference.

Your working is right but take note that the current through 30 ohms is the sum of the current that emanate from the two cell and not the difference.

Kavan Rathod said:
8 years ago

Simple apply node analysis and solve it.

Somashekhar y said:
9 years ago

((va-12)/49)+(va/24)+((va-6)/80) = 0.

((va-12)(1920)+(va)(3920)+(va-6)(1176))/94080 = 0.

va(7016)-30096 = 0.

va = 30096/7016.

va = 4.28.

Please review previous question. Why nodal equation differs? Can any one explain?

((va-12)(1920)+(va)(3920)+(va-6)(1176))/94080 = 0.

va(7016)-30096 = 0.

va = 30096/7016.

va = 4.28.

Please review previous question. Why nodal equation differs? Can any one explain?

(1)

Pente vinodkumar said:
9 years ago

By applying loop analysis at first loop.

15 = 68i'+ 30(i'+i").

15 = 98i'+ 30i".

Now at second loop:

8 = 100i"+30*(i'+i").

8 = 30i'+130i".

After calculation:

i' = 0.145.

i" = 0.028.

So total current in 30 ohm resistance.

i' + i".

0.145 + 0.028.

= 173 mA.

15 = 68i'+ 30(i'+i").

15 = 98i'+ 30i".

Now at second loop:

8 = 100i"+30*(i'+i").

8 = 30i'+130i".

After calculation:

i' = 0.145.

i" = 0.028.

So total current in 30 ohm resistance.

i' + i".

0.145 + 0.028.

= 173 mA.

(4)

Julius said:
9 years ago

Can you please elaborate the solution so that I can understand. Thanks.

Sunilkumar v said:
1 decade ago

Apply nodal analysis for the circuit. After getting Va value, we can find current at 30 ohm.res

(15-VA)/68 - (VA-0)/30 - (VA-8)/100 = 0.

VA = 5.169.

R2 = VA/30.

R2 = 5.169/30.

R2 = 173 mA.

(15-VA)/68 - (VA-0)/30 - (VA-8)/100 = 0.

VA = 5.169.

R2 = VA/30.

R2 = 5.169/30.

R2 = 173 mA.

(2)

Dinesh kumar t said:
1 decade ago

Applying kcl at node A and taking node B as refrence:

(15-va)/68-va/30-(va-8)/100=0

va=5.1689v;

Ir2=va/30=5.1689/30=0.172A=172mA.

(15-va)/68-va/30-(va-8)/100=0

va=5.1689v;

Ir2=va/30=5.1689/30=0.172A=172mA.

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