Electrical Engineering - Branch, Loop and Node Analyses - Discussion

Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 14)
14.
What is the voltage drop across R3?

112 mV
11.25 V
180 mV
1.80 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
16 comments Page 1 of 2.

Somesh said:   9 years ago
For first loop:
12= 68i' + 37(i' + i")
12= 105 i' + 37 i" -----------------------1

For 2nd loop:
4= 90 i" + 37(i' + i")
4= 37 i' + 127 i" ----------------------------2


Solve equations 1 and 2 we get;

i' =114mA,
i" = 2mA,

Vr3 = i" X R,
Vr3 = 2mA X 90,
Vr3 = 180mV.
(1)

ALI said:   1 decade ago
We consider the print between R1, R2 and R3 = VA.

VA-12/86+VA-0/37+VA-4/90 = 0.

(90*37)*VA-(90*37)*12+(86*90)*VA+(37*86)*VA-(37*86)*4/(37*90*86) = 0.

11966 VA-50024 = 0.

11966 VA = 50024.

VA = 4.18.

Krishna agarwal said:   1 decade ago
Dear @Ankitha first consider a voltage V at point b/w R1 R2 R3 and then apply KCL at that point and then solve it for V. Again for voltage across R3 subtract VS2 from V.

ELIAS said:   1 decade ago
By Kirchhoff's current law through nodal analysis.

Solution:

Consider the equation,

(VA-VS1/R1) + (VA/R2) + (VA-VS2/R3) = 0

Fairoz said:   1 decade ago
Super position theorem is very good one to solve this type of network. Because calculation gets very simple.

Bikram said:   9 years ago
I have a question. When we have to use +ve or -ve sign in KCL nodal analysis equation?

Pratik soni said:   1 decade ago
Please help me who to solving the type questions &describe it's steps?

Clex said:   8 years ago
VA.4.18 is right. Further, how can be extracted 180mv?
(1)

Bibhuti said:   1 decade ago
Please can any one describe it step by step?

SKy NeT said:   1 decade ago
Can anyone tell me how to do this problem?


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