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

Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 4)
4.
Find the node voltage VA.

518 mV
5.18 V
9.56 V
956 mV
Explanation:
Reason: The voltage drop through R2 is 5.18v.
Discussion:
6 comments Page 1 of 1.

MUHAMMAD FAROOQ said:   5 years ago
Apply KCL at VA.

Va/68-15/68 + Va/30 + Va/100-8/100 = 0.
Va(1/68 + 1/30 + 1/100) - 15/68-8/100 = 0.
0.05804Va-0.3006 = 0,
0.05804Va = 0.3006,
Va = 0.3006/0.05804.
Va = 5.18.

Shareef said:   7 years ago
Any other way to solve this?

Mani said:   7 years ago
Va (1/68 + 1/30 + 1/100) - 15/68 - 8/100 = 0.
74Va/1275 = - 511/1700.
Va = (511 * 1275) / (1700 * 74).
Va = 5.179V.

Shriki said:   8 years ago
Va/68-15/68 + Va/30 + Va/100-8/100 = 0.
Va(1/68 + 1/30 + 1/100) - 15/68-8/100 = 0.
74Va/1275 = -511/1700.
Va = (511*1275)/(1700*74).
Va = 5.179V.

Using kirchoff's current law at node A,

((Va-15)/68)+((Va/30))+((Va-8)/100)=0

(75Va-1125+170Va+51Va-408/5100)=0

296Va=1533

Va=5.179=5.18 volts

Open the following page:

http://www.indiabix.com/electronics-circuits/ohms-law/

Start simulation. Right click on simulator -> File -> Import -> Paste the following code -> Click Import.

\$ 1 5.0E-6 16.13108636308289 50 5.0 50
v 96 240 96 160 0 0 40.0 15.0 0.0 0.0 0.5
v 416 240 416 160 0 0 40.0 8.0 0.0 0.0 0.5
w 96 160 160 160 0
w 416 160 352 160 0
r 160 160 272 160 0 68.0
r 272 160 352 160 0 100.0
r 272 160 272 240 0 30.0
w 96 240 272 240 0
w 272 240 416 240 0
g 272 240 272 256 0

Then you will see the simulation.