Electrical Engineering - Branch, Loop and Node Analyses - Discussion

Nice explanation, thanks @Antony.

But, the current through R2 is 0.040 @Antony.

Good explanation @Antony.

Even using the super-position theorem, the result remains the same. But we need to apply it with care.
Lets, take the first source, 12V alone and put a short across the 6V supply.
Now, calculate the effective resistance of the circuit.
Eff Res. = 49 + (80*24)/(80+24)
= 67.17 ohm.
Now total current supplied by the source 12V = 12/67.17
= 0.178A.

The current through the 24-ohm resistor (using current divider rule) = main current * resistance of other branch / total resistance.

ie. = 0.178 * 80/(80+24).
= 0.137A ----------------- current through R2 by source 12V.

Now take the same using 2nd source, 6V.

Eff. res. = 49*24/(49+24) + 80.
= 96.109 ohm.

Total current supplied by the 6V source = 6/96.109.
= 0.06 A.

current through R2 resistor = 0.06*49/(49+24).
= 0.040A ---- current through R2 by the source 6 V.

Now add the individual source currents to obtain the resultant current through the R2 resistor = 0.137 + 0.40.
= 0.177A.

@Nawal Kumar.

You made some mistakes. Please refer the above derivation.

Hi all. The sign convention is somewhat a big deal in deciding the answer.
First of all, define the nodes with one as the reference node. Then try to apply KCL at the node where the unknown parameter exists.
* As per conventional rules, the current incoming is considered negative and outgoing is positive to the node.
Now assume the unknown node voltage as the biggest one, so the assumed current would flow away from the node. And here we begin to write the equation.
At the node under consideration,

(VA-12) VA (VA-6)
-------- +----- + ------ = 0
49 24 80

Same as the equation @Mallesh.

After solving the equation you will get the Va value as 4.29V and the current value as 178mA.

@All.
You might have a doubt about the (Va-12)Volts or the (Va-6)Volts, we just have assumed that voltage at Va is the biggest, so the current would flow away from the node. And the current leaving from a node is considered to be positive.

Please can anyone solve complete equation.

By applying superposition theorem.

First 12 volt active and 6 volt replaced by short circuit.

(24*80)/(24+80) = 18.47 ohm.

Current in equation 1.

I1 = 12/(18.47+49) = 0.25 Amp.

Take 6 volt with 12 volt short circuit.

Required v = 49/(49+24)+80 = 80.7 ohm.

I2 = 6/80.7 = 0.074 Amp.

Total current across R2 = 0.25+0.074 = 0.324 Amp.

By applying node analysis

(VA-12)    VA     (VA-6)
-------- +----- + ------  = 0   
  49        24      80

By solving Va = 4.289 v

I thr r2 = 4.289/24 = 178 mA

VA-12/49 +VA/24 +VA-6/80 =0
VA =.178A