I'm trying to apply Nodal Analysis here.I have found that taking reference Nodes and application of Kirchhoff's laws yields some equations and on solving these equations i will get the needed result.But in this question the current sources are not given ,in the online example i just saw there was a current source and 3 equations were formed,in this text book example they have solved it some how,can some one tell me how they have formulated the equations

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The sum of all currents must equate to zero.


All they did in your example is the following

i4 = i1 + i2 + i3


\$i_4 = \frac{20-v}{R4}\$

\$i_1 = \frac{v-50}{R1}\$

\$i_2 = \frac{v}{R2}\$

\$i_3 = \frac{v}{R3}\$

[edit] because of comment about choice of direction of current

To add another example simple2

I randomly chose the direction of the currents. It doesn't matter what direction I choose (the math will work itself out) so long as the the sum of currents and exiting the node is zero.

Looking at my example here, I'm going to say that current entering a node is positive, and current leaving the node is negative. It doesn't have to be this way, it can be the opposite if you want, but I chose this convention. With that said, that means

\$ i_1 + i_2 - i_3 - i_4 =0\$ because current entering a node I've defined as positive and current leaving a node I've defined as negative.

\$i_1 = \frac{50-v}{R1}\$

\$i_2 = \frac{-v}{R2}\$

\$i_3 = \frac{v}{R3}\$

\$i_4 = \frac{v-20}{R4}\$

When you plug these into the equation above, you can isolate v and you'll get v= 31.147V.

So it doesn't matter which direction you think current goes, so long as you define the direction first, then write the equation for the node with your chosen direction, it'll work out.

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  • \$\begingroup\$ Why do you consider current flows from left to right? \$\endgroup\$ – techno Jun 2 '13 at 9:31
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    \$\begingroup\$ Its arbitary. You choose what direction current flows. If you assume current goes left to right, and your calculated current is positive, then your 'guess' was correct. If calculation yeilds a negative value, then your initial guess was wrong. But it doesnt matter which direction current goes, so long as the current entering and exiting a node is zero. \$\endgroup\$ – efox29 Jun 2 '13 at 12:45

First you must realize that there is only one node you don't know the voltage for. They are calling it v, and it is the node at the top center (the bottom node can be your ground reference).

The voltage for the left and right nodes are given by the voltage sources.

Since there is only one unknown, there is only one equation. It is formulated by adding all currents coming into the node. These currents must add zero, otherwise the node would be building up or giving away charge forever!

The current for each branch coming into the node is calculated as per ohm's law: current = voltage/resistance. These are all in terms of v, so you simply proceed to solve for it algebraically.

Now that you have the voltage for all the nodes, you can directly calculate the current flowing through any branch you want, which is what they do at the end of the posted image.

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  • \$\begingroup\$ How can you say that the voltage of the left and right nodes is given by the voltage sources.Exactly which nodes? This question might be little stupid \$\endgroup\$ – techno Jun 2 '13 at 9:30

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