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I need to calculate the voltage at node A at the below circuit. Please tell me whether my approach is right

enter image description here

So, I am assuming that the 1000ohms in series with the voltage sources as internal resistances.

Current through the 500ohm would be, I = 10/500 = 20mA

But we need to consider the internal resistance of 1000ohms right So, terminal voltage, Er = V - IR, Er = 10 - (20mA)*(1000Ohms) = 10-20 = -10V ?

What am i doing wrong here?

Please let me know

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    \$\begingroup\$ If you redraw the circuit with the two 1k resistors on the right in series with the 0.5k resistor you can easily see the voltage divider described by Tabin1000, and the solution is trivial. \$\endgroup\$ – jerry Nov 2 '19 at 6:52
  • \$\begingroup\$ By symmetry, the same value of current flows through each 1k resistor. Let this current be \$I\$. Then solve using KVL, for example, on one of the meshes. \$\endgroup\$ – Chu Nov 2 '19 at 8:14
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You have an voltage divider. When you have two 1000 ohm resistors in parallel like this, then it equals one 10V source with 500 ohm resistor in series with it. When you make voltage divider with two resistors of the same value, then you divide the voltage by 2.

10V / 2 = 5V

It means that on point A you will get 5V.

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  • \$\begingroup\$ I am still unable to get when you say, " When you have two 1000 ohm resistors in parallel like this, then it equals one 10V source with 500 ohm resistor in series with it". I can do it in KCL and KVL. But is there an intuitive way of understanding this \$\endgroup\$ – Newbie Nov 3 '19 at 10:24
  • \$\begingroup\$ If you have x equal resistors in parralel then resistance = R / x \$\endgroup\$ – Tabin1000 Nov 3 '19 at 10:27
  • \$\begingroup\$ But the point is I have a Voltage source in parallel too.. What to do about that voltage source. Please, excuse me for asking naive questions. I am a beginner and would like to understand the answer in simple and intuitive terms for strong basics \$\endgroup\$ – Newbie Nov 3 '19 at 10:44
  • \$\begingroup\$ If you have 2 voltage sources with common ground and the same voltage like that then it works as one source. \$\endgroup\$ – Tabin1000 Nov 3 '19 at 10:55
  • \$\begingroup\$ So, it wont matter even if one of the source is absent, I would still get 5V at node A? Is this correct? Any analogy or intuitive explanation you can provide? It would be really helpful to me. \$\endgroup\$ – Newbie Nov 3 '19 at 10:56

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