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I have this circuit. The voltage source is 4V and the current source is 2A with the 2ohm resistance.

I got the node voltage at the circled node to be +8V from a circuit simulator.

enter image description here

Can someone help me on how to arrive at that +8V on that node intuitively and mathematically please?

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    \$\begingroup\$ Use Ohm's law and you have the answer. What are you able to calculate and where are you stuck? \$\endgroup\$
    – Justme
    Mar 29, 2021 at 16:08
  • \$\begingroup\$ yes. I tried Ohms law. Like, V=IR. Voltage across the resistor would be 2A*2ohms = 4V. So, I thought that there should be a 4V drop across the resistor. And since, one of the resistor is tied to a 4V voltage source and there should be a 4V drop across the resistor, I took like 4V - 4V = 0V at the circled node. This was my confusion. Please clarify \$\endgroup\$
    – Newbie
    Mar 29, 2021 at 16:54
  • \$\begingroup\$ Which way you think the current flows? \$\endgroup\$
    – Justme
    Mar 29, 2021 at 16:59
  • \$\begingroup\$ Counter clockwise \$\endgroup\$
    – Newbie
    Mar 29, 2021 at 17:01

1 Answer 1

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There is 2A flowing counter-clockwise in your circuit, due to the constant current source. Starting at the circle there is a 4V voltage drop through the 2 \$\Omega\$ resistor. The more positive side is at the circle. Continuing counter-clockwise, the battery also has 4V across it. Again, the more positive side is the top, and the more negative on the bottom side of the battery. Since both the resistor and battery are more positive in the same direction, their voltages sum. So the circle is 8V higher than the bottom side of the battery.

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  • \$\begingroup\$ yes. I tried Ohms law. Like, V=IR. Voltage across the resistor would be 2A*2ohms = 4V. So, I thought that there should be a 4V drop across the resistor. And since, one of the resistor is tied to a 4V voltage source and there should be a 4V drop across the resistor, I took like 4V - 4V = 0V at the circled node. This was my confusion. Please clarify \$\endgroup\$
    – Newbie
    Mar 29, 2021 at 16:54
  • \$\begingroup\$ The current flowing through the resistor comes from the constant current source. It is flowing counter-clockwise in the loop of your schematic. So, the voltage drop across the resistor will be more positive at the circle, than at the battery. Continuing counter clockwise, the battery will be more positive on the top than the bottom. So there will be 8V from the circle to the bottom of the battery. \$\endgroup\$ Mar 29, 2021 at 16:59
  • \$\begingroup\$ So, there will be 8V drop across the current source? \$\endgroup\$
    – Newbie
    Mar 29, 2021 at 17:02
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    \$\begingroup\$ Yes, there will be 8V across the current source as well. \$\endgroup\$ Mar 29, 2021 at 17:06
  • \$\begingroup\$ Thank you for the answer \$\endgroup\$
    – Newbie
    Mar 29, 2021 at 17:08

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