There is a Problem in our textbook which asks us to find, using Kirchoff's law, the voltage across component v_x and v_y.


I know that the voltage at node 1 should be 14 volts from [21 + (-7)]=14 (in terms of $$v=\frac{dw}{dq}$$ not in terms just of potential difference.

The voltage across V-x is 9 because the voltage across the two components in series right below it is 5.

Of course, v_y is 14. This said:

Is there going to be any voltage across the rightmost component?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ If you say the voltage at node 1 is 14V, that is most certainly in terms of a potential difference. The implication is that the bottom node in your diagram is ground, which is the assumed point of reference when we speak of a single node having a given voltage. \$\endgroup\$ – Joe Hass Sep 27 '13 at 17:20

I'm going to assume you meant "rightmost" since the leftmost component is a supply.

The voltage across the rightmost component must be the same as any components in parallel. Hence, it has 5V across it.

  • \$\begingroup\$ Ahhhhhh... that absolutely makes sense on the surface of it. So if Applying Kirchoff's voltage law, every node providing a in parallel counts as only one entry departure from the closed surface? \$\endgroup\$ – user1833028 Sep 27 '13 at 5:26

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