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schematic

simulate this circuit – Schematic created using CircuitLab

If the current flows counterclockwise, the current source becomes invalid but if it flows clockwise the voltage source becomes invalid? So what would happen?

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  • \$\begingroup\$ Did you mean to point the current source the other way? \$\endgroup\$ – Ignacio Vazquez-Abrams Apr 4 '14 at 3:30
  • \$\begingroup\$ The magic perfect 1V source has 1V across it (regardless of current), but the magic perfect current source will produce whatever voltage is required to allow +1 Ampere to flow, even a negative voltage, because it's perfect (and magic). \$\endgroup\$ – Spehro Pefhany Apr 4 '14 at 3:31
  • \$\begingroup\$ @IgnacioVazquez-Abrams I don't think so. Does it make more sense if it points the other way? \$\endgroup\$ – dfg Apr 4 '14 at 3:34
  • \$\begingroup\$ @SpehroPefhany But which way would the current flow? \$\endgroup\$ – dfg Apr 4 '14 at 3:36
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    \$\begingroup\$ The current source will always determine the way the current flows (unless the circuit is broken, or series with another current source- those are "undefined" situations), and that's in the direction of the arrow if the current value is positive. \$\endgroup\$ – Spehro Pefhany Apr 4 '14 at 3:37
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You seem to think that you have created a problem, but not in this case. The 1A will flow through current source, there will be 1V across the voltage source. The current source will have the voltage over it that makes this happen, likewise the voltage source will have the current through it to make this happen (which must be 1A).

1A flows through the resistor too. Let's (arbitrarily) define its lefts side as 0V, then its right side must be at +100V, hence the voltage over the current source is 99V.

If you invert the current source the right side of the resistor is at -100V, hence the voltage over the current source is -101V.

Note that voltage sources and current sources are theoretical constructs, just like a line, circle and point. You can create 'impossible' diagrams with them, for instance a shorted voltage source or an open current source, or voltage sources for different voltages in parallel. It makes no sense to ask 'what would happen in such a case' because those idealized components do not exist. We can calculate with them within certain restrictions, outside those restrictions we can't calculate with them. That's all.

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  • \$\begingroup\$ Voltage drop on the resistor could be 100V. \$\endgroup\$ – Vovanium Apr 4 '14 at 15:37
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So what would happen?

This is straightforward - because of the current source, there is a clockwise current of \$1A\$.

By Ohm's law, the voltage across the resistor is

$$V_R = 1A \cdot 100 \Omega = 100V$$

with the rightmost terminal the more positive. By KVL, voltage across the current source is then

$$V_{I1} = 1V - 100V = -99V$$

Thus, the current source is supplying \$99V \cdot 1A = 99W\$ of power to the circuit.

The current through the voltage source is, by inspection, \$-1A\$ (the current enters the more negative terminal) thus, the voltage source supplies \$1V \cdot 1A = 1W \$ of power.

The power delivered to the resistor is

$$p_R = \frac{v^2_R}{R} = \frac{100^2}{100} = 100W$$

which must equal the sum of the power delivered by the sources which is

$$p_{V1} + p_{I1} = 1 + 99 = 100W $$

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Both voltage and current source contribute to the circuit. Current through resistor R1 is sum of current contributed by two sources (super position theorem) . But here Current source is in series with voltage source.

Assuming ideal condition, internal resistance of current source is infinity. Therefore no current from voltage source flows through it.

The only current through resistor R1 is current from current source ie 1A

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    \$\begingroup\$ +1 In general to use superposition, replace all unused voltage sources with a short, and replace all unused current sources with nothing. \$\endgroup\$ – Spehro Pefhany Apr 4 '14 at 4:29
  • \$\begingroup\$ I think you have overcomplicated this analysis. By definition, elements in series have the same current. The current through the ideal current source must be 1A. Therefore the current through the resistor is 1A. It doesn't matter at all how much resistance is in the loop, and there is no "contribution" from the voltage source to be superimposed. \$\endgroup\$ – Joe Hass Apr 11 '14 at 2:31
  • \$\begingroup\$ +1. Joe is right, is it a little deep for this circuit. But the details do become much more important, though, for extended and complicated networks. \$\endgroup\$ – Sean Boddy Apr 11 '14 at 3:33
  • \$\begingroup\$ "Therefore no current from voltage source flows through it. The only current through resistor R1 is current from current source ie 1A." - A very strange idea how this circuit works! It's a serious circuit, so the only one current (determined by the current source 1A) flows through all elements within the circuit, ie. even through the voltage source 1V. Either you like it or not, this source contributes to the total power of 100W dissipated in the resistor R1 = 100 \$\Omega\$ by 1W, while the current source supplies the rest 99W. \$\endgroup\$ – Eric Best Dec 28 '18 at 0:46
  • \$\begingroup\$ Even if the current flows counterclockwise, the current source DOES NOT become invalid! Why? The current in such a serious circuit is ALWAYS given by the current source. The situation will just change this way: the voltage source 1V will dissipate (not supply) the power of 1W due to the opposite current direction flowing through it, the voltage developed over the current source will change its sign and value and this source will supply 101W now - the 100W of which will be dissipated in R1. \$\endgroup\$ – Eric Best Dec 28 '18 at 1:08

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