# Interactions between Ideal Voltage and Current Sources

Last week I asked a question involving a linear network involving ideal current and voltage sources. You guys gave me some great advice as to how to approach those types of problems. However, this example is really stumping me right now as the advice I was given on that post was to look at the circuit from one perspective at a time, first considering one type of source and then the other. However, I am having trouble applying this idea to this circuit diagram.

I was under the impression that if I consider the voltage source first, the current source should be interpreted as an open circuit. This, of course, breaks the circuit and no current can flow. I feel like this problem is more simple than I am making it out to be, but it's hurting my brain.

All that I've done so far is calculated that a circuit involving just the 10 V source and the 1k resistor would draw 10 mA. I just don't understand how adding a 2 mA current source changes this.

Any help would be greatly appreciated. Thanks!!!

Matt

• You started off fine: voltage source acting alone yields no current. Now do the current source acting alone....then apply the superposition rule to find total current. – glen_geek Feb 3 '18 at 22:10
• The problem is really simple and I see no point in considering each P.S. separated. You know the current on the resistor, it is 2mA. From that knowledge, you can calculate very simply the voltage over the resistor and over the current source. – Claudio Avi Chami Feb 4 '18 at 4:50
• glen So that would mean that no current flows when considering the voltage source and 2 mA when considering the current source. So that means 2V drop across the 1k resistor and 8V drop across the current source. So the voltage source provides no power? @Claudio So does that mean that there is two volts across the resistor and only the current source supplies power? Then that power is equal to 0.004 W, it would seem... I am still unclear on where the power comes from, whether the current source has resistance to create a voltage drop across it. There must be a drop of 10V across the circuit – MooshooMatt Feb 5 '18 at 6:01

There's only one path to consider, so don't worry about neglecting the sources in turn. Instead, just focus first on the current source - since there's only one path it will do its best to dictate the current in the path.

That means there is 2mA flowing into the -ve terminal of the 10V, through the 1kR and back to the current source. Does that mean the voltage source is supplying or absorbing power? Just think of a battery - when it supplies power, current leaves the positive terminal and returns to the negative terminal. That's the same arrangement as we have here, therefore the voltage source is supplying power.

That leaves the current source, for which we need the voltage across it. Going around the circuit counterclock-wise we have +10V due to the voltage source and -2V due to the current flowing through the resistor. That leaves 8V to be dropped across the current source, with the most positive terminal being the top.

Like a battery, a current source will normally deliver current out of its positive terminal and return via its negative terminal. In this case the opposite is true, therefore it is absorbing power.

• I understand it a lot more now. Thinking about KVL around the closed loop makes it much more simple. Thanks! – MooshooMatt Feb 26 '18 at 0:32

When a device consumes power, then it will conduct current from a positive voltage to a more negative voltage. The power is then positive.

When a device supplies power, then it will force current to flow from a negative voltage to a more positive voltage. The power is then negative.

If you apply this to your circuit, then you will get $10V$ right of the resistor and $10V - 2mA\cdot 1k\Omega=8V$ on the left. The only device with a negative power consumption is the voltage source, as it will conduct current of $2mA$ from $0V \rightarrow 10V$. The current source conducts $2mA$ from $8V \rightarrow 0V$ and so it will dissipate rather than supply power.