Assumed wrong direction for current through voltage source on branch?

Question

I am trying to solve the following simple circuit to obtain the current flowing through the 5V source. The answer I obtain does not agree with a simulator that I am using, and I think it's because I assume the wrong direction for current through the 19k resistor... Using conventional current flow, here is my approach to the problem:

Calculate the current through the 19k resistor as just 5V over 19k and I get 263uA. I then calculate the current through the 300k resistor as just 20V over 300k. I assume that the current is flowing in from the 5V source and also in from the 19k resistor and outward through the 300k resistor. Therefore using KCL I get:

(current through 19k) + (current through 5V) = current through 300k

263uA + I = 66.67uA

I = -196uA

However when I use the simulator at falstad.com/circuit/ I get a different result. I get like 329uA. Does it actually matter which direction I assume the current is flowing in? Or, is it that maybe the drop across the resistor is actually -5 because of the polarity on the voltage source?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Personally, I find it "singing to my mind" better when I take a moment to redraw a schematic.

^{simulate this circuit – Schematic created using CircuitLab}

Do you see what I've done? It's the same circuit on the right. But I've removed some of the confusing wiring. Also, it's now perfectly clear that one of these nodes (wires) is at \$20\:\textrm{V}\$, too.

Now it is very easy to see what is happening. There is \$20\:\textrm{V}\$ across \$R_2\$. So you know the current in \$R_2\$. Also, there is \$5\:\textrm{V}\$ across \$R_1\$. So you know the current in \$R_1\$. Whatever is left over must be the current in \$V_2\$.

So,

$$\begin{align*} I_{V_2} &= I_{R_2} - I_{R_1}\\\\ &= \frac{20\:\textrm{V}}{R_2} - \frac{15\:\textrm{V}-20\:\textrm{V}}{R_2}\\\\ &= \frac{20\:\textrm{V}}{300\:\textrm{k}\Omega} - \frac{15\:\textrm{V}-20\:\textrm{V}}{19\:\textrm{k}\Omega}\\\\ &= 329.824561\:\mu\textrm{A} \end{align*}$$

I prefer simplifying my understanding of a problem before choosing a method for solving it by hand.

Of course, there are rigorous methods you must also learn to apply "by rote." Computer software in electronic simulators do this all day long. But when you are first learning these things, I think it helps a lot to try and redraw things to help "see" them better.

Answer 2

For conventional current we have this situation :

As you can see 5V source current is equal to blue current plus the red current.

Answer 3

Therefore using KCL I get

This circuit isn't a good candidate for KCL. Using KCL, when there's a voltage source connected to a node you have to form a supernode with the other side of the voltage source.

It would be easier to solve this circuit using KVL, or simply by inspection.

By inspection:

From your ground node, the node on the other side of the 15 V supply is at 15 V. I'll call that node "A".

From node A, the node at the other end of the 5 V supply is at 20 V. I'll call that node "B".

Now you can get the current through the 300 kohm resistor (from right to left) as simply 20 V / 300 kohm.

You can get the current through the 19 kohm resistor (from top to bottom) as simply -5 V / 19 kohms.

Answer 4

You got the first two steps right, but then didn't add the currents properly.

Yes, the current thru the 19 kΩ resistor is (5 V)/(19 kΩ) = 263 µA, and the current thru the 300 kΩ resistor is (20 V)/(300 kΩ) = 67 µA.

With the 300 kΩ resistor open, the current thru the 5 V source is only due to the 19 kΩ resistor. With the 300 kΩ resistor connected, its current adds to what is flowing thru the 5 V source. The current thru the 19 kΩ resistor can't change since the voltage across it stays the same, by definition of what the 5 V source does. All the additional current from the external circuit can only flow thru the 5 V source.

The total current that the 5 V source is sourcing is therefore 330 µA.

Answer 5

^{simulate this circuit – Schematic created using CircuitLab}

$$V_{BC} = V_1 + V_2 = 15V + 5V = 20V$$ $$I_{BC} = \frac{20}{300k}A \approx 66.7uA$$ $$I_{BA} = \frac{5}{19k}A \approx 263uA$$ $$I_{2} = I_{BC} + I_{BA} \approx 330uA$$

You assumed wrong the verse of the currents. Check in the figure above the positive direction of the currents involved.