I'm currently working on a project that involves a circuit with a single voltage source and multiple resistors, and I've hit a snag with my calculations. I could use some guidance from anyone who's adept at solving circuit problems. Here's what I need to figure out:

Calculating Currents (I1, I3, I4, and IA): I'm trying to determine the currents through various parts of the circuit. Specifically, I'm interested in calculating the currents I1, I3, I4, and IA. I understand the basic principles of Ohm's Law and the methods for analyzing simple circuits, but I'm unsure how to apply these to more complex arrangements with multiple resistors in both series and parallel configurations.

Calculating Power in the Voltage Source (VA): Additionally, I need to calculate the power in the voltage source VA, given that VA = 5V. The resistors in the circuit have the following values: R1 = 1 kΩ, R2 = 4 kΩ, R3 = 10 kΩ, and R4 = 5 kΩ. I know that the power can be calculated using P = VI, but I'm not sure how to find the total current supplied by the voltage source or if there's a more direct way to calculate the power with the given values.

The circuit is assumed to be ideal, with no power loss in the wires or components other than the resistors. There's no need to consider any complex impedance; the circuit operates with DC. I'm looking for a methodical approach to solve this, possibly with steps that I can apply to similar problems in the future.

enter image description here

  • \$\begingroup\$ I just looked at Simon's response, looked at your "I'm looking for a methodical approach to solve this, possibly with steps that I can apply to similar problems in the future.", and realized that Simon's answer isn't entirely adaptable as a general approach. It's more Rube-Goldberg to me and buries the lead, so to speak. Which brings me to the question I have. Are you familiar with, or want to become familiar with, the use of KCL for analysis? It's very general and can be applied to a wide variety of circuits. But you may also then need Cramer's Rule and/or simultaneous eq solution methods. \$\endgroup\$ Mar 18 at 21:40
  • \$\begingroup\$ I don't think you get to where you say you want to be without taking on KCL. (A tiny exposure to directed graph and linear algebra ideas -- not to go crazy, but only enough to convince you about how and why KCL and/or KVL works -- might be a nice-to-have at some point, too. But one thing at a time.) \$\endgroup\$ Mar 18 at 21:44
  • \$\begingroup\$ @periblepsis I am familiar with Kirchhoff's laws from High School. How do we apply it here? \$\endgroup\$ Mar 19 at 1:23
  • \$\begingroup\$ I added my thoughts below. \$\endgroup\$ Mar 19 at 2:54

4 Answers 4


I started out writing a brief summary of the steps, but I got caught up in the details, of which there are many. What follows is very comprehensive, and consequently long-winded, but it does contain absolutely every one of my thoughts as I went. My apologies for the length.

It might help to redraw the circuit as follows:


simulate this circuit – Schematic created using CircuitLab

Your own circuit and this one are identical in function and connectivity. The purpose here is to better visualise the current paths, so behaviour is more intuitive. The way I've drawn it makes it very clear that current into and out of node A is split four ways. This isn't so obvious when drawn they way it was originally. My rearrangement will make it trivial to apply Kirchhoff's Current Law (KCL) in a moment.

Next up, current \$I_A\$ seems to be going the wrong way. Remember this: by convention, positive current enters resistors at the higher potential end (the end with the more positive potential), and emerges from the end with lower potential. Intuitively, then, you'd expect all the arrows to be pointing downwards on the right-hand side, since node A is obviously higher in potential (5V higher, due to source \$V_A\$) than node B.

Using a water analogy, if water were flowing in these four pipes, all joined at A, it would be physically impossible for this situation to arise. You can't have water flowing out of all of them, there must be water flowing into that junction from somewhere. This is not a problem, we'll use this as a demonstration of how the maths will still work out fine. Just keep in mind that in electronics we will treat negative currents flowing one way as no different from the same positive amount flowing in the opposite direction. Clearly here we should expect at least one of the values \$I_A\$, \$I_1\$, \$I_3\$ and \$I_4\$ to be negative. You can probably already guess that the negative one is \$I_A\$, but we'll get to that later.

Shown several times in my rendition is the most basic of applications of KCL. In any two terminal device, such as the voltage source, and every resistor there, the current entering one end must equal the current leaving the other end. You can see this is the case for voltage source \$V_A\$, where we see current \$I_A\$ entering the top, and the same current \$I_A\$ exiting at the bottom. That same current would also be seen at all points in the wires between voltage source \$V_A\$ and nodes A and B, since there's nowhere for current to leak in or out of those paths. Thinking of water again, if those were water pipes, there would be no way for water to leak in or out anywhere along that path.

Similarly, resistors R1 and R2 (and their wires) must both carry the same current, \$I_1\$, for the same reason.

KCL applied to current junctions, such as A and B, says that the total amount of current entering the junction must equal the total amount leaving. We don't actually need to know the directions of current at this stage, but we do need to keep true to our annotations. Intentionally or otherwise, the author has shown all currents leaving A, and zero current entering. Algebraically, the equality between total current leaving A, and total entering A, in this case is written:

$$ I_A + I_1 + I_3 + I_4 = 0 $$

Incidentally, if you applied KCL to node B instead, you'd get:

$$ 0 = I_1 + I_3 + I_4 + I_A $$

which is saying exactly the same thing. That makes sense if it's really true that the currents split and merge just as water would in networks of pipes.

Just one small manipulation, to move \$I_A\$ to its own side, to demonstrate where that negative current I mentioned earlier appears:

$$ -I_A = I_1 + I_3 + I_4 $$


$$ I_A = -I_1 -I_3 -I_4 $$

If \$I_1\$, \$I_3\$ and \$I_4\$ are all positive (which they are, we'll see why in a moment), then \$I_A\$ must be negative. A negative \$I_A\$ flowing anticlockwise (as annotated) is the same as a positive \$I_A\$ flowing clockwise.

How do I know that \$I_1\$, \$I_3\$ and \$I_4\$ are all positive? I said earlier that current in a resistor always enters the resistor at the terminal with the higher potential, and I can see that node A, at the top, has the higher potential. 5V higher than B, to be precise. The annotated directions of \$I_1\$, \$I_3\$ and \$I_4\$ are all downwards, in agreement with this principle, and must all therefore be positive.

That's KCL out of the way, and normally we would then move on to applying Kirchhoff's Voltage Law (KVL). However, this problem is so trivial that I'll skip a formal application, and just use a trick (that you asked for). When any independent element or group of elements, a "block" with only two terminals, has both terminals connected directly across a voltage source, the potential difference across that block is fixed, and everything else in the entire circuit can be disregarded in the calculation of that block's current.

We have three blocks here. The first consists of R1 and R2. The junction between them isn't connected to anything outside the block, so there's only one way in for current, and one way out. You could draw a box around them, and see this easily. The same goes for R3, and the third block, R4. Each of these blocks is connected across the same voltage source, \$V_A\$, each has a fixed potential difference across them, and therefore to calculate current through the blocks, they can be treated individually, as follows:


simulate this circuit

The first thing I've done with R1 and R2 is to replace them with a single, equivalent resistance,R5. I have no doubt that you've been taught already that resistors in series are equivalent to a single resistance which is the sum of all individual resistors. In this case:

$$ R_5 = R_2 + R_1 = 4k\Omega + 1k\Omega = 5k\Omega $$

Now you can proceed to calculate currents \$I_1\$, \$I_3\$ and \$I_4\$ in the manner I'm also sure you have been taught, using Ohm's law. The voltage across each resistor is the same in each case, the only thing that changes is resistance:

$$ \begin{aligned} I_1 &= \frac{V}{R_5} \\ \\ &= \frac{5V}{5k\Omega} \\ \\ &= +1mA \end{aligned} $$

$$ \begin{aligned} I_3 &= \frac{V}{R_3} \\ \\ &= \frac{5V}{10k\Omega} \\ \\ &= +0.5mA \end{aligned} $$

$$ \begin{aligned} I_4 &= \frac{V}{R_4} \\ \\ &= \frac{5V}{5k\Omega} \\ \\ &= +1mA \end{aligned} $$

Now you can return to the KCL equation from earlier, and plug in those values:

$$ \begin{aligned} I_A &= -I_1 -I_3 -I_4 \\ \\ &= -1mA -0.5mA - 1mA \\ \\ &= -2.5mA \end{aligned} $$

You now have the currents everywhere in the circuit, and we can move on to calculating power. This is where another gotcha crops up, and it's to do with power in the voltage source \$V_A\$.

To avoid falling into the trap I'm about to explain, first you must understand Passive Sign Convention, which I already alluded to above. When I said that current in a resistor goes from higher potential to lower, that's always true for a resistor, but not all elements obey this rule. For some elements, chiefly voltage sources, current sources, capacitors and inductors, current can flow in either direction, regardless of the polarity of voltage across it.

This can lead to a lot of confusion, especially for beginners, but even seasoned pros can get it wrong. If we take care to label the current through and voltage across any element in accordance with passive sign convention, that is, current direction is annotated to enter the device at the higher potential end, and we take great care with the signs of those values, then our algebra will always be good. So, focussing on the voltage source \$V_A\$, the snippet below left accurately represents the state of affairs as we have calculated them. Middle and right, though, there are problems:


simulate this circuit

On the left, my annotated current direction, with respect to the annotated polarity of the voltage source, is correct, and in accordance with passive sign convention. The way the author drew that arrow in the original schematic is actually quite helpful here, because it means we needn't change anything. Even though our calculated negative value of \$I_A\$ is, in physical reality, +2.5mA of conventional current upwards, this representation obeys passive sign convention, and properly and unambiguously describes the state of affairs.

Above middle and right I've made deliberate errors, which you should identify.

To calculate power \$P\$ in these sources, you must find the product of voltage \$V_A\$ across it, and current \$I_A\$ through it. On the left:

$$ \begin{aligned} P &= V_A \times I_A \\ \\ &= +5V \times -2.5mA \\ \\ &= -12.5mW \end{aligned} $$

This is the correct answer. Voltage source \$V_A\$ is receiving −12.5mW of power from the rest of the circuit. In other words, it is delivering +12.5mW of power to all the resistors. That negative sign is again crucial. If you do the same procedure for the sources middle and right, the sign of the result will be positive, indicating that the sources are receiving energy, which is not the case in this circuit.

By getting the polarity of the voltage source wrong, or the direction and sign of current wrong, you risk getting the wrong sign for power, and you'll think your your battery or capacitor (or whatever) is charging when it is actually discharging, or vice versa. You must always consider passive sign convention. If necessary, to avoid such errors, redraw the relevant section of the circuit again, and re-label everything to be in accordance with passive sign convention, prior to making any calculations.

Just to finish off, I've repeatedly said that resistors will always have positive current flowing into the terminal with the more positive potential. Or, if you want to confuse yourself, you could say that a negative current enters at the terminal with the more negative potential. Both those scenarios are equivalent, and make sense. The product of voltage across and current through a resistor can only ever be positive, because a resistor can only ever receive energy, and heat up in the process.


In comments you wrote:

"I am familiar with Kirchhoff's laws from High School. How do we apply it here?"

First, you identify the nodes. I've taken a green sprayer and circled each of the three nodes in your diagram:

enter image description here

Assigning \$v_0=0\:\text{V}\$ and therefore that \$v_1=V_{_\text{A}}\$, the quick answer from KCL (sum of currents leaving a node must equal sum of current entering a node) is then:

$$\begin{align*} \overbrace{\quad\quad\quad\quad\quad\text{leaving node}\quad\quad\quad\quad\quad}&=\overbrace{\quad\quad\quad\text{entering node}\quad\quad\quad} \\\\ \frac{v_2}{R_1}+\frac{v_2}{R_2}&=\frac{v_0=0\:\text{V}}{R_1}+\frac{v_1=V_{_\text{A}}}{R_2} \\\\ \frac{v_1=V_{_\text{A}}}{R_2}+\frac{v_1=V_{_\text{A}}}{R_3}+\frac{v_1=V_{_\text{A}}}{R_4}+I_{_\text{A}}&=\frac{v_2}{R_2}+\frac{v_0=0\:\text{V}}{R_3}+\frac{v_0=0\:\text{V}}{R_4} \end{align*}$$

The first equation above can be solved for \$v_2\$: \$\quad v_2=V_{_\text{A}}\cdot\frac{R_1}{R_1+R_2}\$.

Using \$v_2\$, now the second equation can be solved for \$I_{_\text{A}}\$: \$\quad I_{_\text{A}}=-V_{_\text{A}}\cdot\left(\frac1{R_1+R_2}+\frac{1}{R_3}+\frac{1}{R_4}\right)\$

The negative sign just means that your arrow points oppositely to the actual direction of the current. Or, put another way, the current is leaving outward from the positive terminal of the voltage supply. Put still another way, the voltage supply is supplying current and therefore is providing power to the circuit rather than accepting power from the circuit. (Which is what one would expect.)

Then simply find that \$i_3=\frac{v_1=V_{_\text{A}}}{R_3}\$, \$i_4=\frac{v_1=V_{_\text{A}}}{R_4}\$, and \$i_1=\frac{v_2=V_{_\text{A}}\cdot\frac{R_1}{R_1+R_2}}{R_1}=\frac{V_{_\text{A}}}{R_1+R_2}\$.

KCL applies widely, too. Not just to this circuit.


R3 is in parallel with Va so Vr3=Va ...(1)

R4 is in parallel with Va so Vr4=Va ...(2)

R1 is in series with R2 and are in parallel with Va so Vr1 + Vr2 = Va ..(3)

From (1) -> Vr3 = Va = R3 * I3 -> I3 = Va/R3 = 5/10k = 0.5mA

From (2) -> Vr4 = Va = R4 * I4 -> I4 = Va/R4 = 5/5k = 1mA

From (3) -> Vr1 + Vr2 = Va = R1 * I1 + R2 * I1 -> I1 = Va/(R1+R2) = 5/5k = 1mA

Ia = I1+I2+I3 = 2.5mA

Power through the voltage source Pa = Va * Ia = 5*2.5 = 12.5mW


The other answers seem to me to be much too compilcated for your simple circuit.


simulate this circuit – Schematic created using CircuitLab

First, R1 and R2 are in series, so could be replaced with a single 5K resistor.

I see that R1/2, VA, R3 and R4 are all in parallel, so will all have the same voltage across them, which is the 5 V from VA. You can then calculate the current in each resistor using Ohm's Law.

For R3, I = E/R = 5/10,000 = 0.5 mA.

Do similar calaculations for the R1/R2 combination and for R4.

You now have I1, I3 and I4. IA will be the sum of those currents, but the current direction arrow on IA is pointing into the positive terminal of VA, so you should report the IA current as negative.


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