Basic Kirchhoff exercise - stumped - Page 1
Forum section: electronics / beginners

author: MuseChaser
author_rank: Contributor
timestamp: March 06, 2022, 08:39:20 pm
content: My pea brain can't figure out how to approach this circuit..I've redrawn the circuit every which way, trying to simplify the multiple resistors into series or parallel, but can't get it. Reduced it to this, which I'm hoping is correct...But still can't figure out how to come up with the answer. The exercise is taken from the AllAboutCircuits worksheets on Kirchhoff's Laws.Thinking of the 4.7k resistor as being in parallel with the series 2.5k (the 1k plus the 1.5k in the original diagram) and 690, and solving for the resistance there, I calculated 4700(3190)/(4700+3190) = 1900.25 ohms total resistance for that part, then added the 250 series resistor left over for a total resistance in the circuit of 2150.25 ohms. Plugging that into I=V/R, 20/2150.25=9.3 mA current. I'm hoping I have that much correct, but my confidence is not real high even there.From here, I just have no idea how to calculate the voltage between A and B. I realize this is kid stuff, but am hoping some kind soul can clear this up for me. Thanks.

author: ledtester
author_rank: Super Contributor
timestamp: March 06, 2022, 08:59:04 pm
content: If you can find the voltage at the junction of the 4.7K, 690, 250 resistors (call this voltage C) then you just have a voltage divider:20V -- 2.5K -+- 690R -- Cwhere the + represents point B.It's not the conventional voltage divider where the lower voltage is zero, but the same idea works with the voltage difference.

author: Picuino
author_rank: Super Contributor
timestamp: March 06, 2022, 09:14:25 pm
content: There are 3 ways.1. Apply mesh method:https://en.wikipedia.org/wiki/Mesh_analysis2. Apply nodal method:https://en.wikipedia.org/wiki/Nodal_analysis3. Reduce the circuit with series and parallel to calculate the currents. Then calculate voltages across resistors.With method 3:a) Equivalent resistance in series of 1000 + 1500 + 690 = 3190 Ohmb) Equivalent resistance in parallel of 3190 // 4700 = 1900 Ohmc) Equivalent resistance in series of 1900 + 250 = 2150 Ohmd) Current of the voltage source: 20 / 2150 = 9.302326 mAe) Current of 3190 Ohm branch (current divider) = 9.302326 mA * 4700/(4700+3190) = 5.541309 mAf) Voltage across 250 ohm: V_250 = 9.302326 mA * 250 = 2.3255815 Vg) Voltage across 690 ohm: V_690 = 5.541309 mA * 690 = 3.8235032 Vh) Voltage between A and B: 2.3255815 V + 3.8235032 V = 6.149V

author: MuseChaser
author_rank: Contributor
timestamp: March 06, 2022, 09:59:39 pm
content: Thank you VERY much! A bit heartening to know that I had at least started out correctly. I need to brush up on current dividers... that's the part I didn't know how to deal with. MUCH appreciated.

author: tooki
author_rank: Super Contributor
timestamp: March 07, 2022, 02:00:04 pm
content: Quote from: MuseChaser on March 06, 2022, 08:39:20 pmFWIW, it’s often helpful to redraw so your voltage source is at the side of the circuit by itself. Here’s how I redrew this circuit, which I think makes it easier to see how the current branches — and that point “A” is actually just circuit ground (0V).

author: tooki
author_rank: Super Contributor
timestamp: March 07, 2022, 02:06:40 pm
content: Just noticed I forgot to draw point A, so I added it and also added the currents. Note that Itotis shown twice, to illustrate that it’s the same thing.

author: Picuino
author_rank: Super Contributor
timestamp: March 07, 2022, 04:32:09 pm
content: With method 2 (nodes):a) Add node Vc in the junction of 250, 4k7 and 640b) Since there is no ground point, we put it in Va, which is more natural: Va = 0c) Write equations of the currents(20 - Vc)/4700 + (Vb - Vc) / 690 = (Vc - Va)/250(20 - Vb) / (1000 + 1500) = (Vb - Vc)/690d) Rearrange equations:20/4700 = Vb(-1/690) + Vc(1/4700 + 1/690 + 1/250)20/2500 = Vb(1/2500 + 1/690) + Vc(-1/690)e) Solve equations (linear system) withCramer's rule:Va = 0V (definition of Va)Vb = 6.148VVc = 2.325V

author: MuseChaser
author_rank: Contributor
timestamp: March 07, 2022, 06:18:44 pm
content: Quote from: Picuino on March 06, 2022, 09:14:25 pmJust to cement this in my mind, I've been working through this example and other similar ones repeatedly trying to make it second nature, and it's starting to come together; I truly appreciate your help and the help of others in this thread (and a couple others).A question on step "e" above, re/ the "4700/(4700+3190)" portion. According to what I've read, that part should be total resistance of that parallel circuit being considered divided by the resistance of the branch, or in this case, 1900/3190, yielding .595611285, which then multiplies I(total), 9.302326mA, to yield 5.540570345mA on that branch of the circuit, and then the voltage over the 690ohm resistor would be .005540570345(690) or 3.822993537931V. 1900/3190 yields a number very close to your 4700/(4700+3190), of course, but I'm not familiar with where that formula came from. Is it accurate?Thank you.

author: Picuino
author_rank: Super Contributor
timestamp: March 07, 2022, 06:52:45 pm
content: The formula is totally accurate.It is quite useful for calculating currents through resistors that are in parallel (shunts). You can read more about it on Wikipedia:https://en.wikipedia.org/wiki/Current_divider

author: MuseChaser
author_rank: Contributor
timestamp: March 07, 2022, 07:46:15 pm
content: Quote from: Picuino on March 07, 2022, 06:52:45 pmI have read that link, and am also currently (no pun intended.....) going through a bunch of exercises and texts related to current divider circuits. I am not being argumentative, at least not intentionally.. I'm not smart or experienced enough to question anyone here. However, I see no mention of being able to take the resistance of one branch or parallel resistor and divide it by the SUM of both branches. All the formulae and text I see instruct one to take the total resistance and divide it by the branch or parallel resistor's .. resistance. What am I missing?I've tried a bunch of different values, and you are, of course, correct... I just can't figure out why. For instance, if a 2k and 12k resistor are parallelled, the total resistance is 2k*12K/(2k+12K)= 1.714K, the current through the 2k resistor would be (1.714k/2k)*I(total) or 85.7% of total current. Similarly, (1.714k/12k) yields 14.3% of total current.Your formula yields the same results, without having to first figure the resistance of the parallel pair....2k/(2k+12k)= 14.2857%12/(2k+12k) = 85.7143%MUCH easier, of course, but I don't understand why it works. Can the formula,[(R1*R2)/(R1+R2)]/R1be simplified toR1/(R1+R2) ?I used to be good at this stuff, a little over four decades ago, but it's been WAY too long and I can't get there algebraically somehow. Help?

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2022, 07:58:01 pm
content: Quote from: MuseChaser on March 07, 2022, 06:18:44 pmIt may help to think of parallel branches in terms of conductance (G) rather than resistance (G = 1/R). In parallel branches the conductances add up, and the total current divides in direct proportion to the conductances (so I1:I2 is as G1:G2).

author: Picuino
author_rank: Super Contributor
timestamp: March 07, 2022, 08:49:57 pm
content: If you have 2 resistors in parallel (R1 and R2) with a total voltage of Vt and total current of It, applying Ohm's law:I_R1 = Vt / R1andVt = It * (R1//R2) = It * R1*R2/(R1+R2)Combining both:I_R1 = [ It * R1*R2/(R1+R2) ] / R1 = It * R2/(R1+R2)

author: Picuino
author_rank: Super Contributor
timestamp: March 07, 2022, 08:52:57 pm
content: Quote from: MuseChaser on March 07, 2022, 07:46:15 pmNot exactly. Could be simplified to R2 / (R1 + R2)Code:[(R1*R2)/(R1+R2)]/R1 =(R1*R2) R1*R2 R2-------- / R1 = ----------- = ----------(R1+R2) (R1+R2)*R1 (R1+R2)

author: MuseChaser
author_rank: Contributor
timestamp: March 07, 2022, 09:00:01 pm
content: Quote from: Picuino on March 07, 2022, 08:49:57 pmGOT IT!!!! Thank you!!!!!

author: Picuino
author_rank: Super Contributor
timestamp: March 07, 2022, 09:01:16 pm
content: The problem with Wikipedia's explanation is that it calls total resistance what is the rest of the resistances (not all the resistances, but all the other resistances). It's a bit of a confusing way of expressing the rule.Maybe you prefer this other explanation:https://www.electronics-tutorials.ws/dccircuits/current-divider.html