Simple circuit, KVL & KCL, different R values: KVL and Ohm's law - Page 1
Forum section: electronics / beginners

author: Sibb
author_rank: Newbie
timestamp: November 22, 2020, 11:42:24 am
content: Hello,notes and circuit as link and attached file:in one exercise (about KVL/KCL), the value of R must be found.After getting all other I values, KLV was used to get R.Equation (step 7. in picture) (150V source, I1 was calculated before, 100V given):-150V+I1*R+100V=025A*R =50 | :25R=2 OhmIf I try to get the value of R (we calculated I, we have voltage of source), I get a different result.R= U : IR= 150 / 25R= 6 OhmMaybe it is not working, because I have two loops and two other resistors in the circuit.Why Ohm's law not working, where did I made a mistake?

author: Paul Rose
author_rank: Regular Contributor
timestamp: November 22, 2020, 04:52:31 pm
content: I like to label my currents with an arrow so I don't forget to keep the signs right.Doesn't matter which way, as long as you don't change your mind halfway throughI labled currents as follows:i1 = current "left to right" through R1i2 = current "down" through R2i3 = current "down" through R3KCL Eq1: i1 = i2 + i3 (current entering = current leaving)KVL Eq2: 150v = i1*R1 - 100v = 0 (or i1*R1 = 50v)KVL Eq3: 100v - i3*5ohm = 0Solve i2 directly because voltage and resistance are given: 100 = i2 * 20 ohm, or i2 = 5ASolve KVL Eq3: i3 = 20ASolve KCL Eq1: i1 - 5A - 20A , which gives i1 = 25ASolve KVL Eq2: 25A*R1 = 50v , which gives R1 = 2ohmDouble checking with Ohm's law:i1*R1 = 50v or 25A*2ohm = 50vi2*R2 = 100v or 5A*20ohm = 100vi3*R3 = 100v or 20A*5ohm = 100vAll checks out.Final sanity check:Parallel combo of r2 and r3 : 5*20/(5+20) = 100/25 = 4 ohmSeries combo of 4ohm and R1 : 4+ 2 = 6ohmTotal current I = V/R : I = 150 / 6 = 25A , which checks out with i1Just a side note:This isn't a bad way to practice learning KVL and KCL, but practically speaking that is overkill for something like this.Mentally note that R2 & R3 are in parallel, so the voltage across both is equal to the given 100v across R2.Parallel combo of 20 and 5 = 4 is easy enough to do mentally also: (20*5)/(20+5) or 100/25 = 4ohm.Finally, dropping 100 volts across the combined 4 ohms is also easy enough to do mentally: 100volts/4ohms = 25amps.So knowing that full current is 25 amps into the parallel combo of R2 and R3, and that current had to come through R1, and that you had to drop 50 volts across R1 to get from supply to the (given) voltage of 100 at the top of R2, the final step is just R = V / I, or 50volts / 25amps = 2ohms.

author: Paul Rose
author_rank: Regular Contributor
timestamp: November 22, 2020, 05:47:52 pm
content: TL;DR. Feel free to skip my long-winded explanation aboveActually looking back at your work, the only problem was in your final check:Quote from: Sibb on November 22, 2020, 11:42:24 amOhm's law is alive an well. 6 ohms is indeed the total resistance seen by the voltage source.But that 6 ohms is the 2 ohms of R1, in series with equivalent parallel resistance of R2||R3 of 4 ohms.

author: pwlps
author_rank: Frequent Contributor
timestamp: November 23, 2020, 09:18:32 am
content: Quote from: Sibb on November 22, 2020, 11:42:24 amShould have been:R= U : IR= (150-100) / 25 -> your mistake is hereR= 2 Ohm

author: Sibb
author_rank: Newbie
timestamp: November 23, 2020, 04:31:15 pm
content: @Paul RoseThank very much for your replay. I went through your calculation and understood everything.I just learned about parallel and series resistors and your answer provided new learned technics. I get 4 ohm for R2,R3 (20*5 : 20+5), that proves R1 must be 2 ohm@pwlpsYes, Ohm's law provided whole resistance of the circuit, not of R. Thanks.