Help understanding Kirchoff's Voltage Law/Current Law - Page 1
Forum section: electronics / beginners

author: Richardcavell
author_rank: Contributor
timestamp: April 30, 2024, 12:39:52 pm
content: Hi.I am a beginner attempting to understand Kirchoff's Voltage/Current Laws. I'll try to describe this circuit, and you tell me if I'm right:There is a cell on the left side producing 15 Volts. The conventional current flows clockwise no matter which of the two vertical resistors the current goes through. One path has the current flow through a 1k resistor then a 2k resistor (so 3k ohms total resistance). The other path has the current flow through a total of 9k resistance. The total voltage from V1 to the negative terminal via the 2k resistor is the same as the voltage as via the 3k and 5k resistors.The current splits at point V1. Some of it goes down the 2k resistor, and the rest down the 8k route.The total resistance of everything to the right of the 1k resistor is 1/R = 1/(1/8k + 1/2k). The total resistance is 1k + that amount.

author: wasedadoc
author_rank: Super Contributor
timestamp: April 30, 2024, 02:44:57 pm
content: Correct as far as you have gone.Don't fall into the trap of "One path has the current flow through a 1k resistor then a 2k resistor (so 3k ohms total resistance). The other path has the current flow through a total of 9k resistance." making you think you can calculate the current through each of those separately and then add them together to get the total current from the cell.

author: bostonman
author_rank: Super Contributor
timestamp: April 30, 2024, 02:57:07 pm
content: This is a simple series/parallel.Take the point farthest from the battery. The 3k and 5k are in series for a total resistance of 8k. Now redraw the circuit so an 8k is in parallel with the 2k. Take the parallel combination of 8k and 2k and that makes 1.6k.Now redraw the circuit again and you have 1k in series with 1.6k and those add to 2.6k.Your Rt is 2.6k and you can take 15v and divide it by 2.6k to get your total current. Now all you need to do is calculate the voltage drops and current 5.77mA.V1 is simply 15v - (5.77mA * 1k) = 9.23v.Since voltage stays the same in a parallel circuit and current divides, you know V1 is 9.23v, so 9.23 / 2k is the current through the 2k.Subtract that current from your total current, and you know the rest of the current going through the 3k and 5k, so you can calculate the voltage drops accordingly.

author: tggzzz
author_rank: Super Contributor
timestamp: April 30, 2024, 03:03:54 pm
content: Do you understand how to formulate and solve simultaneous equations?Assign names to all the components (e.g. R1, R2 etc), and denote the current through Rn as In. Similarly denote the voltage at each node.Then use KCL on each node, and KVL on each loop, e.g. I1=I2+I3, V3=V1+V5, and add Ohms' law e.g. I1=(V5-V3)/R1.Solve the resulting equations. Once you have done that a few times, you will develop an understanding for the patterns, and then be in a position to take "short cuts".I note the OP is good at starting threads, but poor at following up to indicate whether or not they have understood: 10 threads, 3 followups. Doesn't look good.

author: Terry Bites
author_rank: Super Contributor
timestamp: April 30, 2024, 03:06:59 pm
content: This page is very informative. An introduction to network analysis.www.allaboutcircuits.com/textbook/direct-current/chpt-10/what-is-network-analysis/

author: bostonman
author_rank: Super Contributor
timestamp: April 30, 2024, 04:24:22 pm
content: QuoteI was unaware the OP had ten threads going with only three replies. Usually I try closing the thread with the conclusion if one is needed.Thanks for letting us know.