Recent content by boo_lufc

I was thinking Rn is the last resistor so the last branch of any similar circuit would be Rn + Rn-1 in series Then this would be in parallel with Rn-2 + Rn-3 and so on i.e Reqv = 1/ {(1/Rn + Rn-1) + (1/(Rn-2 + Rn-3) + (Rn-4 + Rn-5)} Is this along the right lines at

Is it possible to derive a formula for the equivalent resistance of n such sections in cascade? Show your analysis. Series: Reqv = R1 + R2 +...Rn Parallel: Reqv = 1/{(1/R1) + (1/R2) + (1/R3) +...(1/Rn)}

Ok good, so then to work out the voltage across all resistors using ohms law: VR1 = I1*3 = 3.6V VR2 = (I1 - I2)*2 = -1.6V VR3 = I2*1 = 2V VR4 = (I3 - I2)*R4 = -0.4V VR5 = I3*4 = 6.4V With VR2 and VR4 being negative or positive depending what loop you use

I think i have spotted my mistake, you say you had 2mA for I2, did you also get I1 = 1.2mA and I3 = 1.6mA?

Ok so i then replace I2 with (5*I3 - 6) in the loop equations for loops 1 and 2: Loop1: 2 - 5I1 + 2[5I3 - 6] = 0 2 - 5I1 + 10I3 - 12 = 0 -10 -5I1 + 10I3 = 0 10I3 - 5I1 = 10 -----------------(1) Loop 2: 4 - 4[5I3 - 6] - 2I1 + I3 = 0...

Ok if I take it step by step then should be able to spot mistakes. First I just put the resistor values in and canceled the loop equations as shown: Loop 1: 12 - I1*R1 - (I1 - I2)*R2 - 10 = 0 2 - 3I1 - 2I1 + 2I2 = 0 2 - 5I1 + 2I2 = 0 Loop 2: 10 - (I2 -...

From the equation for loop 3, i rearanged to get I2 = 5*I3 - 6 then subbed this value into the loop equations for lopp 1 and 2 and solved simulatneous equations. Could you confirm if firstly this equation of I2 i have subbed in is correct and then if the method for finding the currents is...

I have got: I1 = 0.782mA I2 = 0.955mA I3 = 1.391mA Do these seem right? When I use these values to work out the current across R2 for example I get the same value but negative and positive depending on what loop eqn I use. This seems like it would be right but in terms of the...

so does this look right: Loop 1: 12 - I1R1 - (I1-I2)R2 - 10 = 0 Loop 3: 6 - (I3-I2)R4 - I3R5 = 0 with loop 2 given in your last post

Would i rewrite as : Loop 1: 12 - I1R1 + (I1 - I2)R2 - 10 = 0 Loop2: 10 - (I1 - I2)R2 - I2R3 + (I2 - I3)R4 - 6 = 0 Loop 3 : 6 - (I2- I3) + I3R5 = 0 ?

ok thanks but where have i gone wrong in the three eqns i posted?

just hoping this cct looks better

i think so but i am unsure about directions because i have not seen 3 voltage sources used before: Here's what i came up with if you could suggest if this is correct: KCL at node 1: -I1 -I2 + I3 = 0 KCL at node 2: -I3 -I4 + I5 = 0 KVL (left) : 12 - (I1*R1) + (I2*R2) -...

I have not seen a problem like this before but my basic knowledge of Kirchoff would lead me to take an equivalent R for the 1k and 4k resistor to make two loops. I would then have I1 coming from the 12V source, then at the node have I3 coming down the branch toward the 10V source and I2 carrying...

1) Using Kirchoff's Laws, find all the currents flowing in the cct. 2) Hence, determine the voltages across all resistors and check that each loop complies with KVL