Crosswords0 min ago
Pit Your Wits Against This One.
11 Answers
Answers
1 to 11 of 11
// Incidentally how long did it take you to arrive at a solution?? // Without putting pen to paper, approx 30 to 40 seconds :0) Why do you ask? My method was to convert the 'delta' network (8Ω, 4Ω and 12Ω) into its equivalent 'star' transformati
16:58 Thu 09th Mar 2023
bobbinwales \Is putting all his homework on here?\
No! This is not student homework or assignment work!!
The way it works is a little like this:
Student attends lecture and is taught certain principles, laws, theorems etc The lecturer will then demonstrate a couple of problems related to the above. Students are then invited at end of the lesson to take problem sheet/s (approx 20 questions with answers provided) to do in their own time. The questions get progressively harder as you work through the sheet/s!
These are the type of questions that could appear on the end of term exam paper. The lecturer/s encourage all students to explore various resources to help provide a solution to the answers already given. Namely, further reading from student library, collaboration with other students, set tutorials, bobbinwales, the internet etc
Note that the answer carries little weighting eg 5%
\ignore the 12 ohms then use normal rule for parallel and in series\
This only applies under certain conditions!!
No! This is not student homework or assignment work!!
The way it works is a little like this:
Student attends lecture and is taught certain principles, laws, theorems etc The lecturer will then demonstrate a couple of problems related to the above. Students are then invited at end of the lesson to take problem sheet/s (approx 20 questions with answers provided) to do in their own time. The questions get progressively harder as you work through the sheet/s!
These are the type of questions that could appear on the end of term exam paper. The lecturer/s encourage all students to explore various resources to help provide a solution to the answers already given. Namely, further reading from student library, collaboration with other students, set tutorials, bobbinwales, the internet etc
Note that the answer carries little weighting eg 5%
\ignore the 12 ohms then use normal rule for parallel and in series\
This only applies under certain conditions!!
Here's my method/working:
Assume the voltage at the top of the circuit is V and at the bottom 0. Denote the resulting voltages at the node to the left and right of the 12 Ω resistor as V1 and V2 respectively. The current flowing through the various resistors will be (by Ohm's Law)
8 Ω (V - V1)/8 Amps
4 Ω (V - V2)/4 Amps
12 Ω (V2 - V1)/12 Amps
14 Ω V1/14 Amps
16 Ω V2/16 Amps
The current through the 4 Ω resistor splits to flow through the 12 and 16 Ω resistors so
(V - V2)/4 = (V2 - V1)/12 + V1/16
which, rearranged, gives
12*V = 19*V2 - 4*V1 ...........(1)
Similarly, the currents through the 8 and 12 Ω resistors combine to flow through the 14 Ω resistor so
(V - V1)/8 + (V2 - V1)/12 = V1/14
and this, rearranged, gives
21*V = 47*V1 - 14*V2 ..........(2)
Solving the simultaneous equations (1) and (2) for V1 and V2 yields
V1 = (21/31)*V and V2 = (24/31)*V
The total current through the circuit, call it I, is given by the sum of the currents through the 14 and 16 Ω resistors so
I = V1/14 + V2/16
= (21/31)*V/14 +(24/31)*V/16
= (3/31)*V Amps
By Ohm's Law the overall resistance is given by
V/I = V/((3/31)*V) = 31/3 = 10.3333 Ω
Assume the voltage at the top of the circuit is V and at the bottom 0. Denote the resulting voltages at the node to the left and right of the 12 Ω resistor as V1 and V2 respectively. The current flowing through the various resistors will be (by Ohm's Law)
8 Ω (V - V1)/8 Amps
4 Ω (V - V2)/4 Amps
12 Ω (V2 - V1)/12 Amps
14 Ω V1/14 Amps
16 Ω V2/16 Amps
The current through the 4 Ω resistor splits to flow through the 12 and 16 Ω resistors so
(V - V2)/4 = (V2 - V1)/12 + V1/16
which, rearranged, gives
12*V = 19*V2 - 4*V1 ...........(1)
Similarly, the currents through the 8 and 12 Ω resistors combine to flow through the 14 Ω resistor so
(V - V1)/8 + (V2 - V1)/12 = V1/14
and this, rearranged, gives
21*V = 47*V1 - 14*V2 ..........(2)
Solving the simultaneous equations (1) and (2) for V1 and V2 yields
V1 = (21/31)*V and V2 = (24/31)*V
The total current through the circuit, call it I, is given by the sum of the currents through the 14 and 16 Ω resistors so
I = V1/14 + V2/16
= (21/31)*V/14 +(24/31)*V/16
= (3/31)*V Amps
By Ohm's Law the overall resistance is given by
V/I = V/((3/31)*V) = 31/3 = 10.3333 Ω
// Incidentally how long did it take you to arrive at a solution?? //
Without putting pen to paper, approx 30 to 40 seconds :0) Why do you ask?
My method was to convert the 'delta' network (8Ω, 4Ω and 12Ω) into its equivalent 'star' transformation (1.3333Ω, 4Ω and 2Ω).
Thus 1.3333Ω is in series with a parallel network which is now made up of a branch containing 4Ω and 14Ω and the other branch consisting of a 2Ω and 16Ω.
The 2 branches, 18Ω in parallel with 18Ω equate to 9Ω which in turn is in series with the 1.3333Ω equal to 10.3333 ohms - Voilá
'Star' resistors can be calculated as follows;
8 x 4 / 24 = 1.3333Ω,
8 x 12 / 24 = 4Ω and 12 x 4 / 24 = 2Ω
See link below;
https:/
The same result can be obtained using the bottom delta network (14Ω, 12Ω and 16Ω).
Without putting pen to paper, approx 30 to 40 seconds :0) Why do you ask?
My method was to convert the 'delta' network (8Ω, 4Ω and 12Ω) into its equivalent 'star' transformation (1.3333Ω, 4Ω and 2Ω).
Thus 1.3333Ω is in series with a parallel network which is now made up of a branch containing 4Ω and 14Ω and the other branch consisting of a 2Ω and 16Ω.
The 2 branches, 18Ω in parallel with 18Ω equate to 9Ω which in turn is in series with the 1.3333Ω equal to 10.3333 ohms - Voilá
'Star' resistors can be calculated as follows;
8 x 4 / 24 = 1.3333Ω,
8 x 12 / 24 = 4Ω and 12 x 4 / 24 = 2Ω
See link below;
https:/
The same result can be obtained using the bottom delta network (14Ω, 12Ω and 16Ω).
ZebuS\Without putting pen to paper, approx 30 to 40 seconds :0) Why do you ask?\
I am amazed and I will tell you for why. I passed on the solutions provided by Etch and Zebu. Both are valid and correct and my student neighbour asked me to convey his many thanks!!
This particular question had been allotted 3 minutes. Student neighbour had said (this bit is funny lol) even if he was given the equations he would struggle to get an answer in that time!
Lo and behold, his Friday morning lecture covered Star to Delta and vice versa transformations as well as Mesh analysis. The students were not expected to have progressed to q15 on the practice sheets where this method is the most applicable in the time allowed.
In our opinion, for sheer brevity, we have awarded BA to Zebu. Thanks from me to Etch for your great contribution too!!
p.s Thanks for the sketch Zebu. It looks like it was drawn with your big toe LOL ;-)
I am amazed and I will tell you for why. I passed on the solutions provided by Etch and Zebu. Both are valid and correct and my student neighbour asked me to convey his many thanks!!
This particular question had been allotted 3 minutes. Student neighbour had said (this bit is funny lol) even if he was given the equations he would struggle to get an answer in that time!
Lo and behold, his Friday morning lecture covered Star to Delta and vice versa transformations as well as Mesh analysis. The students were not expected to have progressed to q15 on the practice sheets where this method is the most applicable in the time allowed.
In our opinion, for sheer brevity, we have awarded BA to Zebu. Thanks from me to Etch for your great contribution too!!
p.s Thanks for the sketch Zebu. It looks like it was drawn with your big toe LOL ;-)
1 to 11 of 11
