Circuit : Current Flow And Voltage Across Two Point

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Homework Help Overview

The discussion revolves around analyzing a circuit to find the voltage across points A and B, as well as determining the missing current (i). The problem involves applying Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) in the context of circuit analysis.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of KCL and KVL, with the original poster attempting to calculate currents and voltages based on assumed directions. Some participants question the correctness of the assumed current directions and the loop chosen for voltage calculations.

Discussion Status

There are various interpretations of the circuit and the calculations involved. Some participants express confidence in the current calculations, while others seek clarification on the assumptions made regarding the circuit diagram and the voltage drops across components. Guidance has been offered regarding the calculations of voltages at points A and B, but no consensus has been reached on the correctness of the original poster's assumptions.

Contextual Notes

Participants note the importance of the circuit diagram and the clarity of the problem statement. There is an ongoing discussion about the implications of the 24V across a resistor and its role in the calculations, as well as the assumption of voltage levels at specific points in the circuit.

gatsbycollege
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Homework Statement


Find Voltage across a and b and the missing current (i)
so this is the circuit
http://img3.imageshack.us/img3/184/circuitsf.jpg

Homework Equations


KCL and KVL
current towards the node is + away is -
Et = E1 + E2 + E3 in series
Et = E1 = E2 = E3 in parallel
Ohms law
E= IR

The Attempt at a Solution


KCL @ node c
3A-i1-i2= 0
i2=6V/3Ohms
i2=2A
3A-2A=i1
i1=1A
KCL @ node d
i2+1A-1A-i3=0
2A+1A-1A=i3
i3=2A
KCL @ node e
i3+i4-i
i4=24V/4Ohms
i4=6A
2A+6A=i
i=8A

KVL @ loop acdeba
E(ab)+i1(8)-6V-i3(12)+12V=0
E(ab)=-8V+6V+24V-12V
E(ab)=10V

so that's what i try..
my question is.. is those current direction is right?.. XD still don't know if i did it right..
so anyone who can help?
and the Voltage across AB.. am i right in having the loop acdeba.. neglecting the resitor which have 24V?
 
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can you please give the proper problem statement and circuit diagram
i can't understand the few basics here such as
'from where the current from 3A current source completing the loop'
'you said you have to find voltages at A 'AND' B and while solving you maid the loop through acdeba which suggest that point a and b are shorted resulting equal voltages at a and b
 
nieur said:
can you please give the proper problem statement and circuit diagram
i can't understand the few basics here such as
'from where the current from 3A current source completing the loop'
'you said you have to find voltages at A 'AND' B and while solving you maid the loop through acdeba which suggest that point a and b are shorted resulting equal voltages at a and b

the diagram is right.
the problem statement.. "Find the voltage across a and b" and find the current "i"

the 3A current source is placed like that.. there's nothing wrong with that

the blue line is the original diagram
and the current i1 , i2 , i3 and i4 is just my assumed direction..i don't know if its correct.
 
i think all current calculations you have done are right
i calculated the voltages at a and b separately 'b' is directly as 12 V and voltage at 'a' as 22
so the difference is 10 and your answer is correct
 
nieur said:
i think all current calculations you have done are right
i calculated the voltages at a and b separately 'b' is directly as 12 V and voltage at 'a' as 22
so the difference is 10 and your answer is correct

ahm can you please show ur computation in doing it separately.. i might have another idea in solving future circuit problems... thx
 
considering voltage at point e is 0V i will call voltage at e as Ve
Vd-Ve=i3*12
so Vd=24
Vc=Vd+6
so Vc=24+6=30
Vc-Va=i1*8
Va=30-8
Va=22
so we have got the voltage at pont a as 22
voltage at b is 12 volt directly

i think this helps
 
nieur said:
considering voltage at point e is 0V i will call voltage at e as Ve
Vd-Ve=i3*12
so Vd=24
Vc=Vd+6
so Vc=24+6=30
Vc-Va=i1*8
Va=30-8
Va=22
so we have got the voltage at pont a as 22
voltage at b is 12 volt directly

i think this helps

am i right that in ur analysis u also neglect the presence of the 24V at current i4
 
first of all 24V is not the source it is just the voltage drop across the resistor and i have used it to calculate the current i4 itself
 

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