Circuit : Current Flow And Voltage Across Two Point

  • #1

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 [Broken]

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 thats what i try..
my question is.. is those current direction is right?.. XD still dont 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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Answers and Replies

  • #2
4
0
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
 
  • #3
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 dont know if its correct.
 
  • #4
4
0
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
 
  • #5
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
 
  • #6
4
0
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
 
  • #7
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
 
  • #8
4
0
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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