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Complex Circuit - Wheatstone Bridge

physics.fisikastudycenter.com,- Discussing how to find the currents in a complex circuit of three loops, and determine the equivalent resistance of a complex circuit

Question
Given a complex circuit contents 5 resistances (a bridge circuit)



Where
R1 = 2 Ω
R2 = 2 Ω
R3 = 2 Ω
R4 = 2 Ω
R5 = 3 Ω
Find:
a) the current in each resistance and the total current through the network
b) the resistance equivalent of the circuit


Discussion
Redraw our circuit above that's look like Jembatan wheatstone (wheatstone bridge circuit), include the currents on resistances and loops directions. Consider that we have six different current in here, but to simplify calculation, we could have three currents that are I1, I2,and I3. The others, we will express in the sum or defference of the three currents, I1, I2,and I3. See our drawing below


Remember the kirchoff law for current and kirchoff law for voltage:
Σ ε + Σ IR = 0
Σ Iin = Σ Iout

Apply Σ ε + Σ IR = 0 to each loop one by one, and we'll have three equations as below:

Loop 1
Σ ε + Σ IR = 0
−12 + I1 R1 + (I1− I3)R4 = 0
Continue with simplify this calculation and finally we have our first equation
2 I1 − I3 = 6 .................(Eq. 1)

Loop 2
Σ ε + Σ IR = 0
−12 + I2 R2 + (I2 + I3)R5 = 0
Finish it, and then we have the second equation:
5 I2 + 3 I3 = 12 .................(Eq. 2)

Loop 3
Σ ε + Σ IR = 0
I1 R1 + I3 R3 − I2R2 = 0
Finish it, and now we have the third equation:
I1 + I3 = I2 ..................(Eq.3)

Combine equation 3 and equation 2, we have equation 4:
I1 + I3 = I2 ..................(Eq.3)
5 I2 + 3 I3 = 12 .................(Eq. 2)
5 (I1 + I3) + 3 I3 = 12 , proceed,..and we get
5 I1 + 8 I3 = 12 ....................(Eq. 4)

Finally by combining eq. 4 and eq. 1:
2 I1 − I3 = 6 | x 8|
5 I1 + 8 I3 = 12 | x 1|

16 I1 − 8 I3 = 48
5 I1 + 8 I3 = 12
------------------------- +
21 I1 = 60
I1 = 60 / 21 Ampere

It's easy enough to find I2 and I3 cause know we have I1. And the results:
I3 = −6 / 21 Ampere
I2 = 54/21 Ampere

The negative value of I3 indicates that its direction is opposite to our initial assumption.

The total current of our circuit is :
I total = I1 + I2
I total = 60 /21 + 54 /21 = 114 / 21 Ampere

b) The equivalent resistance of the circuit:

R equivalent = 12 volt / (114/21 Ampere)
= 2.21 Ω

Is the calculation above correct?
We could check by calculating the potential difference between point A and point B that we have to get 12 volt as the result.



VAB = I1 R1 + (I1 − I3)R4
VAB = (60/21)(2) + (60/21 − [− 18/17])(2)
VAB = 120/21 + 132/21 = 252/21 = 12 volt

Prepared by fisikastudycenter.com
Literatures
-Young and Freedman, University Physics with Modern Physics, Direct Current Circuits
-Halliday and Resnick, Fundamentals of Physics

 

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