## Complex Circuit - Wheatstone Bridge

- Category: Enrichments
- Written by fisikastudycenter

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

R_{1} = 2 Ω

R_{2} = 2 Ω

R_{3} = 2 Ω

R_{4} = 2 Ω

R_{5} = 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 I_{1}, I_{2},and I_{3}. The others, we will express in the sum or defference of the three currents, I_{1}, I_{2},and I_{3}. See our drawing below

Remember the kirchoff law for current and kirchoff law for voltage:

Σ ε + Σ IR = 0

Σ I_{in} = Σ I_{out}

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 + I_{1} R_{1} + (I_{1}− I_{3})R_{4} = 0

Continue with simplify this calculation and finally we have our first equation

2 I_{1} − I_{3} = 6 .................(Eq. 1)

**Loop 2**

Σ ε + Σ IR = 0

−12 + I_{2} R_{2} + (I_{2} + I_{3})R_{5} = 0

Finish it, and then we have the second equation:

5 I_{2} + 3 I_{3} = 12 .................(Eq. 2)

**Loop 3**

Σ ε + Σ IR = 0

I_{1} R_{1} + I_{3} R_{3 } − I_{2}R_{2} = 0

Finish it, and now we have the third equation:

I_{1} + I_{3} = I_{2} ..................(Eq.3)

Combine equation 3 and equation 2, we have equation 4:

I_{1} + I_{3} = I_{2} ..................(Eq.3)

5 I_{2} + 3 I_{3} = 12 .................(Eq. 2)

5 (I_{1} + I_{3}) + 3 I_{3} = 12 , proceed,..and we get

5 I_{1} + 8 I_{3} = 12 ....................(Eq. 4)

Finally by combining eq. 4 and eq. 1:

2 I_{1} − I_{3} = 6 | x 8|

5 I_{1} + 8 I_{3} = 12 | x 1|

16 I_{1} − 8 I_{3} = 48

5 I_{1} + 8 I_{3} = 12

------------------------- +

21 I_{1} = 60

I_{1} = ^{60} / _{21} Ampere

It's easy enough to find I_{2} and I_{3} cause know we have I_{1}. And the results:

I_{3} = ^{−6 }/ _{21} Ampere

I_{2} = ^{54}/_{21} Ampere

The negative value of I_{3} indicates that its direction is opposite to our initial assumption.

The total current of our circuit is :

I _{total} = I_{1} + I_{2}

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.

V_{AB} = I_{1} R_{1} + (I_{1 }− I_{3})R_{4}

V_{AB} = (^{60}/_{21})(2) + (^{60}/_{21} − [−^{ 18}/_{17}])(2)

V_{AB} = ^{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*