# Mesh Analysis - Assignment Urgent

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#### GilbertSam

##### New Member
Hi All,

I have some querry here. Based on the attached, it shows that the bridge network is balanced circuit since the left and right mesh of bridge network resistor are the same. 6k Ohm and 4k Ohm. Can I say that the voltage across 3k Ohm is 0 volt or 0 amp?

Thanks a lot,

Best Regards,
Gilbert

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#### Electronworks

##### New Member
The voltage across the 3k will be 0V and the current flowing in it will be 0A.

The voltage at the junctions of the 2 potential dividers made up of the 4k on the bottom and the 6k on the top will be the same, hence, regardless of the voltage at these junctions, the voltage ACROSS the 3k will be 0.

#### GilbertSam

##### New Member
Hi ElectronWorks,

Thanks for reply. If I need to use the mesh analysis to find the current I0, can I removed the 3K ohm since there is no voltage / current on it by using only 2 mesh?
1 mesh for the bridge network ( from 6k to 6k to 4k to 4k ) and another is from voltage source to 2k to left 6k to left 4k .

Thanks a lot,

Best Regards,
Gilbert

#### Electronworks

##### New Member
Hi

You make a good observation. If there is no voltage across the 3k, it can effectively be removed from the circuit. In an ideal world, I would like to do mesh analysis with the 3k in situ, but this makes it too hard (!).

Therefore, ignoring the 3k, you have 2 strings of 10k (6k + 4k). These 2 are in parallel (so = 5k), so you have a load on the 2k resistor of 5k.

Your total load on the 30V source is 2k + 5k = 7k, so your current is 30/7k = 4.28mA.

Alternatively...

Current going into the 2k resistor is Io. Voltage across the bridge, lets call Vb.

Current into each leg of the bridge is Vb/10k

Using nodal analysis: (30-Vb)/2k = Vb/10k + Vb/10k = Vb/5k

Therefore, 30/2k = Vb(1/5k + 1/2k)

From this you can work out Vb (= 21.428V), then either divide (30 - 21.428V) by 2k to get Io, or divide 21.428V by 5k. Both ways gives you 4.28mA.

Thanks for the refresher - years since I did that!

#### Electronworks

##### New Member
.... alternatively, treat the (2x10k) strings as a resistor in a potential divider. Your voltage across the bridge is then:

5k/(5k+2k) x 30V = 21.428V.

You can work out the current from there..

Now I will let go of this problem...

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