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Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141
When a wheatstone bridge is balanced there is no current flowing in the ammeter.
The 'beauty' of a wheatstone bridge is that it compares the ratio of the resistors on the left with the ratio of the resistors on the right and the actual supply voltage is unimportant, it does not need to be stable or accurate. And the meter only has to be sensitive, not accurate.
Is this a coursework question or are you going to make one, in which case there are many more points to consider. e.g. Meter protection. 'Coarse' and 'Fine' ballance. The choice of potentiometer. Switched reference resistors for a wide range of measurement. Use an ac signal for measuring capacitance and inductance etc.
Due to electronic meters old wheatstone bridge instruments are cheap, I sold a beauty for 99p on eBay :( So if you're going to build one, buy one off ebay for the case, meter, components etc.
Registered Member #2008
Joined: Tue Mar 03 2009, 05:11AM
Location: USA, Frederick, MD
Posts: 118
It was a question that I couldn't answer when I tried to understand someone else's circuit. Now I learned a litle more about circuits like that. The mesh currents. I found examples on how problems like this can be solved.
I didn't quite get the Norton and Thevenin theorems. Thank you for help.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Mesh currents will get you there, but it's a little heavy duty, it's a bit of a sledgehammer to crack this fairly simple nut. Doing the Thevnin thang on the potential dividers can be done real fast on the back of an envelope
As a hint, the left-hand arm looks like a voltage of V1 * 6000 / (6000+10) coming through an impedance of (6000*10) / (6000+10), that is the voltage of the potential divider coming from an impedance of the potential divider arms in parallel.
Now you have turned the left hand arm into a stiff voltage, with a resistor in series. Do the same with the right arm. Now you have two stiff voltages connected by three series resistors. Easy peasy.
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How do I find current going through two halves of the Wheatstone bridge?
