I Measuring Low Resistance: Why Meter Bridge is the Preferred Choice

AI Thread Summary
The meter bridge is preferred over the post office box for measuring low resistance due to its superior sensitivity and accuracy. The post office box, which operates as a Wheatstone bridge, is limited in its measurement range and primarily used for identifying short circuits rather than measuring resistance. Its design does not effectively utilize the meter's sensitivity because it is affected by the bridge components. Additionally, the meter bridge allows for a more precise measurement technique by enabling current to pass through the meter for full-scale deflection before measuring the voltage drop across an unknown low resistance. Overall, the meter bridge is more effective for low resistance measurements compared to the post office box.
phymath7
Messages
48
Reaction score
4
TL;DR Summary
Why post office box can't be used to measure high or low resistance?
Why meter bridge is chosen over post office box to measure low resistance?
 
Physics news on Phys.org
What is "post office box" (or "vox")?
 
Ah. According to the Wikipedia article the Post Office Box (origin UK) is in fact a Wheatstone bridge circuit.
 
phymath7 said:
TL;DR Summary: Why post office box can't be used to measure high or low resistance?

Why meter bridge is chosen over post office box to measure low resistance?
Who says it can't be used to measure high or low resistance ?
The post office box was two arms of a Wheatstone bridge.
It was used to find short circuits, not open breaks in the line.
 
phymath7 said:
TL;DR Summary: Why post office box can't be used to measure high or low resistance?

Why meter bridge is chosen over post office box to measure low resistance?
I think the measurement range will be limited by the available ratio arms and standard resistances. In addition, a bridge does not make best use of the meter sensitivity because it is shunted by the bridge components. There is a sensitive method for low resistances where we first pass a current through the meter to give full scale deflection and then measure the drop in reading when the meter is shunted by the unknown low resistance.
(Sorry, I recently disposed of my copy of "Telephony", by Herbert and Proctor, which would have given the definitive answer).
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top