Question #97562
Design a simple ohmmeter and explain the conditions of the ohmmeter circuit when the terminal to which the unknown resistance is connected "open" and "close".
1
Expert's answer
2019-11-04T06:31:42-0500

A simple ohmmeter can be made of 2 constant resistors, 1 variable resistor, a voltage source, and a galvanometer, or a voltmeter. One puts the unknown resistance to x, and adjusts the variable resistor 2 so that the current through the galvanometer is 0 or the voltage between the terminals of the Wheatstone bridge. As soon as 0 is obtained, find the unknown resistance:


Rx=R3R2R1.R_x=\frac{R_3R_2}{R_1}.


If the terminals where the x is supposed to be are opened, the ammeter's readings are


IG=ER1R1R2+R2R3+R1R3.I_G=\frac{ER_1}{R_1R_2+R_2R_3+R_1R_3.}

If IG=0,I_G=0, you should record the value of R2R_2 and calculate the unknown resistance according to the first equation.

If there is a voltmeter instead of G, its readings would be


VG=ER1R1+R2.V_G=\frac{ER_1}{R_1+R_2}.

If VG=0,V_G=0, you should record the value of R2R_2 ​and calculate the unknown resistance.

Any other readings of IG,VGI_G, V_G mean that the Wheatstone's bridge is unbalanced, and the terminals of RxR_x are closed.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS