Derivation of the equation for an ohmmeter circuit. Given the circuit shown in t

Question

Derivation of the equation for an ohmmeter circuit. Given the circuit shown in the accompanying figure. Symbols: V: A fixed voltage source (often 1.5 volts) G: Galvanometer, the current through which will measure the unknown resistor. R_m: Internal resistance of the galvanometer G. R_x: The resistance of the unknown resistor (not yet shown) R: A variable resistance used to protect the galvanometer and to calibrate the ohmmeter. I_f: The current required by the galvanometer for full-scale deflection. I_m: The current through the galvanometer in use, when R_x is connected between the X's. Steps to be taken in when using an ohmmeter. and in obtaining the equation for this ohmmeter circuit. Step 1: Short X to X and adjust variable resistance R so that the galvanometer reads l_f. For the equation, solve for R in terms of V, R_m and I_f. Thus, zero external resistance (the short) corresponds to maximum readable current (I_f). (If zero resistance allows more current, you risk the meter; if less, then the upper end of the meter's capability is never used, since R_x will not be negative.) Step 2: With R determined and remaining fixed, connect the unknown resistance, R_x. between the terminals x-x. This will result in a current I_m. In use, the meter face would indicate R; for the exercise, solve for R_x in terms of V, R_m, I_f, and I_m (eliminate R). This is the theoretical formula you plot in the experiment:

Explanation / Answer

Step 1: For shorted X-X, the total equivalent resistance in circuit = Rm + R

by Ohm's law, V = If*(Rm+R) so R = V/If - Rm

Step 2: the new equivalent resistance in circuit = Rm + R + Rx

by Ohm's Law, V = Im*(Rm + R + Rf) so eleminating R and solving for Rx we get Rx = V(Im-1- If-1)

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