A simple ohmmeter is made by connecting a 1.50-V flashlight battery in series with a resistor R and a 1.00-mA ammeter, as shown in the figure. R is adjusted so that when the circuit elements are shorted together the meter deflects its full-scale value of 1.00 mA. We have to calculate the values of the resistances across the terminal which would give a deflection of (a) 10%, (b) 50%, and (c) 90% of full scale. (d) If the ammeter has a resistance of 18.5 and the internal resistance of the battery is negligible, we have to calculate the value of R.

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In the figure circuit diagram of an ohmmeter has been drawn. When the clip leads are shorted together, it is given that the 1.0-mA ammeter shows full scale deflection. That is a 1.0 mA current is flowing through the circuit. Assuming that the ammeter and the battery have negligible resistances, the value of the resistance R will have to be

(b)

If the ammeter shows 10% deflection of the full scale, the current in the circuit will be 0.1 mA. Therefore, the external resistance, , across the leads will be as given by the relation

(c)

If the ammeter shows a 50% deflection of the full scale, the current in the circuit will be 0.5 mA. The resistance, , will be given by the relation

(d)

If the ammeter shows a 90% deflection of the full scale, the current in the circuit will be 0.9 mA. The resistance, , will be given by the relation

(e)

If the ammeter has a resistance of 18.5 and the internal resistance of the battery is negligible, the required value of R will be given by the equation