Final answer:
An electronic circuit to measure resistance can be made by connecting a galvanometer modeled as a 2.4kΩ resistor in series with a high-value resistor to detect a full-scale deflection with a 100μA current at 10V, which is the maximum measurable voltage without exceeding the available ±15V.
Step-by-step explanation:
To design an electronic circuit that can measure a resistance in the range of 0−100kΩ with ±5% accuracy and display the measurement on a 100μA full-scale meter movement, one can follow the principles used in creating a voltmeter out of a galvanometer. The circuit should be configured such that the meter movement, modeled by a 2.4kΩ resistor, is connected in series with another high-value resistor. This high-value resistor will determine the maximum voltage range that the meter can measure.
As a first step, we consider that the maximum measurable voltage is 10V, and this should result in a full-scale deflection of the meter movement. Given that the meter movement is equivalent to a 2.4kΩ resistor, we need to calculate the total resistance required to achieve a full-scale deflection with a 100μA current, remembering that we should not exceed the maximum available voltage of ±15V in the circuit. The total resistance can be found using Ohm's law: V = IR, where I is the current (100μA) and V is the voltage (10V).
Once the total resistance is known, subtract the meter movement's resistance (2.4kΩ) from this value to find the resistance of the series resistor needed. For fine-tuning accuracy, variable resistors (potentiometers) could be included in the design to account for component tolerances and calibration. Finally, proper protection such as fuses should be placed to secure the meter against excessive currents.