Final answer:
To design a variable voltage divider that can adjust from 10 V to 100 V with a 10 mA current, a 9 kΩ potentiometer in series with a 1 kΩ fixed resistor should be used, all in accordance with Ohm's Law.
Step-by-step explanation:
Designing a Variable Voltage Divider
To design a variable voltage divider that provides an adjustable voltage source ranging from 10 V to 100 V using a 1 to 120 V source with a current of 10 mA, Ohm's Law can be used. The potentiometer in this case functions as the variable resistor within the voltage divider circuit. By adjusting the resistance of the potentiometer, we can vary the output voltage across it. To obtain the maximum output voltage at the maximum resistance setting, and the minimum voltage at the minimum resistance (zero) setting, we need to calculate the total resistance and the resistance required for the potentiometer in series with a fixed resistor.
Using Ohm's Law (V = IR), where V is the voltage, I is the current, and R is the resistance, we can determine the total resistance needed for the maximum voltage:
Vmax = Imax × Rtotal
We know that Vmax is 100 V and Imax is 10 mA, which translates to 0.01 A:
Rtotal = Vmax / Imax = 100 V / 0.01 A = 10,000 Ω
In order to achieve a minimum of 10 V output at the minimum resistance, which is zero for the potentiometer, a fixed resistor is required in series. The fixed resistor (Rfixed) value is calculated by subtracting the potentiometer's minimum resistance from the total resistance needed for the minimum voltage output.
Using Ohm's Law again:
Vmin = Imin × Rfixed
Given that Vmin is 10 V and Imin is 10 mA:
Rfixed = Vmin / Imin = 10 V / 0.01 A = 1,000 Ω
The potentiometer, therefore, must have a maximum resistance value of:
Rpot_max = Rtotal - Rfixed = 10,000 Ω - 1,000 Ω = 9,000 Ω
Thus, a 9 kΩ potentiometer in series with a 1 kΩ fixed resistor will meet the design requirements for the variable voltage divider circuit.