Solving Circuits with a Potentiometer: Calculate R & Rx for 1A-10A

[mircobot]

Homework Statement

There's an 110V source and two resisters in series. The first resister, R, is a fixed resistor. The second resister, Rx, is a variable resister (a potentiometer). The potentiometer is to be designed to adjust current I from 1A to 10A. Calculate the values of R and Rx to achieve this.

Homework Equations

The Attempt at a Solution

I have no clue where to start. I have never dealt with circuit problems that involved a potentiometer before.

 

[CEL]

The maximum current will flow trough the circuit when the potentiometer is at its minimum value and the minimum current when it is at its maximum value.

 

[piercas]

First, we need to understand what a potentiometer is and how it works in a circuit. A potentiometer is a type of variable resistor that allows us to adjust the resistance in a circuit by changing its position. It consists of a fixed resistor with a movable contact point, which can be moved along the resistor to change the resistance value.

In this circuit, we have an 110V source and two resistors in series. The first resistor, R, is a fixed resistor with a known resistance value. The second resistor, Rx, is the potentiometer and its resistance value is what we need to calculate. Our goal is to adjust the potentiometer to achieve a current range of 1A to 10A.

To solve this circuit, we can use Ohm's law (V=IR) and the equation for calculating total resistance in a series circuit (Rt = R1 + R2).

First, we need to determine the total resistance of the circuit at 1A and at 10A. At 1A, the total resistance (Rt) can be calculated as:

Rt = V/I = 110V/1A = 110Ω

At 10A, the total resistance (Rt) can be calculated as:

Rt = V/I = 110V/10A = 11Ω

Now, we need to find the value of Rx that will give us a total resistance of 110Ω at 1A and 11Ω at 10A. This can be done by setting up two equations using Ohm's law and the equation for total resistance in a series circuit:

Equation 1: 110Ω = R + Rx (at 1A)
Equation 2: 11Ω = R + Rx (at 10A)

We now have two equations and two unknowns (R and Rx). We can solve these equations simultaneously using algebra to find the values of R and Rx.

Subtracting Equation 2 from Equation 1, we get:

99Ω = 0 + Rx (since R is cancelled out)

Therefore, Rx = 99Ω.

Now, we can plug this value of Rx into either Equation 1 or 2 to find the value of R. Let's use Equation 1:

110Ω = R + 99Ω
R = 110Ω - 99