Using Kirchhoff's laws for the loops ADCA and BCDB write an expression of the resistance of the PRT in terms of L1, L and the value of the standard resistor.
Circuit provided:
imgur image: mQYi7RH.png (add imgur . com / before it).
To find the expression for the resistance (R) of the Potentiometer (PRT) in terms of L1, L, and the value of the standard resistor, we need to analyze the circuit using Kirchhoff's laws.
First, let's label the resistors in the circuit. The resistor connected between B and C will be labeled as R1, and the resistor connected between A and D will be labeled as R2.
Now, we can proceed with analyzing the circuit using Kirchhoff's laws:
1. Loop ADCA:
- Start at A and move clockwise around the loop.
- As you move across the standard resistor (R), you encounter a voltage drop equal to I * R, where I is the current in the loop.
- After that, you encounter the resistance L1 in series with the resistor R2.
- Finally, you reach point A.
- Applying the Kirchhoff's voltage law in this loop, the sum of the voltages in a closed loop is zero.
- We can write the expression as: -IR + L1 * I + R2 * I = 0
2. Loop BCDB:
- Start at B and move clockwise around the loop.
- As you move across the resistor R1, you encounter a voltage drop equal to I * R1, where I is the current in the loop.
- After that, you encounter the resistance L in series with the Potentiometer (PRT) resistance R.
- Finally, you reach point B.
- Applying the Kirchhoff's voltage law in this loop, the sum of the voltages in a closed loop is zero.
- We can write the expression as: -I * R1 + L * I + R = 0
Now, we have a system of two equations with two unknowns: R and I. Our goal is to find the resistance R of the Potentiometer in terms of L1, L, and the value of the standard resistor.
To solve this system of equations, we can use algebraic manipulation to eliminate I:
From the first equation: -IR + L1 * I + R2 * I = 0
Rearrange to find I in terms of R: I * (L1 + R2 - R) = IR
Divide both sides by (L1 + R2 - R): I = (IR) / (L1 + R2 - R)
Substitute this value of I into the second equation:
-(IR1) + L * (IR) + R = 0
Group the terms with I: I * (L * R - R1) + R = 0
Simplify: IR * (L - R1) = -R
Now, we can substitute the value of I from the first equation into the second equation:
[(IR) / (L1 + R2 - R)] * (L - R1) = -R
Simplify and rearrange to isolate R:
(L - R1) * R = -R * (L1 + R2 - R)
Expand the equation: L * R - R * R1 = -LR1 - RR2 + R * R
Collect similar terms: L * R - R * R1 + RR2 - R * R = -LR1
Combine like terms: L * R + R * (R2 - R1) = -LR1
Finally, isolate R by dividing both sides by (L + R2 - R1):
R = -(LR1) / (L + R2 - R1)
Thus, the expression for the resistance (R) of the Potentiometer (PRT) in terms of L1, L, and the value of the standard resistor is:
R = -(LR1) / (L + R2 - R1)