The algebraic sum of the currents into any junction is zero. Our goal is to find the current I 2 through the resistor R thus we express all currents as functions of I 2. The resistance of a wire is a linear function of its length thus we can express the resistances of the resistors R x and R y express as:
Where l stands for the length of the whole potentiometer and a is a constant of proportionality. The expression for the resistance of the potentiometer reads The voltage across the resistor R is given as a product of the resistance of the resistor and the current through this resistor. We know the total voltage across the circuit hence we can express the emf (electromotive force) as a sum of drops of voltage across resistors R x and R y.įrom the expression for the total voltage we obtain the current I 2. That helps to express the current I 1 as a function of I 2. The resistors R and R x are in parallel connection thus there must be equal drop of voltage across them. The current I is a sum of currents I 1 and I 2. We need to find the current I 2 through the resistor R therefore we express all currents as functions of I 2. Let us label all the currents through different loops of the circuit in the picture. The resistance of a conductor is a linear function of its length thus the resistance of the part of the potentiometer is directly proportional to the length of the resistance wire used in this part of the potentiometer. We can find the resistances of each part directly. Let us redraw the circuit with the potentiometer divided by the sliding contact into two parts with resistances R x and R y (see Illustrating picture).
0 kommentar(er)
