Potential divider and potentiometer principle

[nutzweb]

does anyone knows how to do it? help and pls explain to me... thanks!

1) The slide-wire of a simple potentiometer is 2 meters long and has a resistance of 5 ohms. The emf of the working battery is 6 V and its internal resistance is 0.20 W. What resistance must be added to the working-battery circuit in order for the wire to have a direct reading of 1 mV/mm?

2) A slide-wire potentiometer is balanced against a 1.0182-V standard cell when the slide wire is set at 40.2 cm out of a total length of 100 cm. For an unknown source, the setting is 13. 8 cm. What is the emf of the unknown?

3) Under what condition is a voltmeter more desirable than a potentiometer for the measurement of voltages?

4) Prove that the potentiometer measures the emf of the test cell and not its TPD (hint: use the equation TPD=EMF-Ir).

 

[Astronuc]

Assuming the resistance is uniform along the wire, then the resistance per unit length is given by 5 ohms / 2 m = 2.5 ohms/m.

Now remember V = IR or V/m = I * R/m.

Also remember 1 mV/mm = 1 V/m.

I think you can handle it now.

 

[thepatientmental]

1) To solve this problem, we can use the potential divider principle, which states that the voltage across a resistor in a series circuit is proportional to its resistance. In this case, we can set up a potential divider with the 2-meter long slide-wire as one resistor and the added resistance as the other. We know that the desired reading is 1 mV/mm, so we can convert this to an overall resistance of 1 ohm (1 mV/mm = 1 V/m = 1 ohm). We also know that the total voltage is 6 V and the internal resistance is 0.20 ohms. Using the potential divider equation (V1/V2 = R1/R2), we can set up the following equation: (1/6) = (5/R2 + 0.20). Solving for R2, we get a resistance of 0.833 ohms. Therefore, the added resistance must be 0.833 ohms for the wire to have a direct reading of 1 mV/mm.

2) In this problem, we can use the potentiometer principle, which states that the ratio of the lengths of the wire on either side of the balancing point is equal to the ratio of the emfs of the two cells. Using this principle, we can set up the following equation: (40.2/100) = (1.0182/Vunknown). Solving for Vunknown, we get an emf of 0.5485 V. Therefore, the unknown source has an emf of 0.5485 V.

3) A voltmeter is more desirable than a potentiometer for the measurement of voltages when the voltage being measured is constantly changing or fluctuating. A potentiometer requires manual adjustment to balance the circuit, whereas a voltmeter can continuously measure the voltage without any adjustments. Additionally, a voltmeter has a higher accuracy and precision compared to a potentiometer.

4) The equation TPD=EMF-Ir represents the total potential difference (TPD) in a circuit, where EMF is the electromotive force and Ir is the voltage drop across the internal resistance. In a potentiometer, the balancing point is achieved when the TPD of the test cell is equal to the TPD of the standard cell. This means that the EMF of the test cell is equal to the EMF of the standard cell, as