Two cylindrical cans with insulating sides and conducting...

Two cylindrical cans with insulating sides and conducting end caps are filled with water, attached to the circuitry shown in Fig. $\mathbf{P} 25.69,$ and used to determine salinity levels. The cans are identical, with radius $r=5.00 \mathrm{~cm}$ and length $L=3.00 \mathrm{~cm} .$ The battery supplies a potential of $10.0 \mathrm{~V},$ has a negligible internal resistance, and is connected in series with a resistor $R=15.0 \Omega .$ The left cylinder is filled with pure distilled water, which has infinite resistivity. The right cylinder is filled with a saltwater solution. It is known that the resistivity of the saltwater solution is determined by the relationship $\rho=\left(s_{0} / s\right) \Omega \cdot \mathrm{m},$ where $s$ is the salinity in parts per thousand $(\mathrm{ppt})$ and $s_{0}=6.30$ ppt. (a) The ammeter registers a current of $484 \mathrm{~mA}$. What is the salinity of the saltwater solution? (b) The left cylinder acts as a capacitor. Use Eq. (24.19) for its capacitance. How much charge is present on its upper plate? Note that pure water has a dielectric constant of $80.4 .$ (c) At what rate is energy dissipated by the saltwater? (d) For what salinity level would the $15.0 \Omega$ resistor dissipate half the power supplied by the battery?

Key Concepts

Ohm’s Law

This principle relates voltage, current, and resistance in an electrical circuit (V = IR). In problems where a known voltage results in a measured current, Ohm’s law allows the determination of the resistance of an element in the circuit, which is essential for calculating properties like resistivity when geometry is taken into account.

Resistivity and Geometric Resistance

Resistivity is an intrinsic property of a material that quantifies how strongly it opposes the flow of electric current. The resistance of a conductor with uniform cross?section is given by R = ?L/A, where L is the length and A is the cross-sectional area. This concept links the material property with the physical dimensions, allowing for the determination of unknown parameters such as salinity when the resistivity depends on the salt concentration.

Series Circuit Analysis

In a series circuit, the same current flows through all components while the total supplied voltage is distributed across each element in proportion to its resistance. Understanding series circuit behavior is crucial for determining voltage drops, current distribution, and for analyzing how individual resistances, such as that of the saltwater-filled can and an additional resistor, affect overall circuit function.

Capacitance of a Parallel?Plate Capacitor with a Dielectric

The capacitance of a parallel?plate capacitor is given by C = ???A/d, where ? is the dielectric constant, ?? is the permittivity of free space, A is the area of the plates, and d is the separation between them. This concept is important when using a physical system (such as a water-filled cylinder with conducting end caps) as a capacitor, as the dielectric properties of the medium (here, water) directly affect the amount of charge that can be stored for a given potential difference.

Energy Dissipation in Electrical Circuits

This concept involves calculating the power dissipated as heat in resistive components using relations such as P = I²R or P = V²/R. Understanding how energy is converted into heat within resistive elements is key to evaluating the efficiency of circuits and analyzing how varying parameters like salinity (which alters resistance) impact power distribution.

Relationship Between Salinity and Resistivity

In some systems, resistivity depends inversely on the solute concentration; here, resistivity is given as a function of salinity. This relationship bridges the gap between the chemical composition of a solution and its electrical properties, allowing for the determination of salt concentration through electrical measurements in a circuit.

Transcript

00:01 In this question we have this electrical circuit and the left side is filled with distilled water.

00:09 They are both insulating containers but arrange in a like a parallel plate capacitor manner.

00:18 Okay, so the left capacity, the left thing is like a still with the water and then the right thing is filled with salt water.

00:27 So there are four parts in this question.

00:30 In part a, you are given that the current is 484 millie -ampeer, which is 0 .4 -84 ampers.

00:45 We want to find the salinity of the salt water, okay? so we can find the total resistance in the circuit, which is just emf divided by i -10 divide by 0 .4 is 4.

01:01 This is 20 .7 oms and then the resistance of the salt water.

01:08 Okay, it's just 20 .7 minus 15.

01:12 Okay, this is 5 .66 oms actually.

01:16 Okay.

01:17 And then the resistance of the salt water, you can use row l over a to represent to calculate, okay, and the area is pi r square.

01:33 Okay, so l here is 0 .03 meters, r is 0 .05 meters, okay? so we can calculate, and then we are given that the resistivity of the salt water is a function of s -not, by s okay so s not is the constant salinity okay and then s is the salinity of the salt water okay in parts per 3 ,000 okay so we can calculate s and this is s not l divide by r s high r square so putting the numbers we have 6 .3 .3 times 0 .03 divide by pi 0 .05 square times 5 .66.

02:38 Calculate this you get 4 .25 parts per thousand.

02:43 Okay.

02:48 Then in part b, we want to calculate the charge stored in the left electrical component because the whole thing acts like a capacitor.

02:58 Okay...

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