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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?

Step by step solution

01

Calculate the resistance of the saltwater solution

Using Ohm's law, \(V = IR\), we can rearrange the equation to solve for the resistance: \(R = V / I\). The total resistance is given by \(10.0 V / 484 mA = 20.67 惟\). As the circuit also contains a known 15 惟 resistor, the resistance of the saltwater solution, \(R_{sw}\) can be calculated by subtracting the known resistance from the total resistance: \(R_{sw} = 20.67 惟 - 15 惟 = 5.67 惟\).

02

Calculate the salinity of the saltwater

We can utilize the relationship \(蟻 = R * (A / L)\) where \(A\) and \(L\) are the cross section area and the length of the cylindrical can respectively, to find the resistivity of the salt water. Using the given dimensions, we find \(A = 蟺 (0.05 m)^2 = 0.00785 m^2\) and \(L = 0.03 m\). Plugging these values in gives \(蟻 = 5.67 惟 * (0.00785 m^2 / 0.03 m) = 1.42 惟.m\). Since we are told \(蟻 = (s_0 / s) * 惟.m\), we solve for \(s = s_0 / 蟻\) = 6.30 ppt / 1.42 惟.m = 4.44 ppt.

03

Calculate the charge on the upper plate of the capacitor

The distilled water acts like a capacitor and the charge on it can be calculated from \(Q = CV\) where \(V\) is the voltage across it, which is the same as the battery potential - \(10.0 V\), and \(C\) is the capacitance calculated using the formula \(C = 魏蔚_0 ( A / d)\) where \(魏\) is the dielectric constant of water - \(80.4\), \(蔚_0\) is the permittivity of free space - \(8.85 脳 10^-12 F/m\), \(A\) is the cross section area of the can and \(d\) is the distance between the plates - \(0.03 m\). Substituting these values into the equation and calculating gives \(C = 80.4 * (8.85 脳 10^-12 F/m) * (0.00785 m^2 / 0.03m) = 184.37 nF\). Substituting for \(C\) in the earlier equation and solving gives \(Q = 10.0 V * 184.37 nF = 1.84 渭C\).

04

Calculate the rates of energy dissipation

The power dissipated by the salt water can be calculated using \(P = I^2R\). Substituting the values for \(I = 0.484 A\) and \(R = 5.67 惟\), we have \(P = (0.484 A)^2 * 5.67 惟 = 1.33 W\).

05

Determine the salinity for half power dissipation

For the power dissipated by the 15 惟 resistor to be half the power supplied by the battery, we must have \(P_R = (1/2)IV\), where \(I\) is the current in the circuit and \(V\) is the battery potential. We rearrange to solve for the current, \(I = (2P_R) / V\) = \(2 * (0.484 A)^2 * 15 惟 / 10 V = 0.70 A\). We use Ohm's law to calculate the total resistance at this current, \(R_T = V / I\) = 10V / 0.70 A = 14.29 惟. Subtracting the known 15 惟 resistance from this gives the saltwater resistance \(R_sw' = -0.71 惟\). Since resistance cannot be negative, this means that it's not possible for the 15 惟 resistor to ever dissipate half the total power in this circuit.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistance

In electric circuits, resistance measures how much an object opposes the flow of electric current. It is a crucial factor when designing circuits. Ohm's law relates resistance to the voltage and current in a circuit.

• Measured in ohms (惟).
• Calculated using the formula: \( R = \frac{V}{I} \), where \( V \) is voltage and \( I \) is current.
• In our exercise, the saltwater's resistance is found by calculating total resistance and subtracting known resistances. This method isolates the unknown resistance element.

Remember, changes in material or configuration, like adding salt to water, can significantly alter resistance. This property is utilized in measuring salinity.

Ohm's Law

Ohm's law is a fundamental principle in electric circuits, establishing a relationship between voltage, current, and resistance.

• Expressed by the formula: \( V = IR \).
• Where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
• It allows calculation of one variable if the other two are known.

In practice, as shown in the exercise, Ohm鈥檚 law helps to calculate the total resistance in series circuits. It aids in analyzing complex circuits, enabling prediction of how current and voltage distribution occurs.

Capacitance

Capacitance is the ability of a system to store an electric charge, crucial in many electronic applications.

• Measured in farads (F).
• Calculated using the formula for a parallel plate capacitor: \( C = \kappa \varepsilon_0 \frac{A}{d} \).
• \(\kappa\) is the dielectric constant, \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area, and \( d\) is the separation between plates.

In this exercise, the distilled water between the cans acts as a capacitor. It stores energy and demonstrates how dielectrics like water influence capacitance. Energy storage in circuits is critical for operations like smoothing power delivery.

Salinity Measurement

Salinity measurement in solutions can significantly impact electrical properties like resistivity.

• Salinity is often measured in parts per thousand (ppt).
• Higher salinity reduces resistivity, allowing more current flow.
• In our scenario, resistivity is inversely related to salinity: \( \rho = \frac{s_0}{s} \cdot \Omega \cdot \text{m} \).

Detecting salinity using electric current is practical in various fields, including environmental science and food processing. Understanding this relationship helps predict how saline solutions behave in different electrical applications.

Energy Dissipation

Energy dissipation refers to the conversion of electrical into other forms of energy, such as heat.

• Measured in watts (W).
• Calculated as \( P = I^2R \), where \( I \) is current and \( R \) is resistance.
• Reflects the efficiency and safety of a circuit design.

In the exercise, energy dissipation in the saltwater shows how energy converts to heat in resistive elements. This concept is vital in managing thermal conditions in electronics, ensuring they operate at safe and sustainable temperatures.