Question

A potential difference of \(V=0.500 \mathrm{~V}\) is applied across a block of silicon with resistivity \(8.70 \cdot 10^{-4} \Omega \mathrm{m}\). As indicated in the figure, the dimensions of the silicon block are width \(a=2.00 \mathrm{~mm}\) and length \(L=15.0 \mathrm{~cm} .\) The resistance of the silicon block is \(50.0 \Omega\), and the density of charge carriers is \(1.23 \cdot 10^{23} \mathrm{~m}^{-3}\) Assume that the current density in the block is uniform and that current flows in silicon according to Ohm's Law. The total length of 0.500 -mm-diameter copper wire in the circuit is \(75.0 \mathrm{~cm},\) and the resistivity of copper is \(1.69 \cdot 10^{-8} \Omega \mathrm{m}\) a) What is the resistance, \(R_{w}\) of the copper wire? b) What are the direction and the magnitude of the electric current, \(i\), in the block? c) What is the thickness, \(b\), of the block? d) On average, how long does it take an electron to pass from one end of the block to the other? \(?\) e) How much power, \(P\), is dissipated by the block? f) In what form of energy does this dissipated power appear?

Short answer

Answer: The power dissipated in the silicon block appears in the form of heat energy.

Step-by-step solution

a) Calculating the resistance of the copper wire

First, we will find the resistance of the copper wire using the resistivity formula: \(R_w = \frac{\rho_w \cdot L_w}{A_w}\) We know that the resistivity of copper (\(\rho_w\)) is \(1.69 \cdot 10^{-8} \Omega m\), and the total length of the copper wire (\(L_w\)) is \(75.0 cm = 0.75 m\). The wire's cross-sectional area (\(A_w\)) can be found using the formula for the area of a circle, as the wire is cylindrical. The diameter of the wire is \(0.500 mm = 5.00 \cdot 10^{-4} m\), so its radius (\(r_w\)) is half of that: \(r_w = 2.50 \cdot 10^{-4} m\) Now we can find the area: \(A_w = \pi r_w^2 = \pi (2.50 \cdot 10^{-4})^2\) Plug all the values into the resistivity formula to find the wire resistance: \(R_w = \frac{1.69 \cdot 10^{-8} \cdot 0.75}{\pi (2.50 \cdot 10^{-4})^2} \approx 0.000169 \Omega\) So the resistance of the copper wire is approximately \(0.169 \times 10^{-3} \Omega\).

b) Calculating the direction and magnitude of electric current in the block

We are given that the resistance of the silicon block is \(50.0 \Omega\) and the potential difference (V) across the block is \(0.5 V\). According to Ohm's Law, the electric current (i) can be found using the formula: \(i = \frac{V}{R_{total}}\) The total resistance (\(R_{total}\)) includes both the resistance of the silicon block and the resistance of the copper wire: \(R_{total} = 50.0 + 0.000169 \approx 50.0 \Omega\) By Ohm's Law, the electric current (i) is: \(i = \frac{0.5}{50} = 0.01 A\) So the magnitude of the electric current in the block is \(0.01 A\). The direction of the current in the block is the same as the direction of current flow from the positive terminal of the voltage source to the negative terminal.

c) Calculating the thickness of the silicon block

Now we will find the thickness (b) of the silicon block using the resistivity formula. We have the block's resistivity, width, and length, so we can rewrite the formula in terms of its thickness: \(R_{block} = \frac{\rho_{block} \cdot L}{A_{block}}\) Now, \(A_{block} = a \cdot b\). Therefore, \(R_{block} = \frac{\rho_{block} \cdot L}{a \cdot b}\) Plugging in the given values, we have: \(50 \Omega = \frac{8.7 \times 10^{-4} \Omega m \cdot 0.15 m}{0.002 m \cdot b}\) Rearrange the equation and solve for b: \(b = \frac{8.7 \times 10^{-4} \Omega m \cdot 0.15 m}{50 \Omega \cdot 0.002 m} \approx 1.3 \times 10^{-3} m\) So the thickness of the silicon block is approximately \(1.3 mm\).

d) Calculating the average time an electron takes to pass through the block

We first need to find the drift velocity (\(v_d\)) of the electrons using the formula: \(v_d = \frac{I}{n \cdot q \cdot A_{block}}\) Where n is the density of charge carriers (\(1.23 \cdot 10^{23} m^{-3}\)), and q is the charge of an electron (\(1.6 \times 10^{-19} C\)). We already know the cross-sectional area of the block, and the current in the block is \(0.01 A\). So, \(v_d = \frac{0.01}{1.23 \cdot 10^{23} \cdot 1.6 \times 10^{-19} \cdot (2 \times 10^{-3} \cdot 1.3 \times 10^{-3})} \approx 2.4 \times 10^{-5} m/s\) Now we can find the average time (t) an electron takes to pass through the block using the formula: \(t = \frac{L}{v_d}\) \(t = \frac{0.15}{2.4 \times 10^{-5}} \approx 6.25 \times 10^3 s\) So on average, it takes an electron approximately \(6.25 \times 10^3\) seconds to pass from one end of the block to the other.

e) Calculating the power dissipated by the silicon block

Now we will find the power (P) dissipated by the silicon block using the formula: \(P = I^2 \cdot R_{block}\) Plug in the current (\(0.01 A\)) and the resistance of the silicon block (\(50 \Omega\)): \(P = (0.01)^2 \cdot 50 \approx 5 \times 10^{-3} W\) So, the power dissipated by the silicon block is approximately \(5 mW\).

f) Identifying the form of energy from dissipated power

The energy dissipated as power in the silicon block is in the form of heat. This is because, when the block resists the flow of the electric current, the kinetic energy of moving electrons gets converted into heat energy. This increases the temperature of the block and causes it to dissipate heat to its surroundings.

Key concepts

Ohm's Law

One of the foundational principles in resistive circuits is Ohm's Law. This law describes the relationship between voltage, current, and resistance in an electrical circuit. It is expressed by the formula: \(V = IR\). Here, \(V\) represents voltage measured in volts, \(I\) is the current in amperes, and \(R\) is the resistance in ohms.
Ohm's Law helps us understand how electrical potential difference drives current through a resistor. If you know any two of the three values—voltage, current, or resistance—you can calculate the third. For example, the exercise uses Ohm's Law to determine the electric current in a silicon block given its voltage and resistance.
Ohm's Law is vital in analyzing circuits to predict how they'll behave under different electrical loads. It's a simple yet powerful tool that provides insight into the flow of electricity in resistive materials.

Electric Current

Electric current is the flow of electric charge in a conductor. It is typically carried by moving electrons in a wire or conductor. In the exercise, the current is calculated using Ohm's Law, where a known voltage and resistance give the current value as \(0.01\) A.
Current is measured in amperes (A) and can flow in various conductive materials like metals or semiconductors. The direction of current is conventionally from positive to negative, although the electrons technically flow the opposite way.
Understanding the magnitude and direction of electric current is crucial, especially in designing circuits and ensuring safety. It's what powers devices, drives motors, and lights up bulbs. Changes in current can affect how well a device functions, making it an essential concept in electronics and physics.

Resistivity

Resistivity is a material property that indicates how much a material opposes the flow of electric current. It's denoted by the symbol \(\rho\) and is measured in ohm meters (\(\Omega \cdot m\)).
In the exercise, silicon has a resistivity of \(8.70 \times 10^{-4} \Omega \cdot m\), which impacts the block's resistance. Resistivity depends on factors like temperature and the material's structure. For instance, metals generally have lower resistivity compared to semiconductors like silicon.
Knowing the resistivity helps in determining the right material for electrical applications, enabling engineers to select materials with suitable conductive properties. The thickness and dimensions of an object also factor into its overall resistance when combined with resistivity.

Power Dissipation

Power dissipation in electrical circuits refers to the conversion of electrical energy into heat. This is often unwanted as it can lead to energy losses and heating of components.
Power dissipation is calculated using the formula \(P = I^2R\), where \(I\) is the current and \(R\) is the resistance. In the exercise, the power dissipation in the silicon block was calculated to be 5 mW.
Understanding power dissipation is critical when designing circuits since excessive heat can damage components. Engineers strive to optimize circuits to reduce power loss, thus improving efficiency and performance.