A 160 - μF capacitor charged to 450 V is discharged through a 31.2 -k Ωresistor. (a) Find the ti

A 160 - μF capacitor charged to 450 V is discharged through...

A 160 - $\mu$ F capacitor charged to 450 V is discharged through a 31.2 -k $\Omega$ resistor. (a) Find the time constant. (b) Calculate the temperature increase of the resistor, given that its mass is $2.50 \mathrm{g}$ and its specific heat is $1.67 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot^{\circ} \mathrm{C}}$ noting that most of the thermal energy is retained in the short time of the discharge. (c) Calculate the new resistance, assuming it is pure carbon. (d) Does this change in resistance Seem significant?

A 160 - $\mu$ F capacitor charged to 450 V is discharged through a 31.2 -k $\Omega$ resistor. (a) Find the time constant. (b) Calculate the temperature increase of the resistor, given that its mass is $2.50 \mathrm{g}$ and its specific heat is $1.67 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot^{\circ} \mathrm{C}}$
noting that most of the thermal energy is retained in the short time of the discharge. (c) Calculate the new resistance, assuming it is pure carbon. (d) Does this change in resistance Seem significant?

Key Concepts

RC Time Constant

The RC time constant is a fundamental concept in circuits comprising a resistor and a capacitor. It is defined as the product of the resistance (R) and the capacitance (C) and describes the rate at which the capacitor charges or discharges. In transient circuit analysis, the time constant determines how quickly a capacitor will release its stored energy through the resistor, with the voltage decaying exponentially over time.

Energy Stored in a Capacitor

This concept refers to the energy that can be stored in the electric field of a capacitor. The energy is given by the formula ½CV², where C is the capacitance and V is the voltage across the capacitor. This stored energy becomes significant in transient events, such as discharges, where it is quickly converted into other forms, typically heat, as it is released through circuit components like resistors.

Specific Heat Capacity and Heat Transfer

Specific heat capacity is a measure of the amount of energy required to raise the temperature of a unit mass of a substance by one degree Celsius. In the context of electrical circuits, when a resistor dissipates electrical energy as heat, the specific heat capacity, along with the mass of the resistor, dictates how much its temperature will increase. This relationship is given by the equation ?T = Q/(m·c), where Q is the energy transferred, m is the mass, and c is the specific heat.

Thermal Effects on Resistive Materials

Thermal effects on resistance refer to the change in a material’s electrical resistance as its temperature changes. In conductors or semiconductors like carbon, the resistance may vary due to temperature-dependent changes in resistivity. This concept is crucial when evaluating the performance of a resistor after it has dissipated a significant amount of energy, as the temperature rise can alter its resistance value, affecting circuit behavior.

Transcript

00:01 So in the first part of this question, we're asked to find the time constant of this rc circuit.

00:09 And we know the value of r &c, the resistance, and the capacitance.

00:14 So to solve for tau, which represents the time constant, simply multiply r &c, and that will end up giving us the value of 4 .99 seconds.

00:27 So here is our time constant.

00:29 And in the next part, we're asked to find by what amount the temperature increases in this resistor.

00:37 And we're given that the mass of the resistor is 2 .5 grams.

00:42 And the specific heat of the material of the resistor is 1 .6 .7 kilojoules per kilogram celsius.

00:51 So what we're first going to do is find the energy that's stored in the capacitor.

00:56 We can do this through this equation.

01:00 Where q, which represents the charge, is equal to one -half times v squared, voltage squared times the capacitance.

01:11 And when we plug in for the values that we have here, for 50 volts and 160 micro -phirids, what we end up getting is an energy stored in the capacitor of 16 .2 joules.

01:25 So now that we know the energy stored, we can use this amount to solve for the temperature change.

01:34 So the equation for thermal energy is as follows.

01:39 So we have q representing the energy or the charge stored equals the mass times the specific heat times the change in temperature.

01:52 So when we're plugging in for the values here, and i'll just rearrange so that we have delta t, which would then be the energy divided by the mass times the specific heat...

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