/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 111 A \(4 \mu \mathrm{F}\) condenser... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(4 \mu \mathrm{F}\) condenser is charged to \(400 \mathrm{~V}\) and then its plates are joined through a resistance. The heat produced in the resistance is : (a) \(0.64 \mathrm{~J}\) (b) \(0.32 \mathrm{~J}\) (c) \(0.16 \mathrm{~J}\) (d) \(1.28]\)

Short Answer

Expert verified
The heat produced in the resistance is \(0.32 \mathrm{~J}\).

Step by step solution

01

Identify the given values

First, let's identify the values given in the problem. We have a capacitance \( C = 4 \mu \mathrm{F} \) (microfarads) and a voltage \( V = 400 \mathrm{~V} \).
02

Formula for Stored Energy in a Capacitor

The energy \( U \) stored in a capacitor is given by the formula: \[ U = \frac{1}{2} C V^2 \]
03

Substitute the given values into the formula

Substitute \( C = 4 \times 10^{-6} \mathrm{~F} \) (since microfarads need to be converted to farads) and \( V = 400 \mathrm{~V} \) into the formula: \[ U = \frac{1}{2} \times 4 \times 10^{-6} \mathrm{~F} \times (400 \mathrm{~V})^2 \]
04

Calculate the energy

Calculate \( U \) using the substituted values: \[ U = \frac{1}{2} \times 4 \times 10^{-6} \times 160000 = \frac{1}{2} \times 640 \times 10^{-6} \] Perform the multiplication: \[ U = 320 \times 10^{-6} \times 160000 = 0.32 \mathrm{~J} \]
05

Heat produced in the resistance

Since all the stored energy is dissipated as heat when the plates are joined through a resistance, the heat produced in the resistance is equal to the energy initially stored in the capacitor: \( 0.32 \mathrm{~J} \).

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Key Concepts

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

Electrical Circuits
An electrical circuit is the pathway through which electricity flows. Think of it like a water channel system that directs the flow of water. In this specific exercise, the electrical circuit consists of a capacitor, which is charged and then connected through a resistance. This resistance controls the flow of electric charge, playing a vital role in how energy is distributed within the circuit.

Key components of electrical circuits include:
  • Voltage Source: Provides the electric potential energy (like the 400 V in our exercise).
  • Capacitor: Stores and releases electrical energy.
  • Resistor: Limits or regulates the flow of electrical current.
Understanding these components can help you see how electric energy flows and transforms in a circuit, similar to water moving through pipes with taps regulating its flow.
Heat Dissipation
Heat dissipation in electrical circuits refers to the conversion of electrical energy into thermal energy, often due to resistance. When the capacitor discharges, its stored energy is released, flowing through a resistor. This flow generates heat, similar to how friction heats up your hands when you rub them together.

In the given problem, all of the capacitor's stored energy is eventually converted into heat by the resistor. This is why the heat produced equals the energy initially stored in the capacitor, which is calculated to be 0.32 Joules. Understanding heat dissipation is crucial for managing electronic devices because excessive heat can damage components.
Capacitor Discharge
Capacitor discharge refers to the process of releasing stored electrical energy from the capacitor into the circuit. Normally, capacitors store electrical energy when connected to a power supply and discharge when the circuit is closed via a resistor, as in our example.

The discharge rate of a capacitor depends on the circuit's resistance and capacitance. As the capacitor releases its energy, voltage across it decreases exponentially, and current follows the same pattern. This gradual discharge creates a situation where all stored energy flows through the resistor, converting entirely to heat. Hence, controlling how a capacitor discharges ensures efficient energy use in electronic applications.
Energy Conversion
Energy conversion in this context means the transformation of electrical energy into another form, namely heat. Energy in capacitors is stored in the electric field between their plates and can be discharged to perform work like lighting a bulb or, in this case, heating a resistor.

The conversion occurs when stored electrical energy is released, creating a current through the resistor. The resistor, due to its inherent properties, converts this electrical energy into thermal energy. This transformation is a fundamental principle of physics, where energy changes form but the total energy remains conserved. Understanding energy conversion helps predict how energy will behave in practical situations, ensuring efficient system designs.

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Most popular questions from this chapter

There are 45 number of cells with internal resistance of each cell is \(0.5 \Omega\). To get the maximum current through a resistance of \(2.5 \Omega\), one can use \(m\) rows of cells, each row having \(n\) cells. The values of \(m\) and \(n\) are : (a) \(m=3, n=15\) (b) \(m=5, n=9\) (c) \(m=9, n=5\) (d) \(m=15, n=3\)

Choose the correct option : Three bulbs of \(100 \mathrm{~W}, 200 \mathrm{~W}\) and \(40 \mathrm{~W}\) are connected in series to the main supply of \(200 \mathrm{~V}\). The current will be : (a) maximum through \(100 \mathrm{~W}\) (b) maximum through \(200 \mathrm{~W}\) (c) maximum through \(40 \mathrm{~W}\) (d) same in all

A fuse wire of circular cross-section and having diameter of \(0.4 \mathrm{~mm}\), allows \(3 \mathrm{~A}\) of current to pass through it. But if another fuse wire of same material and circular cross-section and having diameter of \(0.6 \mathrm{~mm}\) is taken, then the amount of current passed through the fuse is : (a) \(3 \mathrm{~A}\) (b) \(3 \times \sqrt{\frac{3}{2}} \mathrm{~A}\) (c) \(3 \times\left(\frac{3}{2}\right)^{3 / 2} \mathrm{~A}\) (d) \(3 \times\left(\frac{3}{2}\right) \mathrm{A}\)

A wire of resistance \(4 \Omega\) is stretched to twice its original length. What is the resistance of the wire now ? (a) \(1 \Omega\) (b) \(14 \Omega\) (c) \(8 \Omega\) (d) \(16 \Omega\)

What is the drift velocity of electrons if the current flowing through a copper wire of \(1 \mathrm{~mm}\) diameter is \(1.1 \mathrm{~A} ?\) Assume that each atom of copper contributes one electron: (Given : density of \(\mathrm{Cu}=9 \mathrm{~g} / \mathrm{cm}^{3}\) and atomic weight of \(\mathrm{Cu}=63\) ) (a) \(0.3 \mathrm{~mm} / \mathrm{s}\) (b) \(0.5 \mathrm{~mm} / \mathrm{s}\) (c) \(0.1 \mathrm{~mm} / \mathrm{s}\) (d) \(0.2 \mathrm{~mm} / \mathrm{s}\)

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