/*! 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 34 A series combination of a \(12-\... [FREE SOLUTION] | 91Ó°ÊÓ

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A series combination of a \(12-\mathrm{k} \Omega\) resistor and an unknown capacitor is connected to a \(12-V\) battery, One second after the circuit is completed, the voltage across the capacitor is \(10 \mathrm{~V}\). Determine the capacitance of the capacitor.

Short Answer

Expert verified
Based on the given information and the formula for capacitor discharge, the capacitance value C has been found to be \(C = 1/((12*10^-3)* -ln(2/12))\) Farads.

Step by step solution

01

Write the given information

From the exercise we know that the resistance \(R\) is \(12k \Omega\), the initial voltage \(V_0\) is \(12 V\), the voltage across the capacitor one second (\(t = 1 s\)) after the circuit is completed, \(V_c, t=1\), is \(10 V\). The task is to determine the value of the capacitor \(C\).
02

Substitute into the formula

Plug the known values into the formula: \(10 = 12(1 - e^{-1/(12*10^-3*C)})\) where C is in Farads. Now the equation should be solved for C.
03

Isolate the exponential term

First, divide both sides by 12. \(10/12 = 1 - e^{-1/(12*10^-3*C)}\). Then, subtract the result on the left side from 1: \(e^{-1/(12*10^-3*C)} = 1 - 10/12\). Hence, \(e^{-1/(12*10^-3*C)} = 2/12\).
04

Decipher the capacitor value

To find the exponent in the \( e \) , take the natural logarithm (ln) of both sides: \( -1/(12*10^-3*C) = ln(2/12) \). Multiply both sides by \( -1 \): \(1/(12*10^-3*C) = -ln(2/12)\). Now, find C by rearranging the formula: \(C = 1/\((12*10^-3) * -ln(2/12))\).

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Key Concepts

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

RC circuit
An RC circuit is a basic electrical circuit that consists of a resistor (R) and a capacitor (C) connected in series or parallel. In this case, we have a series RC circuit connected to a voltage source, a 12-V battery. The key function of an RC circuit is to control the flow of electrical current over time, particularly affecting how quickly a capacitor charges or discharges.

To better understand this:
  • The resistor controls the rate at which the current flows in the circuit.
  • The capacitor stores and releases energy in the form of electrical charge.
This combination plays a significant role in timing applications, filters, and other electronic applications where specific time delays or filtering actions are required. When the switch in an RC circuit is closed, the capacitor begins to charge up through the resistor, following a predictable pattern known as exponential decay.
Exponential decay
Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. In an RC circuit, the voltage across a charging or discharging capacitor follows an exponential decay pattern.

For a charging capacitor, the formula is:\[ V(t) = V_0 (1 - e^{-t/RC}) \]where:
  • \(V(t)\) is the voltage across the capacitor at time \(t\).
  • \(V_0\) is the initial voltage provided by the battery.
  • \(R\) is the resistance in ohms.
  • \(C\) is the capacitance in farads.
  • \(e\) is the base of the natural logarithm, approximately equal to 2.71828.
The rate of decay or charging depends on the time constant \(\tau = RC\). This is the time it takes for the voltage to reach approximately 63.2% of its final value when charging, or to drop to about 36.8% of its initial value when discharging.

Understanding this exponential relationship is crucial for calculating the characteristics of RC circuits and predicting how quickly capacitors will charge or discharge over time.
Natural logarithm
The natural logarithm, denoted as \(\ln\), is a mathematical function used to determine the power to which the base \(e\) must be raised to produce a given number. In the context of RC circuits, the natural logarithm helps solve equations where the exponential decay function is involved.

For example, in solving the equation\[ e^{-t/RC} = \text{a constant} \]we apply the natural logarithm to both sides to isolate terms and solve for variables like the resistance \(R\) or capacitance \(C\).

The natural logarithm has practical applications in calculating time constants in RC circuits because:
  • It simplifies the manipulation of exponential equations, allowing easy extraction of time-dependent variables.
  • It provides insight into the decay behavior within the involved circuit elements.
By using \(\ln\), equations that contain exponential components become linear and more manageable, which is especially useful in deriving solutions for phenomena like the voltage across a capacitor in an RC circuit.
Voltage across capacitor
When analyzing an RC circuit, one critical variable to understand is the voltage across the capacitor. This voltage changes over time as the capacitor charges or discharges.

Initially, when the circuit is completed, the voltage across an uncharged capacitor starts at zero but rises quickly as the capacitor stores energy. Eventually, it reaches a maximum value, equivalent to the battery's voltage if charging completes.
  • The specific voltage at any point in time is given by the formula: \[ V(t) = V_0 (1 - e^{-t/RC}) \]
  • This represents a part of an exponential curve, reflecting how swiftly the voltage increases in the beginning and slows down as it approaches the supply voltage.
By understanding the voltage's behavior, students can gain important insights into how RC circuits function, how long it takes for changes to settle, and the nature of time-dependent processes in electronics.

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Most popular questions from this chapter

An electric eel generates electric currents through its highly specialized Hunter's organ, in which thousands of disk-shaped cells called electrocytes are lined up in series, very much in the same way batteries are lined up inside a flashlight. When activated, each electrocyte can maintain a potential difference of about \(150 \mathrm{mV}\) at a current of \(1 \mathrm{~A}\) for about \(2.0 \mathrm{~ms}\). Suppose a grown electric eel has \(4.0 \times\) \(10^{3}\) electrocytes and can deliver up to 300 shocks in rapid series over about 1 s. (a) What maximum electrical power can an electric eel generate? (b) Approximately how much energy does it release in one shock? (c) How high would a mass of \(1 \mathrm{~kg}\) have to be lifted so that its gravitational potential energy equals the energy released in 300 such shocks?

A series \(R C\) circuit has a time constant of \(0.960 \mathrm{~s}\). The battery has an emf of \(48.0 \mathrm{~V}\), and the maximum current in the circuit is \(0.500 \mathrm{~mA}\). What are (a) the value of the capacitance and (b) the charge stored in the capacitor \(1.92 \mathrm{~s}\) after the switch is closed?

A charged capacitor is connected to a resistor and a switch as in Active Figure \(18.17\), The circuit has a time constant of \(1.5 \mathrm{~s}\). After the switch is closed, the charge on the capacitor is \(75 \%\) of its initial charge. (a) Find the time it takes to reach this charge. (b) If \(R=250 \mathrm{k} \Omega\), what is the value of \(C ?\)

(a) You need a \(45-\Omega\) resistor, but the stockroom has only \(20-\Omega\) and \(50-\Omega\) resistors. How can the desired resistance be achicved under these circumstances? (b) What can you do if you need a \(35-\Omega\) resistor?

Two \(1.50-\mathrm{V}\) batteries-with their positive terminals in the same direction-are inserted in seties into the barrel of a flashlight. One battery has an internal resistance of \(0.255 \Omega\), the other an internal resistance of \(0.153 \Omega\). When the switch is closed, a current of \(0.600 \mathrm{~A}\) passes through the lamp. (a) What is the lamp's resistance? (b) What fraction of the power dissipated is dissipated in the batteries?

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