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Chapter 4: Problem 9

What is the output voltage of a \(3.0000-V\) lithium cell in a digital wristwatch that draws \(0.300 \mathrm{~mA}\), if the cell's internal resistance is \(2.00 \Omega ?\)

Short Answer

Expert verified
The output voltage is \(2.9994\,V\).

Step by step solution

01

Understanding the Problem

We need to find the output voltage across the terminals of a lithium cell. The cell has a nominal voltage of \(3.0000\,V\), draws a current of \(0.300\,\text{mA}\), and has an internal resistance of \(2.00\,\Omega\). The formula for the output voltage is \(V_{\text{out}} = V_{\text{nominal}} - I \cdot r_{\text{internal}}\).
02

Convert Current to Amperes

The given current is \(0.300\,\text{mA}\). Convert this current into amperes: \[ 0.300\,\text{mA} = 0.300 \times 10^{-3}\,\text{A} = 0.0003\,\text{A} \].
03

Calculate Voltage Drop

Using Ohm's Law, calculate the voltage drop across the internal resistance. The formula is \(V = I \cdot R\). Substitute \(I = 0.0003\,\text{A}\) and \(R = 2.00\,\Omega\): \[ V = 0.0003\,\text{A} \times 2.00\,\Omega = 0.0006\,\text{V} \].
04

Calculate Output Voltage

Subtract the voltage drop from the nominal voltage to find the output voltage: \( V_{\text{out}} = 3.0000\,\text{V} - 0.0006\,\text{V} = 2.9994\,\text{V} \).

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Key Concepts

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

Understanding Internal Resistance
Internal resistance is a critical concept when discussing batteries and electrical circuits. Imagine every battery as having a small resistor inside it. This is what we call internal resistance.
It restricts the flow of electric current within the battery and can reduce the efficiency of a device that the battery powers.
When current flows through the battery, the internal resistance causes a small drop in voltage. This means that not all the energy from the battery's chemical reaction is available at the terminals.
  • Lower internal resistance is desirable as it allows more current to flow with less voltage drop.
  • Higher internal resistance can lead to greater energy losses inside the battery.
Understanding this concept allows you to predict how real-life batteries will perform, especially in small devices like digital wristwatches.
Exploring Voltage Drop
Voltage drop is a reduction in voltage across a component or conductor in an electrical circuit. Here, it happens within the battery due to its internal resistance.
Using Ohm's Law, which is defined as \( V = I \cdot R \), we can calculate the voltage drop.
In our example, the voltage drop across the internal resistance of the lithium cell is calculated by multiplying the current (\(0.0003\,\text{A}\)) with the internal resistance (\(2.00\,\Omega\)). This gives us a voltage drop of \(0.0006\,\text{V}\).
  • This slight reduction can impact the device's overall performance.
  • In high precision devices, even minor voltage drops can cause significant effects.
Understanding voltage drop is crucial for designing efficient circuits and choosing appropriate power sources.
Calculating Voltage for a Digital Wristwatch
Calculating the voltage available for a digital wristwatch involves understanding both the nominal voltage and the effects of voltage drop.
We start with the nominal voltage of the battery, which is given as \(3.0000\,\text{V}\).
After determining the voltage drop caused by the internal resistance (\(0.0006\,\text{V}\)), we can find the actual output voltage available for the wristwatch by subtracting this drop from the nominal voltage. The formula used here is:- \( V_{\text{out}} = V_{\text{nominal}} - V_{\text{drop}} \)
  • Substituting the values, we have \(2.9994\,\text{V}\) as the output voltage.
  • This is the effective voltage that powers the digital watch.
Understanding these calculations ensures the watch operates correctly without unexpectedly draining the battery.

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Most popular questions from this chapter

A \(500-\Omega\) resistor, an uncharged \(1.50-\mu \mathrm{F}\) capacitor, and a \(6.16-\mathrm{V}\) emf are connected in series. (a) What is the initial current? (b) What is the \(R C\) time constant? (c) What is the current after one time constant? (d) What is the voltage on the capacitor after one time constant?

(a) What is the terminal voltage of a large 1.54-V carbon-zinc dry cell used in a physics lab to supply \(2.00 \bar{A}\) to a circuit, if the cell's internal resistance is \(0.100 \Omega ?\) (b) How much electrical power does the cell produce? (c) What power goes to its load?

(a) What is the internal resistance of a \(1.54-\mathrm{V}\) dry cell that supplies \(1.00 \mathrm{~W}\) of power to a \(15.0-\Omega\) bulb? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

An electronic apparatus may have large capacitors at high voltage in the power supply section, presenting a shock hazard even when the apparatus is switched off. A "bleeder resistor" is therefore placed across such a capacitor, as shown schematically in Figure, to bleed the charge from it after the apparatus is off. Why must the bleeder resistance be much greater than the effective resistance of the rest of the circuit? How does this affect the time constant for discharging the capacitor?

(a) Given a 48.0-V battery and \(24.0-\Omega\) and \(96.0-\Omega\) resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.
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