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Chapter 19: Problem 28

. A complete series circuit consists of a 12.0 \(\mathrm{V}\) battery, a 4.70\(\Omega\) resistor, and a switch. The internal resistance of the battery is 0.30\(\Omega .\) The switch is open. What does an ideal voltmeter read when placed (a) across the terminals of the battery, (b) across the resistor, (c) across the switch? (d) Repeat parts (a), (b), and (c) for the case when the switch is closed.

Short Answer

Expert verified
(a) 12.0 V, 0 V, 0 V (open switch); (a) 11.28 V, 11.28 V, 0 V (closed switch).

Step by step solution

01

Understand the Circuit and Open Switch

When the switch is open, no current flows through the circuit. Therefore, the voltage across the battery terminals should equal the electromotive force (emf) of the battery, which is the same for an open circuit. Similarly, no current through the resistor or switch allows us to consider their influence on voltage drop.
02

Calculate Voltage Across the Battery Terminals (Open Switch)

For an open switch, the voltmeter reading across the battery terminals will be equal to the emf of the battery since no current flows, resulting in no voltage drop across the internal resistance.Voltage across battery (a):\( V = 12.0 \, \text{V} \)
03

Calculate Voltage Across the Resistor (Open Switch)

With the switch open, no current flows in the circuit, so there is no voltage drop across the resistor.Voltage across resistor (b):\( V = 0.0 \, \text{V} \)
04

Calculate Voltage Across the Switch (Open Switch)

Similarly, since the circuit is open, there is no voltage drop over the switch.Voltage across switch (c):\( V = 0.0 \, \text{V} \)
05

Understanding the Circuit and Closed Switch

When the switch is closed, the circuit is complete, and current can flow. Use Ohm's law to determine the current in the circuit considering both the internal resistance and the resistor.
06

Determine the Total Resistance (Closed Switch)

The total resistance in the circuit is the sum of the resistor and the internal resistance of the battery:\( R_{\text{total}} = 4.70 \, \Omega + 0.30 \, \Omega = 5.00 \, \Omega \)
07

Calculate the Current in the Circuit (Closed Switch)

Using Ohm's Law, calculate the current:\( I = \frac{V}{R_{\text{total}}} = \frac{12.0 \, \text{V}}{5.00 \, \Omega} = 2.4 \, \text{A} \)
08

Calculate Voltage Across Battery Terminals (Closed Switch)

The voltage across the terminals of the battery will be the emf minus the voltage drop due to the internal resistance:Voltage across battery (a):\( V = 12.0 \, \text{V} - I \times 0.30 \, \Omega = 12.0 \, \text{V} - 0.72 \, \text{V} = 11.28 \, \text{V} \)
09

Calculate Voltage Across the Resistor (Closed Switch)

Using Ohm's Law for the resistor:Voltage across resistor (b):\( V = I \times 4.70 \, \Omega = 2.4 \, \text{A} \times 4.70 \, \Omega = 11.28 \, \text{V} \)
10

Calculate Voltage Across the Switch (Closed Switch)

In a closed switch, it is similar to a wire with no resistance leading to zero voltage drop.Voltage across switch (c):\( V = 0.0 \, \text{V} \)

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle used in electric circuits. It is defined by the equation \( V = I \times R \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. This law helps us understand how voltage, current, and resistance interact in an electrical circuit.

Understanding Ohm's Law is crucial for analyzing circuits because it allows us to calculate unknown values if we know the other two. For example:
  • If you know the resistance and the voltage, you can find the current by rearranging the formula to \( I = \frac{V}{R} \).
  • If you have the current and the resistance, you can calculate the voltage using \( V = I \times R \).
  • Knowing the voltage and the current allows you to find the resistance with \( R = \frac{V}{I} \).
Using Ohm's Law in circuits with switches, as in this exercise, shows how current flow changes with an open or closed switch, impacting the calculations of voltages across components.
Voltmeter
A voltmeter is an instrument used to measure the voltage, or potential difference, between two points in an electrical circuit. It is designed to have a very high internal resistance so that it does not draw significant current from the circuit, ensuring accurate voltage readings.

When using a voltmeter:
  • Connect it parallel to the component you wish to measure. This method of connection prevents the voltmeter from affecting the component's operation.
  • With an open switch, measure the voltage across the terminals of a battery to see its electromotive force (emf). The voltmeter will display the battery's full voltage because no current flows.
  • With a closed switch, the voltmeter shows the actual voltage across a component due to current, taking into account any voltage drops.
In the described exercise, the voltmeter helps to determine the voltage across various components both when the circuit is open and closed, illustrating its critical role in circuit analysis.
Electrical Resistance
Electrical resistance is a measure of how much a material opposes the flow of electric current. It is denoted in ohms (\( \Omega \)). Every component in a circuit presents some resistance, influencing the current flow and voltage distribution.

Key points to understand about resistance include:
  • The total resistance in a series circuit is the sum of individual resistances. For example, in this exercise, the resistance of the resistor and the internal resistance of the battery combine to determine the total resistance.
  • A higher resistance signifies that the material or component restricts the flow of electrons more strongly.
  • The amount of voltage drop across a component is directly proportional to its resistance when a certain current flows through it, according to Ohm's Law (\( V = I \times R \)).
In practical applications, resistance influences how a circuit behaves with both open and closed switches, affecting both current flow and voltage measurements.
Current Flow
Current flow in an electrical circuit is the movement of electric charge, usually carried by moving electrons in a wire. It is measured in amperes (\( A \)) and is driven by a voltage source, such as a battery.

Considerations for current flow:
  • In an open circuit (like when the switch is open), no current flows, meaning voltage readings across certain elements will be zero except at the source terminals.
  • Once the circuit is closed, current begins to flow. The amount of current is determined by the total resistance in the circuit and the voltage, calculated using Ohm's Law (\( I = \frac{V}{R} \)).
  • Understanding current flow helps predict how much power is used by electrical components, as power (\( P \)) is calculated by \( P = I \times V \).
This concept is essential for analyzing how a circuit behaves when a switch is moved from an open to a closed position, revealing changes in both voltage and resistance characteristics.

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Most popular questions from this chapter

A 6.00 \(\mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a 4500\(\Omega\) resistor and a 500 \(\mathrm{V}\) emf source with negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor, (b) the voltage drop across the resistor, (c) the charge on the capacitor, and (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants), what are the values of the preceding four quantities?

Electricity through the body, I. A person with a body resistance of 10 \(\mathrm{k} \Omega\) between his hands accidentally grasps the terminals of a 14 \(\mathrm{kV}\) power supply. (a) If the internal resistance of the power supply is \(2000 \Omega,\) what is the current through the person's body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the situation just described to be 1.00 \(\mathrm{mA}\) or less?

Electrical safety. This procedure is not recommended! You'll see why after you work the problem. You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver having a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of 1.0 \(\mathrm{k} \Omega\) . (a) If you accidentally touch the "hot" wire connected to the 120 \(\mathrm{V}\) line, how much current will pass through your body? Is this amount enough to be dangerous? (The maximum safe current is about 5 \(\mathrm{mA}\). (b) How much electrical power is delivered to your body?

Lightbulbs. The wattage rating of a lightbulb is the power it consumes when it is connected across a 120 potential difference. For example, a 60 W lightbulb consumes 60.0 \(\mathrm{W}\) of electrical power only when it is connected across a 120 \(\mathrm{V}\) potential difference. (a) What is the resistance of a 60 \(\mathrm{W}\) lightbulb? (b) Without doing any calculations, would you expect a 100 \(\mathrm{W}\) bulb to have more or less resistance than a 60 \(\mathrm{W}\) bulb? Calculate and find out.

In an ionic solution, a current consists of \(\mathrm{Ca}^{2+}\) ions (of charge \(+2 e )\) and \(\mathrm{Cl}^{-}\) ions (of charge \(-e )\) traveling in opposite directions. If \(5.11 \times 10^{18} \mathrm{Cl}^{-}\) ions go from \(A\) to \(B\) every 0.50 min, while \(3.24 \times 10^{18} \mathrm{Ca}^{2+}\) ions move from \(B\) to \(A\), what is the current (in mA) through this solution, and in which direction \((\) from \(A\) to \(B\) or from \(B\) to \(A)\) is it going?
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