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An \(86-\Omega\) resistor and a \(67-\Omega\) resistor are connected in series across a battery. The voltage across the \(86-\Omega\) resistor is \(27 \mathrm{V}\). What is the voltage across the \(67-\Omega\) resistor?

Short Answer

Expert verified
The voltage across the 67-Ohm resistor is approximately 21.04 V.

Step by step solution

01

Understand Ohm's Law

Ohm's Law is given by the formula: \( V = IR \), where \( V \) is the voltage across the resistor, \( I \) is the current through the resistor, and \( R \) is the resistance. This law will be useful to determine the current in the circuit since the resistors are in series.
02

Find the Current in the Circuit

The voltage across the \(86-\Omega\) resistor is given as \(27\, \text{V}\). Using Ohm's Law, calculate the current: \[ I = \frac{V}{R} = \frac{27}{86} \] Thus, the current through the circuit is \(I = 0.314 \text{ A}\) (approximately).
03

Calculate the Voltage Across the 67-Ohm Resistor

Now that we have the current in the circuit, use it to find the voltage across the \(67-\Omega\) resistor using Ohm's Law:\[ V = IR = 0.314 \times 67 \] This results in \( V = 21.038 \text{ V} \) (approximately).

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Key Concepts

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

Series Circuit
A series circuit is a basic configuration in electronics where components are connected end-to-end. This means the current, which is the flow of electric charge, moves through each component without branching. This setup is quite common in various electrical devices and is characterized by certain properties:
  • All components in a series circuit share the same current. This is a crucial detail because it implies that if you know the current in one part of the circuit, you know the current throughout the entire circuit.
  • The total voltage across the series circuit is the sum of the voltages across each component. As such, understanding how each component contributes to the overall voltage is vital in circuit design and analysis.
  • If one component fails or is removed, the entire circuit is interrupted, which will stop the flow of current unless there is a bypass.
These characteristics make series circuits different from parallel circuits, where the voltage remains constant across all components.
Voltage Calculation
Voltage calculation in a series circuit isn't just about measuring energy difference across a component. It's also about understanding the role of each component in dividing total voltage. By applying Ohm's Law, we can calculate these differences. Ohm's Law states: \[ V = IR \]where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. When dealing with a series circuit as explained in the exercise, you know the current through all components is consistent. This makes calculating the voltage across individual components straightforward once the current is found.
  • First, you need to determine the current using the given voltage and resistance values with \( I = \frac{V}{R} \).
  • Then, you apply this current to other components in the circuit to find their respective voltage drops. This is done through \( V = IR \).
This approach highlights how intricacies in voltage measurement can be unravelled by systematically applying circuit laws.
Electrical Resistance
Electrical resistance is the property of a material that opposes the flow of electric current, converting electric energy into heat. In simple terms, it's like friction for electricity. Let's see why it is important in series circuits:
  • Resistance determines how much current flows through a component when a voltage is applied across it. Higher resistance means less current flows when compared to lower resistance at the same voltage.
  • In a series circuit, total resistance is the sum of the individual resistances. This total affects the current flow through the circuit. For resistors in our example, it can be calculated as:\[ R_{total} = R_1 + R_2 = 86 + 67 = 153 \, \Omega \]
The concept of resistance not only helps in calculating current and voltage but also impacts how we design and choose components for electrical applications.
Physics Education
Physics education can be both challenging and rewarding, in particular when studying electricity and circuits. Students encounter concepts such as Ohm's Law, resistance, and circuit types like series and parallel. Understanding these fundamental concepts through hands-on exercises, such as the one provided, is crucial for grasping the theoretical principles:
  • Applying theory to practise helps reinforce learning. As seen in the exercise, calculations based on Ohm’s Law demonstrate practical applications of theoretical concepts. It turns abstract numbers into something more tangible.
  • Solving real problems encourages critical thinking and analytical skills, allowing students to tackle complex physics problems with confidence.
  • Learning collaboratively and discussing with peers can also enhance comprehension, turning challenges into opportunities for shared discovery.
Ultimately, a solid physics education forms the foundation for future studies in engineering, electronics, and various other sciences.

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Most popular questions from this chapter

A coffee-maker contains a heating element that has a resistance of 14 \(\Omega .\) This heating element is energized by a \(120-\mathrm{V}\) outlet. What is the current in the heating element?

How many time constants must elapse before a capacitor in a series \(R C\) circuit is charged to \(80.0 \%\) of its equilibrium charge?

A large spool in an electrician's workshop has \(75 \mathrm{m}\) of insulation- coated wire coiled around it. When the electrician connects a battery to the ends of the spooled wire, the resulting current is 2.4 A. Some weeks later, after cutting off various lengths of wire for use in repairs, the electrician finds that the spooled wire carries a \(3.1-\) A current when the same battery is connected to it. What is the length of wire remaining on the spool?

Two wires have the same cross-sectional area and are joined end to end to form a single wire. One is tungsten, which has a temperature coefficient of resistivity of \(\alpha=0.0045\left(\mathrm{C}^{\circ}\right)^{-1} .\) The other is carbon, for which \(\alpha=-0.0005\left(\mathrm{C}^{\circ}\right)^{-1} .\) The total resistance of the composite wire is the sum of the resistances of the pieces. The total resistance of the composite does not change with temperature. What is the ratio of the lengths of the tungsten and carbon sections? Ignore any changes in length due to thermal expansion.

The rear window defogger of a car consists of thirteen thin wires (resistivity \(\left.=88.0 \times 10^{-8} \Omega \cdot \mathrm{m}\right)\) embedded in the glass. The wires are connected in parallel to the \(12.0-\mathrm{V}\) battery, and cach has a length of \(1.30 \mathrm{m}\). The defogger can melt \(2.10 \times 10^{-2} \mathrm{kg}\) of ice at \(0^{\circ} \mathrm{C}\) into water at \(0^{\circ} \mathrm{C}\) in two minutes. Assume that all the power delivered to the wires is used immediately to melt the ice. Find the cross-sectional area of each wire.

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