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Three resistors, \(25,45,\) and \(75 \Omega,\) are connected in series, and a \(0.51-\mathrm{A}\) current passes through them. What is (a) the equivalent resistance and (b) the potential difference across the three resistors?

Short Answer

Expert verified
(a) 145 Ω; (b) 73.95 V

Step by step solution

01

Understand Series Connection

In a series connection, resistors are connected end-to-end, and the total or equivalent resistance is simply the sum of their individual resistances.
02

Calculate Equivalent Resistance

To find the equivalent resistance \(R_{eq}\) in a series circuit, sum the individual resistances: \(R_{eq} = R_1 + R_2 + R_3\). Plug in the given resistances: \(R_{eq} = 25 + 45 + 75 = 145 \Omega\).
03

Recall Ohm's Law

Ohm's Law relates current (\(I\)), voltage (\(V\)), and resistance (\(R\)) in a circuit: \(V = IR\). This law will help us find the potential difference across the resistors.
04

Calculate Potential Difference Across Resistors

Using Ohm's Law, find the voltage across the equivalent resistance: \(V = I \times R_{eq}\). Plug in the values: \(V = 0.51\,\text{A} \times 145\,\Omega = 73.95\,\text{V}\).

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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 electrical circuits. It explains the relationship between voltage, current, and resistance. According to this principle, the voltage across a circuit is the product of the current flowing through it and the resistance it encounters. This can be noted as:

\[ V = I \times R \]Where:
  • \(V\) is the voltage (measured in volts)
  • \(I\) is the current (measured in amperes)
  • \(R\) is the resistance (measured in ohms)
The law provides a way to calculate the electrical potential difference, as shown in the preceding exercise where the potential difference was found using known current and resistance.

Practically, Ohm's Law helps in determining how much voltage is necessary for a particular current to pass through a given resistance in any circuit. This understanding is critical in designing and analyzing circuits for desired outcomes.
Series Circuits
In a series circuit, all components are connected one after the other, forming a single path for the electric current.

The total or equivalent resistance of a series circuit is simply the sum of all individual resistances. This property makes it easy to calculate the combined resistance:

\[ R_{eq} = R_1 + R_2 + R_3 + \ldots + R_n \]Some key features of series circuits include:
  • The same current flows through every component, so if one component fails, it interrupts the whole circuit.
  • The total voltage across the circuit is the sum of the individual voltage drops across each component.
In the exercise, the series connection of resistors resulted in an equivalent resistance of \(145 \,\Omega\), showing how resistances add up in such configurations. This characteristic is vital for calculating total resistance and understanding how circuits respond to changes in any individual component's resistance.
Resistor Calculations
In electrical circuits, resistors play a crucial role by providing resistance to the flow of electric current. Calculations involving resistors are fundamental when determining equivalent resistance in different circuit configurations. For series circuits, as in the exercise:

- **Summation of Resistances**: To find the total resistance, add each resistor's resistance: \( R_{eq} = R_1 + R_2 + R_3 \)- **Impact on Current and Voltage**: With a fixed current, as resistances increase, the voltage across the whole circuit also increases due to Ohm's Law.When working with resistors, it is important to:
  • Know the arrangement (series, parallel, or a combination).
  • Correctly identify and sum resistances for calculations.
  • Apply Ohm’s Law where necessary to find other unknowns in the circuit.
In the original problem, the series arrangement required calculation of the total resistance to find the potential difference effectively and accurately using these principles.

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Most popular questions from this chapter

Help with problems of this kind is available in Interactive Solution 20.5120 .51 at . A coffee cup heater and a lamp are connected in parallel to the same \(120-\mathrm{V}\) outlet. Together, they use a total of \(111 \mathrm{~W}\) of power. The resistance of the heater is \(4.0 \times 10^{2} \Omega\). Find the resistance of the lamp.

A resistor has a resistance \(R\), and a battery has an internal resistance \(r\). When the resistor is connected across the battery, ten percent less power is dissipated in \(R\) than would be dissipated if the battery had no internal resistance. Find the ratio \(r / R\).

A fax machine uses \(0.110\) A of current in its normal mode of operation, but only \(0.067 \mathrm{~A}\) in the standby mode. The machine uses a potential difference of \(120 \mathrm{~V}\). (a) In one minute, how much more charge passes through the machine in the normal mode versus the standby mode, and (b) how much more energy is used?

A toaster uses a Nichrome heating wire. When the toaster is turned on at \(20^{\circ} \mathrm{C}\), the initial current is \(1.50 \mathrm{~A}\). A few seconds later, the toaster warms up and the current now has a value of \(1.30 \mathrm{~A}\). The average temperature coefficient of resistivity for Nichrome wire is \(4.5 \times 10^{-4}\left(\mathrm{C}^{0}\right)^{-1} .\) What is the temperature of the heating wire?

A lightning bolt delivers a charge of \(35 \mathrm{C}\) to the ground in a time of \(1.0 \times 10^{-3} \mathrm{~s}\). What is the current?

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