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Chapter 20: Problem 47

Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: \(2.0 \Omega\) and \(4.0 \mathrm{W}, 12.0 \Omega\) and \(10.0 \mathrm{W},\) and \(3.0 \Omega\) and \(5.0 \mathrm{W}\) (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in (a)?

Short Answer

Expert verified
The greatest voltage is 15.504 V and the power delivered by the battery is 14.143 W.

Step by step solution

01

Recognize the limitations

The maximum power that each resistor can handle is given. We must ensure that the voltage applied across each resistor does not exceed this limit.
02

Calculate current for each resistor's limit

For each resistor, use the formula for power: \(P = I^2 R\). We will find the maximum permissible current for each resistor before it surpasses its power limit. - For the first resistor (\(2.0 \Omega\), \(4.0 \mathrm{W}\)): \[ I_1 = \sqrt{\frac{P_1}{R_1}} = \sqrt{\frac{4.0 \mathrm{W}}{2.0 \Omega}} = \sqrt{2.0} = 1.41 \mathrm{A} \]- For the second resistor (\(12.0 \Omega\), \(10.0 \mathrm{W}\)): \[ I_2 = \sqrt{\frac{P_2}{R_2}} = \sqrt{\frac{10.0 \mathrm{W}}{12.0 \Omega}} = \sqrt{0.833} = 0.912 \mathrm{A} \]- For the third resistor (\(3.0 \Omega\), \(5.0 \mathrm{W}\)): \[ I_3 = \sqrt{\frac{P_3}{R_3}} = \sqrt{\frac{5.0 \mathrm{W}}{3.0 \Omega}} = \sqrt{1.667} = 1.29 \mathrm{A} \]
03

Determine the limiting current

The smallest current calculated from Step 2 will dictate the greatest current allowable in the series.- The limiting current \(I_{\text{lim}}\) is the smallest among \(I_1, I_2, I_3\), which is \(0.912 \mathrm{A}\).
04

Calculate total resistance

Since the resistors are in series, the total resistance \(R_T\) is the sum of individual resistances:\[ R_T = R_1 + R_2 + R_3 = 2.0 \Omega + 12.0 \Omega + 3.0 \Omega = 17.0 \Omega \]
05

Calculate maximum allowable voltage

Using Ohm's Law: \(V = IR\), with the limiting current and total resistance: \[ V_{\text{max}} = I_{\text{lim}} \cdot R_T = 0.912 \mathrm{A} \cdot 17.0 \Omega = 15.504 \mathrm{V} \]
06

Calculate total power delivered by the battery

The power delivered by the battery can be found using the formula \(P = IV\). With \(I_{\text{lim}}\) and \(V_{\text{max}}\): \[ P_{\text{battery}} = I_{\text{lim}} \cdot V_{\text{max}} = 0.912 \mathrm{A} \cdot 15.504 \mathrm{V} = 14.143 \mathrm{W} \]

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Key Concepts

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

Resistors
Resistors are fundamental components in electrical circuits. They hinder the flow of electric current, and each resistor has a specific resistance value measured in ohms (\(\Omega\)). In a series circuit, resistors are arranged one after another, forming a single path for the current to flow. This configuration affects the total resistance and current throughout the circuit.

Some important points to remember about resistors in series:
  • The total resistance is the sum of all individual resistances. This means that the overall resistance increases with each added resistor.
  • This total resistance impacts the current, as a higher resistance typically means less current flow.
  • Each resistor has a maximum power rating, indicating the maximum power it can handle before overheating or burning out.
Ohm's Law
One of the cornerstones of electricity is Ohm's Law. It clearly articulates the relationship between voltage (\(V\)), current (\(I\)), and resistance (\(R\)) in an electrical circuit: \(V = IR\). This formula is vital when analyzing circuits, as it offers a straightforward method to calculate any one of these three variables.

How Ohm's Law applies in our series circuit:
  • Helps us determine the voltage across a resistor when we know the current and resistance.
  • Essentially assists in calculating the maximum allowable voltage when working within a resistor's power limits.
  • In a series circuit, the current through each resistor is the same, making Ohm's Law particularly useful for calculating voltage drops across each component.
Power Rating
Power rating is a critical aspect of resistors in a circuit. It is defined as the maximum amount of power a resistor can handle, expressed in watts (\(W\)). Exceeding this rating can lead to resistor failure due to excessive heat.
  • The power rating is calculated using the formula: \(P = I^2 R\), where \(P\) is the power, \(I\) is the current, and \(R\) is the resistance.
  • In a series circuit, ensuring each resistor does not surpass its power rating is crucial to avoid circuit damage.
  • In practice, the resistor with the lowest power limit often determines the maximum operable current within the series circuit, as it will reach its limit first.

Understanding these ratings helps to determine the safe parameters for operating a circuit, ensuring longevity and reliability of its components.
Circuit Analysis
Circuit analysis involves evaluating a circuit to understand how its components interact. This process is essential for determining voltages, currents, and power distributions in a circuit.
  • For a series circuit, the total resistance is straightforward, being the sum of all resistors.
  • Knowing the total resistance allows us to use Ohm's Law to find other missing values, such as total voltage or current limits.
  • The overall analysis helps identify potential issues, such as a resistor that could exceed its power rating, enabling adjustments to prevent circuit failure.

This methodology allows us to systematically determine the operating conditions and ensure they conform to safety standards while allowing the circuit to function efficiently. Understanding circuit analysis leads to informed decisions when designing or troubleshooting circuits.

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Most popular questions from this chapter

According to Equation \(20.7,\) an ac voltage \(V\) is given as a function of time \(t\) by \(V=V_{0} \sin 2 \pi f t,\) where \(V_{0}\) is the peak voltage and \(f\) is the frequency (in hertz). For a frequency of \(60.0 \mathrm{Hz}\), what is the smallest value of the time at which the voltage equals one-half of the peak value?

Two resistances, \(R_{1}\) and \(R_{2},\) are connected in series across a \(12-\mathrm{V}\) battery. The current increases by \(0.20 \mathrm{A}\) when \(R_{2}\) is removed, leaving \(R_{1}\) connected across the battery. However, the current increases by just \(0.10 \mathrm{A}\) when \(R_{1}\) is removed, leaving \(R_{2}\) connected across the battery. Find (a) \(R_{1}\) and \((b) R_{2}\).

A battery delivering a current of 55.0 A to a circuit has a terminal voltage of \(23.4 \mathrm{V}\). The electric power being dissipated by the internal resistance of the battery is \(34.0 \mathrm{W}\). Find the emf of the battery.

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.

In Section 12.3 it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wire. Suppose that the resistance of a platinum resistance thermometer is \(125 \Omega\) when its temperature is \(20.0^{\circ} \mathrm{C}\). The wire is then immersed in boiling chlorine, and the resistance drops to \(99.6 \Omega\). The temperature coefficient of resistivity of platinum is \(\alpha=3.72 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1} .\) What is the temperature of the boiling chlorine?
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