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Chapter 10: Problem 29

Your car's 30.0-W headlight and 2.40-kW starter are ordinarily connected in parallel in a 12.0 -V system. What power would one headlight and the starter consume if connected in series to a 12.0 -V battery? (Neglect any other resistance in the circuit and any change in resistance in the two devices.)

Short Answer

Expert verified
The total power consumed by the system in series is done by the calculation at Step 3. This can be accomplished by having the total resistance for this system as well as the given voltage of the battery.

Step by step solution

01

Calculate the resistance of the starter and headlight in parallel

By using the formula for power \( P = \frac{V^2}{R} \), where \( P \) is power, \( V \) is voltage, and \( R \) is resistance, we can easily calculate the resistance for both headlight and starter. For the headlight it is \( R_{headlight} = \frac{V^2}{P} = \frac{(12.0)^2}{30.0} \) and for the starter it is \( R_{starter} = \frac{V^2}{P} = \frac{(12.0)^2}{2400.0} \).
02

Find the total resistance in series

When resistors are connected in series, their total resistance \( R_{total} \) is simply the sum of their resistances: \( R_{total} = R_{headlight} + R_{starter} \).
03

Calculate the power consumed in series

The power consumed by the system in series is given by the same formula as before: \( P_{total} = \frac{V^2}{R_{total}} \). In this case, \( V \) is still the battery voltage (12.0 V), and \( R_{total} \) is the total resistance we just calculated. Substitute these values into the formula to find the total power consumed by the system in series.

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Key Concepts

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

Electric Power Calculation
Electric power is a crucial concept in understanding how circuits function. Power in an electrical context is the rate at which energy is consumed or converted in a circuit. For any electrical component, power (P) can be calculated using the formula \( P = \frac{V^2}{R} \), where \( V \) is the voltage across the component and \( R \) is its resistance.
In the exercise given, the headlight and starter are each powered by a 12.0 V source. When these components operate in a parallel circuit with the battery, they independently consume power.
  • The headlight, with its power rating of 30.0 W, and the starter, rated at 2.40 kW (2400 W), have their resistances derived from the formula \( R = \frac{V^2}{P} \).
  • Once the components are linked in this manner, we can then calculate the effective energy conversion (power consumption) in different configurations, such as series and parallel.
  • Changing from parallel to series involves altering how voltage and current distribute through the components, significantly affecting the power each consumes.
Resistance in Series and Parallel
Resistance behaves differently when components are connected in series versus parallel.
  • In series circuits, total resistance is the sum of individual resistances: \( R_{total} = R_{1} + R_{2} + \ldots + R_{n} \). Therefore, the circuit's overall resistance increases, which in turn affects current flow throughout the circuit.
  • For parallel circuits, the total resistance decreases and can be calculated using \( \frac{1}{R_{total}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \ldots + \frac{1}{R_{n}} \). The characteristic feature of parallel circuits is that they provide multiple paths for current to travel, thus reducing overall resistance.
  • In the problem, the resistive values derived from both components, when summed for a series connection, yield a greater total resistance than when the devices operate individually in parallel.
Knowing the differences in resistance configuration helps to predict and calculate circuit behavior under various conditions.
Ohm's Law
Ohm's Law provides a foundational formula for understanding the relationship between voltage, current, and resistance in electrical circuits. The law is typically expressed as \( V = I \times R \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
This relationship underpins much of electric circuit analysis, including the exercise in question:
  • Knowing the values of any two of the three variables allows us to solve for the third.
  • Within the context of our series circuit, with a constant voltage of 12.0 V and the total resistance calculated previously, one can infer the total current flowing using Ohm's law, \( I = \frac{V}{R_{total}} \).
  • The calculated current informs how much power is consumed, linking it back to the power equation \( P = I \times V \).
Understanding Ohm鈥檚 Law is essential for comprehensively analyzing and solving circuits, particularly when managing different configurations like those presented in the exercise.

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Most popular questions from this chapter

A \(2.00-\) and a \(7.50-\mu \mathrm{F}\) capacitor can be connected in series or parallel, as can a 25.0 - and a \(100-\mathrm{k} \Omega\) resistor. Calculate the four \(R C\) time constants possible from connecting the resulting capacitance and resistance in series.

Some light bulbs have three power settings (not including zero), obtained from multiple filaments that are individually switched and wired in parallel. What is the minimum number of filaments needed for three power settings?

An 1800-W toaster, a 1400-W speaker, and a 75-W lamp are plugged into the same outlet in a 15 -A fuse and 120-V circuit. (The three devices are in parallel when plugged into the same socket.) (a) What current is drawn by each device? (b) Will this combination blow the \(15-\mathrm{A}\) fuse?

A homemade capacitor is constructed of 2 sheets of aluminum foil with an area of 2.00 square meters, separated by paper, \(0.05 \mathrm{mm}\) thick, of the same area and a dielectric constant of \(3.7 .\) The homemade capacitor is connected in series with a \(100.00-\Omega\) resistor, a switch, and a \(6.00-\mathrm{V}\) voltage source. (a) What is the \(R C\) time constant of the circuit? (b) What is the initial current through the circuit, when the switch is closed? (c) How long does it take the current to reach one third of its initial value?

Find the resistance that must be placed in parallel with a \(60.0-\Omega\) galvanometer having a 1.00 -mA sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 25.0 - A full-scale reading. Include a circuit diagram with your solution.
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