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Chapter 19: Problem 74

An \(1800 \mathrm{~W}\) toaster, a \(1400 \mathrm{~W}\) electric frying pan, and a 75 W lamp are plugged into the same electrical outlet in a \(20 \mathrm{~A}, 120 \mathrm{~V}\) circuit. (Note: When plugged into the same outlet, the three devices are in parallel with each other across the \(120 \mathrm{~V}\) outlet. \()\) (a) What current is drawn by each device? (b) Will this combination blow the circuit breaker?

Short Answer

Expert verified
(a) Toaster: 15 A, Frying pan: 11.67 A, Lamp: 0.625 A. (b) Yes, the circuit breaker will blow.

Step by step solution

01

Calculate the current for the toaster

The power equation for electrical appliances is given by \( P = IV \), where \( P \) is power, \( I \) is current, and \( V \) is voltage. Rearranging for current, we have \( I = \frac{P}{V} \). For the toaster with \( P = 1800 \mathrm{~W} \) and \( V = 120 \mathrm{~V} \), the current is \( I = \frac{1800}{120} = 15 \mathrm{~A} \).
02

Calculate the current for the frying pan

Similarly, use the formula \( I = \frac{P}{V} \). For the frying pan with \( P = 1400 \mathrm{~W} \), the current is \( I = \frac{1400}{120} \approx 11.67 \mathrm{~A} \).
03

Calculate the current for the lamp

For the lamp, with \( P = 75 \mathrm{~W} \), the current is \( I = \frac{75}{120} = 0.625 \mathrm{~A} \).
04

Find the total current drawn from the outlet

Since the devices are in parallel, the total current is the sum of the individual currents: \[ I_{\text{total}} = 15 + 11.67 + 0.625 = 27.295 \mathrm{~A} \].
05

Determine if the circuit breaker will blow

A circuit breaker will blow if the total current exceeds its rating. Here, the circuit is rated for \( 20 \mathrm{~A} \). Since \( 27.295 \mathrm{~A} > 20 \mathrm{~A} \), the circuit breaker will blow.

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Key Concepts

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

Current Calculation
To understand how current calculations work for each electrical device, we use the formula that relates power, current, and voltage. This formula is: \[ P = IV \] where:
  • \( P \) is the power in watts (W),
  • \( I \) is the current in amperes (A),
  • \( V \) is the voltage in volts (V).
The current \( I \) can be calculated by rearranging the formula to: \[ I = \frac{P}{V} \]Using this, you can find out how much current each device draws from the outlet. For instance, if you have an 1800 W toaster and a 120 V outlet, you calculate the current as \( I = \frac{1800}{120} = 15 \) A. This calculation is repeated for each device plugged in.
Circuit Breaker
A circuit breaker acts as a safety device, designed to protect an electrical circuit from damage caused by excess current. It automatically interrupts the current flow if it exceeds the breaker's rating, thus preventing potential hazards like fire or electrical shock. In a 20 A rated circuit, if the total current drawn by all devices exceeds this value, the breaker will trip. In our scenario, the combined current of a toaster, frying pan, and lamp is calculated to be approximately 27.3 A. Since this exceeds the 20 A limit of the circuit breaker, it will indeed trip, cutting off the power to prevent damage.
Parallel Circuits
Devices connected in a parallel circuit, each have their own direct path to both ends of the power supply. This setup ensures each device gets the full voltage of the circuit. The advantages of parallel connection include consistent voltage supply across all devices, and failure of one device does not affect the others. When calculating the total current drawn in parallel circuits, you sum up all individual currents. Each device operates independently, drawing its own current as calculated, without affecting the voltage experienced by other devices in the circuit.
Electrical Power
Electrical power quantifies the rate at which energy is used or produced by a device. It is represented in watts (W). Power is calculated by the formula: \[ P = IV \] This indicates how much current an electrical device uses when it is plugged into a voltage source. A higher power rating implies more energy consumption over time. For example, a device with a power rating of 1800 W will use 1800 joules of energy per second when connected to a 120 V outlet, highlighting its efficiency and energy requirement. Understanding power ratings helps users manage energy consumption effectively.

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Most popular questions from this chapter

An iron wire has a temperature coefficient of resistivity of \(+0.0050\left(\mathrm{C}^{\circ}\right)^{-1} .\) When it is placed into boiling water, it has a resistance of \(1000 \Omega\). What is its resistance when it is placed into an ice-water bath?

A typical small flashlight contains two batteries, each having an emf of \(1.5 \mathrm{~V},\) connected in series with a bulb having resistance \(17 \Omega\). (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for \(5.0 \mathrm{~h}\), what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes because this changes the temperature of the filament and hence the resistivity of the filament wire.

When you connect an unknown resistor across the terminals of a 1.50 V AAA battery having negligible internal resistance, you measure a current of 18.0 mA flowing through it. (a) What is the resistance of this resistor? (b) If you now place the resistor across the terminals of a \(12.6 \mathrm{~V}\) car battery having no internal resistance, how much current will flow? (c) You now put the resistor across the terminals of an unknown battery of negligible internal resistance and measure a current of 0.453 A flowing through it. What is the potential difference across the terminals of the battery?

The portion of a nerve cell that conducts signals is called an axon. Many of the electrical properties of axons are governed by ion channels, which are protein molecules that span the axon's cell membrane. When open, each ion channel has a pore that is filled with fluid of low resistivity and connects the interior of the cell electrically to the medium outside the cell. In contrast, the lipid-rich cell membrane in which ion channels reside has very high resistivity. Assume that a typical open ion channel spanning an axon's membrane has a resistance of \(1 \times 10^{11} \Omega .\) We can model this ion channel, with its pore, as a 12 -nm-long cylinder of radius \(0.3 \mathrm{nm}\). What is the resistivity of the fluid in the pore? A. \(10 \Omega \cdot \mathrm{m}\) B. \(6 \Omega \cdot \mathrm{m}\) C. \(2 \Omega \cdot \mathrm{m}\) D. \(1 \Omega \cdot \mathrm{m}\)

A 14 gauge copper wire of diameter \(1.628 \mathrm{~mm}\) carries a current of \(12.5 \mathrm{~mA} .\) (a) What is the potential difference across a \(2.00 \mathrm{~m}\) length of the wire? (b) What would the potential difference in part (a) be if the wire were silver instead of copper, but all else was the same?
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