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Chapter 26: Problem 69

A resistor \(R_1\) consumes electrical power \(P_1\) when connected to an emf \(\varepsilon\). When resistor \(R_2\) is connected to the same emf, it consumes electrical power \(P_2\). In terms of \(P_1\) and \(P_2\), what is the total electrical power consumed when they are both connected to this emf source (a) in parallel and (b) in series?

Short Answer

Expert verified
(a) Parallel: \( P_1 + P_2 \); (b) Series: \( \frac{\varepsilon^2}{R_1 + R_2} \)

Step by step solution

01

Understanding Power Dissipation

The power dissipated by a resistor is given by the formula \( P = \frac{\varepsilon^2}{R} \). For resistor \( R_1 \), this can be written as \( P_1 = \frac{\varepsilon^2}{R_1} \), and for \( R_2 \), \( P_2 = \frac{\varepsilon^2}{R_2} \).
02

Calculating Power in Parallel Connection

When resistors \( R_1 \) and \( R_2 \) are connected in parallel, the total power consumed \( P_{\text{parallel}} \) is the sum of the power dissipated by each resistor. Therefore, \( P_{\text{parallel}} = P_1 + P_2 \).
03

Calculating Equivalent Resistance in Series

For a series connection, the total resistance \( R_{\text{series}} \) is the sum of the resistances of the individual resistors: \( R_{\text{series}} = R_1 + R_2 \).
04

Calculating Power in Series Connection

Using the equivalent resistance from Step 3, the total power consumed in series \( P_{\text{series}} \) is calculated using the power formula: \( P_{\text{series}} = \frac{\varepsilon^2}{R_{\text{series}}} \). Substitute \( R_{\text{series}} = R_1 + R_2 \) into the formula to get \( P_{\text{series}} = \frac{\varepsilon^2}{R_1 + R_2} \).

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Key Concepts

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

Resistors in Series
When resistors are arranged in a series, they are connected end-to-end, forming a single path for the current to flow. This configuration is straightforward and has a crucial property: the total resistance of the series is simply the sum of the individual resistances.

For two resistors, this is calculated as:
  • \( R_{\text{series}} = R_1 + R_2 \)
This means all the resistors in the series have the same current passing through them because there is only one path for the current.

However, the voltage across each resistor can be different unless the resistors are identical. The total voltage across the series is the sum of the voltages across each resistor. This property leads to interesting implications when calculating power dissipation in series resistors.
Resistors in Parallel
In parallel circuits, resistors are connected so that each one has the same voltage across it. Each resistor provides a separate path for current to flow.

To find the total resistance in a parallel circuit, you use the reciprocal formula:
  • \( \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \)
After finding the reciprocal of the sum, you get the equivalent resistance for the parallel configuration.

The fantastic part of parallel circuits is that they can lead to a total resistance value that's smaller than the smallest resistor in the network. This allows each path to consume its share of the power, providing a more balanced load across different components.
Power Dissipation in Resistors
When resistors consume electrical energy, they convert it into heat. This process is referred to as power dissipation.

The basic formula to determine the power dissipated by a resistor is:
  • \( P = \frac{\varepsilon^2}{R} \)
For a series connection, calculating the power dissipation requires using the total series resistance:
  • \( P_{\text{series}} = \frac{\varepsilon^2}{R_1 + R_2} \)
In parallel circuits, the power dissipation is simply the sum of the power dissipated by each resistor:
  • \( P_{\text{parallel}} = P_1 + P_2 \)
This approach shows the clear difference in power consumption between series and parallel circuits, emphasizing the importance of choosing the right configuration for specific electrical needs.

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Most popular questions from this chapter

A 2.36-\(\mu\)F capacitor that is initially uncharged is connected in series with a 5.86-\(\Omega\) resistor and an emf source with \(\varepsilon =\) 120 V and negligible internal resistance. (a) Just after the connection is made, what are (i) the rate at which electrical energy is being dissipated in the resistor; (ii) the rate at which the electrical energy stored in the capacitor is increasing; (iii) the electrical power output of the source? How do the answers to parts (i), (ii), and (iii) compare? (b) Answer the same questions as in part (a) at a long time after the connection is made. (c) Answer the same questions as in part (a) at the instant when the charge on the capacitor is one-half its final value.

A 1500-W electric heater is plugged into the outlet of a 120-V circuit that has a 20-A circuit breaker. You plug an electric hair dryer into the same outlet. The hair dryer has power settings of 600 W, 900 W, 1200 W, and 1500 W. You start with the hair dryer on the 600-W setting and increase the power setting until the circuit breaker trips. What power setting caused the breaker to trip?

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 nm. What is the resistivity of the fluid in the pore? (a) 10 \(\Omega\) \(\cdot\) m; (b) 6 \(\Omega\) \(\cdot\) m; (c) 2 \(\Omega\) \(\cdot\) m; (d) 1 \(\Omega\) \(\cdot\) m.

The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (a) What is the current in the heating element? Is 12-gauge wire large enough to supply this current? (b) What is the resistance of the dryer's heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?

A 6.00-\(\mu\)F capacitor that is initially uncharged is connected in series with a 5.00-\(\Omega\) resistor and an emf source with \(\varepsilon =\) 50.0 V and negligible internal resistance. At the instant when the resistor is dissipating electrical energy at a rate of 300 W, how much energy has been stored in the capacitor?
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